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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08u7 | How many triplets $(a, b, c)$ of positive integers less than or equal to $2010$ are there for which $a + b + c$ is a multiple of each of $a$, $b$, $c$? We regard such triplets as ordered triplets (i.e., two such triplets are considered different if the same set of three numbers appear in a different order.) | [
"Since the condition of the problem is symmetric with respect to $a$, $b$, $c$, let us assume, to begin with, that $a \\le b \\le c$ is satisfied. We consider the effect of permuting the order of $a$, $b$, $c$ later. Let us determine the form that $(a, b, c)$ has to take if $a+b+c$ is a multiple of each of $a$, $b$... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics"
] | null | proof and answer | 9045 | |
0788 | Prove that for all integers $k > 2$, there exists $k$ distinct positive integers $a_1, \dots, a_k$ such that
$$
\sum_{1 \le i < j \le k} \frac{1}{a_i a_j} = 1.
$$ | [
"We will construct such $a_1, \\dots, a_k$ by induction on $k$.\n\nBase case: $k = 3$.\nLet $a_1 = 1$, $a_2 = 2$, $a_3 = 3$. Then\n$$\n\\sum_{1 \\le i < j \\le 3} \\frac{1}{a_i a_j} = \\frac{1}{1 \\cdot 2} + \\frac{1}{1 \\cdot 3} + \\frac{1}{2 \\cdot 3} = \\frac{1}{2} + \\frac{1}{3} + \\frac{1}{6} = 1.\n$$\n\nInduc... | India | IMOTC Practice Test 1 | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
095f | Problem:
Fie $a, b, c$ şi $d$ numere reale, $a < b < c < d$. Să se arate că $(a + b + c + d)^2 > 8(a c + b d)$. | [
"Solution:\n\nNotăm $b - a = x$, $c - b = y$, $d - c = z$, $x, y, z \\in \\mathbb{R}_{+}^{*}$.\nObţinem $b = a + x$, $c = a + x + y$, $d = a + x + y + z$.\n\nInegalitatea dată ia forma $(4a + 3x + 2y + z)^2 > 8(a(a + x + y) + (a + x)(a + x + y + z))$.\n\nUltima inegalitate este echivalentă cu:\n$$\n16a^2 + 9x^2 + 4... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
003r | En un tablero cuadriculado de $10$ filas por $500$ columnas, comenzando por la casilla inferior izquierda y siguiendo el modelo que sugiere la figura, Joaquín pintó de negro $2006$ casillas.
Luego Agustina pintó de azul las casillas que tenían dos o más casillas vecinas de color negro. ¿Cuánta... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Español | proof and answer | 2007 | |
02vn | Problem:
Existem 2017 pessoas em uma festa. Em um determinado momento, cada uma delas dá um presente para um outro convidado (é possível que um convidado receba mais de um presente). Mostre que podemos encontrar um grupo de 673 pessoas na festa de modo que quaisquer duas delas não trocaram presentes entre si. | [
"Solution:\n\nNossa estratégia será classificar os convidados da festa em grupos de três cores de modo que pessoas associadas a uma mesma cor não trocaram presentes entre si. Pelo menos uma dessas cores terá 673 pessoas, pois caso contrário, o total de pessoas será no máximo $672 \\cdot 3=2016<2017$. Bastará então ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0b93 | Consider a complex number $z$, $z \neq 0$ and the real sequence
$$
a_n = \left| z^n + \frac{1}{z^n} \right|, \quad n \ge 1.
$$
a) Show that if $a_1 > 2$, then
$$
a_{n+1} < \frac{a_n + a_{n+2}}{2}, \text{ for all } n \in \mathbb{N}^*.
$$
b) Prove that if there exists $k \in \mathbb{N}^*$ such that $a_k \le 2$, then $a_1... | [
"a)\nWe easily notice that\n$$\n\\begin{aligned}\n2|z^{n+1} + \\frac{1}{z^{n+1}}| &< |z + \\frac{1}{z}| \\cdot |z^{n+1} + \\frac{1}{z^{n+1}}| \\\\\n& = |z^n + \\frac{1}{z^n} + z^{n+2} + \\frac{1}{z^{n+2}}| \\leq |z^n + \\frac{1}{z^n}| + |z^{n+2} + \\frac{1}{z^{n+2}}|.\n\\end{aligned}\n$$\n\n\nAlternative Solution. ... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
05cx | The website of a company has multiple pages, which may point to each other. A list containing each of these pages exactly once is called *natural*, if whenever one page points to another, it is written before the other page.
The webmaster of the company has a natural list of pages. Then she reorders the pages, writing ... | [
"We write $x \\to y$ whenever the page $x$ points to the page $y$.\n\na. Consider any two pages $a$ and $b$ with $a \\to b$. By the assumption, $b$ is not pointed to by any other page. Thus $b$ is located in the new list in the group of pages pointed to by page $a$. We consider two cases.\n* If $a$ is not pointed t... | Estonia | Estonian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | a. Yes. b. No. | |
0ehm | Problem:
a) Dani so prvi štirje členi neskončnega aritmetičnega zaporedja:
$$
-\frac{3}{2}, \frac{1}{2}, \frac{5}{2}, \frac{9}{2}
$$
Izračunaj, katero število je dvestoti člen tega zaporedja in kolikšna je vsota prvih dvestotih členov zaporedja.
b) Med števili $\frac{1}{2}$ in $\frac{9}{2}$ vrinemo tri števila, tako ... | [
"Solution:\n\na.\nIzračunamo diferenco aritmetičnega zaporedja:\n$$\nd = \\frac{1}{2} - \\left(-\\frac{3}{2}\\right) = 2.\n$$\nIzračunamo dvestoti člen zaporedja:\n$$\na_{200} = a_1 + (200-1) \\cdot d = -\\frac{3}{2} + 199 \\cdot 2 = -\\frac{3}{2} + 398 = \\frac{793}{2}.\n$$\nIzračunamo vsoto prvih dvestotih členov... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | a) Two hundredth term: 793/2. Sum of first two hundred terms: 39500. b) Inserted triples: (sqrt(3)/2, 3/2, 3*sqrt(3)/2) and (-sqrt(3)/2, 3/2, -3*sqrt(3)/2). | |
08ih | Problem:
The numbers $a_{1}, a_{2}, \ldots, a_{n}$ are the first $n$ positive integers with the property that the number $8 a_{k}+1$ is a perfect square for every $k=1,2, \ldots, n$. Find the sum $S_{n}=a_{1}+a_{2}+\ldots+a_{n}$. | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | S_n = n(n+1)(n+2)/6 | |
0629 | Problem:
Man finde alle Quadrupel positiver ganzer Zahlen ($m, n, p, q$) mit der Eigenschaft
$$
p^{m} q^{n} = (p+q)^{2} + 1
$$ | [
"Solution:\n1. Hilfssatz: Sind $k, l, a$ positive ganze Zahlen mit $a = (k^{2} + l^{2} + 1)/(k l)$, folgt $a = 3$.\n\nBeweis des Hilfssatzes: Zu gegebenem Wert von $a$ betrachten wir ein Paar $(k, l)$ mit $k^{2} + l^{2} + 1 = a k l$ und minimaler Summe $k + l$ unter allen derartigen Paaren. Ohne Einschränkung ist $... | Germany | 1. IMO-Auswahlklausur | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [(1, 2, 2, 5), (1, 2, 13, 5), (2, 1, 5, 2), (2, 1, 5, 13)] | |
06cd | There are $n$ vertices on a plane. A graph $G$ is formed by connecting every two vertices by exactly one edge, and painting all edges, each in one colour, chosen from $k$ given colours.
a. For $n = 6$ and $k = 3$, show that it is possible to paint the edges so that the graph contains no closed circuit painted in one c... | [
"a. We can colour the edges as shown. Since the subgraph formed by edges of the same colour connects all 6 vertices but there are only 5 edges, it is a tree and hence has no closed circuit.\n\n\n\nb. There are $\\binom{n}{2} \\ge \\frac{n(n-1)}{2}$ edges in total. By the pigeonhole principl... | Hong Kong | HKG TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0eku | Problem:
Kvocient dolžin katete $a$ in hipotenuze $c$ v pravokotnem trikotniku je $3:4$.
a) Pod katerim kotom se sekata simetrali ostrih kotov?
b) Pod katerim kotom seka simetrala kota $\alpha$ nasprotno kateto? | [
"Solution:\n\na) Kot med simetralama ostrih kotov je enak $\\varphi = 180^\\circ - \\frac{\\alpha}{2} - \\frac{\\beta}{2}$. Ker v pravokotnem trikotniku velja, da je $\\alpha = 90^\\circ - \\beta$, je kot $\\varphi$ enak\n$$\n\\varphi = 180^\\circ - \\frac{90^\\circ - \\beta}{2} - \\frac{\\beta}{2} = 180^\\circ - 4... | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a) 45°; b) 114° 18′ | |
06ja | There are 100 squares in a $10 \times 10$ grid table. Each square is painted in one colour. What is the maximum number of colours that may be used, so that in each row, and also in each column, the number of different colours used does not exceed five? | [
"The maximum number of colours is 41.\n\nAn example of 41 colours is given below, where the numbers represent different colours.\n\n| 1 | 1 | 1 | 1 | 1 | 1 | 2 | 3 | 4 | 5 |\n|---|---|---|---|---|---|---|---|---|---|\n| 1 | 1 | 1 | 1 | 1 | 6 | 7 | 8 | 9 | 1 |\n| 1 | 1 | 1 | 1 | 10 | 11 | 12 | 13 | 1 | 1 |\n| 1 | 1 ... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 41 | |
08xd | How many 25-tuples $(x_1, x_2, \dots, x_{25})$ of non-negative integers are there which satisfy the following identity?
$$
x_1^2 + x_2^2 + \dots + x_{25}^2 = 2 + x_1x_2 + x_2x_3 + \dots + x_{24}x_{25}.
$$ | [
"29900\n\nLet $x_0 = 0$. Then, the given identity can be rewritten as:\n$$\n\\begin{aligned}\n& x_0^2 + x_1^2 + \\dots + x_{25}^2 = 2 + x_0x_1 + x_1x_2 + \\dots + x_{24}x_{25} + x_{25}x_0 \\\\\n& \\iff (x_0 - x_1)^2 + (x_1 - x_2)^2 + \\dots + (x_{24} - x_{25})^2 + (x_{25} - x_0)^2 = 4.\n\\e... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 29900 | |
0cx4 | Find all triples $(x, y, z)$ of positive integers such that
$$
3^{x} + 4^{y} = 5^{z}
$$ | [
"Looking $\\bmod 4$ we get $(-1)^{x} \\equiv (\\bmod 4)$, hence $x$ is even. Looking $\\bmod 3$ we get $1 \\equiv (-1)^{z} (\\bmod 3)$, hence $z$ is even. Let $x = 2m$, $z = 2n$. We have\n$$\n4^{y} = (5^{n} - 3^{m})(5^{n} + 3^{m})\n$$\nhence $5^{n} + 3^{m} = 2^{a}$, $5^{n} - 3^{m} = 2^{b}$, $a + b = 2y$, $y > 0$. T... | Saudi Arabia | SAMC | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (x, y, z) = (2, 2, 2) | |
06o0 | Find all functions $f: \mathbb{R} \to \mathbb{R}$ satisfying the following condition: for any real numbers $x$ and $y$, the number $f(x + f(y))$ is equal to $x + f(y)$ or $f(f(x)) + y$. | [
"$f(x) = x + c$ for some constant $c$.\n\nWe first check that all such functions are solutions. Indeed, if $f(x) = x + c$, then\n$$\nf(x + f(y)) = f(x + y + c) = x + y + 2c = f(f(x)) + y\n$$\nfor any $x, y \\in \\mathbb{R}$.\n\nNext, we prove that these are the only solutions. Replacing $x$ by $x - f(y)$, the given... | Hong Kong | Hong Kong Team Selection Test 2 | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = x + c for some real constant c | |
0dox | A subset $S$ of the set $M = \{1, 2, \dots, p-1\}$, where $p$ is a prime number of the kind $12n + 11$ is *essential*, if the product $\Pi_s$ of all elements of the subset is not less than the product $\overline{\Pi}_s$ of all other elements of the set. The difference $\Delta_s = \Pi_s - \overline{\Pi}_s$ is called the... | [
"Answer: 2.\nLet's consider the subset $S = \\{(p+1)/2, \\dots, p-2, p-1\\}$. Then\n$$\n\\Pi_s = (-1)^{\\frac{p-1}{2}} \\left(\\frac{p-1}{2}\\right)! = - \\left(\\frac{p-1}{2}\\right)! \\equiv a \\pmod{p},\n$$\n$$\n\\overline{\\Pi}_s = \\left(\\frac{p-1}{2}\\right)! \\equiv -a \\pmod{p}.\n$$\nBy the Wilson's theore... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | 2 | |
024s | Problem:
João e Maria têm, cada um, um jarro grande com um litro de água. No primeiro dia, João coloca $1\ \mathrm{ml}$ da água do seu jarro no jarro da Maria. No segundo dia, Maria coloca $2\ \mathrm{ml}$ da água do seu jarro no jarro do João. No terceiro dia, João coloca $3\ \mathrm{ml}$ da água do seu jarro no jarr... | [
"Solution:\n\nInicialmente, o volume de água no jarro da Maria é $1\\ \\mathrm{l} = 1000\\ \\mathrm{ml}$. Depois de 200 dias, o volume é o mesmo, acrescido do que é colocado por João e diminuído do que ela tirou para colocar no do João, ou seja,\n$$\n\\begin{aligned}\n1000 & + 1 - 2 + 3 - 4 + \\cdots + 199 - 200 \\... | Brazil | Nível 2 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 900 | |
077d | Problem:
Let $n \geq 3$ be an integer and let $1 < a_{1} \leq a_{2} \leq a_{3} \leq \cdots \leq a_{n}$ be $n$ real numbers such that $a_{1} + a_{2} + a_{3} + \cdots + a_{n} = 2n$. Prove that
$$
a_{1} a_{2} \cdots a_{n-1} + a_{1} a_{2} \cdots a_{n-2} + \cdots + a_{1} a_{2} + a_{1} + 2 \leq a_{1} a_{2} \cdots a_{n}
$$ | [
"Solution:\n\nWe use Chebyshev's inequality. Observe\n$$\n\\begin{aligned}\n& n\\left(a_{1} a_{2} \\cdots a_{n-1} + a_{1} a_{2} \\cdots a_{n-2} + \\cdots + a_{1} + 1\\right) \\\\\n& \\quad = \\left(a_{1} a_{2} \\cdots a_{n-1} + a_{1} a_{2} \\cdots a_{n-2} + \\cdots + a_{1} + 1\\right)\\left(\\left(a_{n} - 1\\right)... | India | INMO | [
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0j29 | Problem:
John needs to pay 2010 dollars for his dinner. He has an unlimited supply of 2, 5, and 10 dollar notes. In how many ways can he pay? | [
"Solution:\n\nLet the number of 2, 5, and 10 dollar notes John can use be $x$, $y$, and $z$ respectively. We wish to find the number of nonnegative integer solutions to $2x + 5y + 10z = 2010$.\n\nConsider this equation modulo $2$. Because $2x$, $10z$, and $2010$ are even, $5y$ must also be even, so $y$ must be even... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 20503 | |
0dnu | Problem:
Нека су $a, b$ и $c$ позитивни реални бројеви за које важи $a+b+c=1$. Доказати:
$$
a \sqrt{2 b+1}+b \sqrt{2 c+1}+c \sqrt{2 a+1} \leqslant \sqrt{2-\left(a^{2}+b^{2}+c^{2}\right)}
$$
| [
"Solution:\n\nКвадрирањем обе стране и коришћењем једнакости $1-\\left(a^{2}+b^{2}+c^{2}\\right)=2(ab+bc+ca)$ неједнакост из задатка се своди на\n$$\n\\begin{aligned}\nL= & 2 a^{2} b+2 b^{2} c+2 c^{2} a+ \\\\\n& 2 a b \\sqrt{(2 b+1)(2 c+1)}+2 b c \\sqrt{(2 c+1)(2 a+1)}+2 c a \\sqrt{(2 a+1)(2 b+1)} \\\\\n\\leqslant ... | Serbia | 11. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
0dky | An $8 \times 8$ chessboard is initially colored in a checkerboard pattern. Dark, semi-transparent L-shaped tiles, each covering three cells, can be placed on the board. To make the entire board a uniform shade, each white square must be covered by exactly one more layer of tiles than each black square. What is the mini... | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 32 | |
0e2f | Problem:
Za neničelna realna števila $x$, $y$ in $z$ velja $3x + 2y = z$ in $\frac{3}{x} + \frac{1}{y} = \frac{2}{z}$. Dokaži, da je vrednost izraza $5x^2 - 4y^2 - z^2$ vedno celo število. | [
"Solution:\n\n1. način\nV drugo enačbo vstavimo $z = 3x + 2y$ in odpravimo ulomke. Dobimo $x^2 + 3xy + 2y^2 = 0$ oziroma $(x + y)(x + 2y) = 0$. Od tod sledi $x = -y$ ali $x = -2y$.\n\nV prvem primeru je $x = -y$ in $z = -y$, zato je $5x^2 - 4y^2 - z^2 = 5y^2 - 4y^2 - y^2 = 0$.\n\nV drugem primeru je $x = -2y$ in $z... | Slovenia | 54. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
01hb | Denote by $d(n)$ the number of (positive) divisors of a positive integer $n$. Prove that there are infinitely many positive integers $n$ such that $\lceil \sqrt{3} \cdot d(n) \rceil$ divides $n$. | [
"Note that $\\lfloor \\sqrt{3} \\cdot 8 \\rfloor = 13$. Therefore all numbers with $8$ divisors that are divisible by $13$ satisfy the condition. There are infinitely many of those, for example, all numbers in the form $13p^3$, where $p$ is a prime different from $13$."
] | Baltic Way | Baltic Way 2020 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0bi9 | a) Prove that for any real numbers $a$ and $b$ the following inequality holds:
$$
(a^2 + 1)(b^2 + 1) + 50 \ge 2(2a + 1)(3b + 1).
$$
b) Find all positive integers $n$ and $p$ such that:
$$
(n^2 + 1)(p^2 + 1) + 45 = 2(2n + 1)(3p + 1).
$$ | [
"a) The given condition rewrites as $(ab-6)^2 + (a-2)^2 + (b-3)^2 \\ge 0$, obviously true.\n\nb) Similarly, we obtain $(np-6)^2 + (n-2)^2 + (p-3)^2 = 5$, hence the numbers $(np-6)^2$, $(n-2)^2$, and $(p-3)^2$ are equal to $0$, $1$, and $4$, in some order. By inspection, we find $(n, p) \\in \\{(2, 4), (2, 2)\\}$."
... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | Part (a): holds for all real numbers. Part (b): the positive integer solutions are (2, 2) and (2, 4). | |
0ali | Problem:
If $x + y + x y = 1$, where $x, y$ are nonzero real numbers, find the value of
$$
x y + \frac{1}{x y} - \frac{y}{x} - \frac{x}{y}
$$ | [
"Solution:\nObserve that\n$$\n\\begin{aligned}\nx y + \\frac{1}{x y} - \\frac{y}{x} - \\frac{x}{y} &= \\frac{(x y)^2 + 1 - x^2 - y^2}{x y} \\\\\n&= \\frac{\\left(x^2 - 1\\right)\\left(y^2 - 1\\right)}{x y} \\\\\n&= \\frac{(x + 1)(y + 1)(x - 1)(y - 1)}{x y} \\\\\n&= (x y + x + y + 1) \\frac{(x y - x - y + 1)}{x y}\n... | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | 4 | |
0dkt | Given an acute triangle $ABC$ inscribed in $(O)$ with $\angle A = 60^\circ$ and symmedian point $L$. The tangents at $B, C$ of $(O)$ intersect $CA, AB$ at $E, F$. Prove that the line $OL$, two circumcircles of triangles $AEF$ and $BOC$ pass through the same point. | [
"Let $D$ be the midpoint of the minor arc $BC$ of $(O)$ and $P$ be the intersection point of the tangents at $B, C$ of $(O)$. We will show that $D$ and $P$ lie on $(AEF)$.\n\nIndeed, we have\n$$\n\\angle AEP = \\angle AEB = \\angle BAC - \\angle ABE = \\angle ODB - \\angle ADB = \\angle ADO\n$$\nso $(ADP)$ passes t... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellane... | null | proof only | null | |
0fxn | Problem:
Sei $n$ eine natürliche Zahl. Jedes Feld eines $n \times n$-Quadrates enthält eines von $n$ verschiedenen Symbolen, sodass jedes der Symbole in genau $n$ Feldern steht. Zeige, dass eine Zeile oder eine Spalte existiert, die mindestens $\sqrt{n}$ verschiedene Symbole enthält. | [
"Solution:\n\nWir nummerieren die Symbole von 1 bis $n$. Sei $a_{i}$ bzw. $b_{i}$ die Anzahl verschiedener Symbole in der $i$-ten Zeile bzw. Spalte, und sei $u_{k}$ bzw. $v_{k}$ die Anzahl Zeilen bzw. Spalten, in denen das Symbol $k$ vorkommt. Es gelten dann die Gleichungen\n$$\n\\sum_{i=1}^{n} a_{i}=\\sum_{k=1}^{n... | Switzerland | SMO Finalrunde | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
07rp | The incentre of triangle $ABC$ is $I$, and $H$ is the foot of the perpendicular from $I$ on $AB$. The perpendicular from $H$ on $BC$ meets $BC$ at $E$, and it meets the bisector of $\angle ABC$ at $D$. The perpendicular from $A$ on $BC$ meets $BC$ at $F$. Prove that $\angle BAC = 2\angle EFD$. | [
"Let $G$ on $BC$ be the foot of the perpendicular from $I$ on $BC$, then\n$$\n|IG| = |IH|\n$$\nis the radius of the incircle of $\\triangle ABC$. Because $AI$ is the angle bisector of the angle at $A$, we have $2\\angle HAI = \\angle BAC$.\n\n\n\nBecause $BH$ and $BG$ are tangent to the inc... | Ireland | Irish | [
"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 | |
0jm1 | Problem:
In quadrilateral $A B C D$, we have $A B=5$, $B C=6$, $C D=5$, $D A=4$, and $\angle A B C=90^{\circ}$. Let $A C$ and $B D$ meet at $E$. Compute $\frac{B E}{E D}$. | [
"Solution:\n\nAnswer: $\\sqrt{3}$\n\nWe find that $A C=\\sqrt{61}$, and applying the law of cosines to triangle $A C D$ tells us that $\\angle A D C=120^{\\circ}$. Then $\\frac{B E}{E D}$ is the ratio of the areas of triangles $A B C$ and $A D C$, which is $\\frac{(5)(6)}{(4)(5) \\frac{\\sqrt{3}}{2}}=\\sqrt{3}$."
] | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | sqrt(3) | |
0js4 | Problem:
Let $a$, $b$ and $c$ be positive real numbers such that
$$
\begin{aligned}
a^{2}+a b+b^{2} & =9 \\
b^{2}+b c+c^{2} & =52 \\
c^{2}+c a+a^{2} & =49
\end{aligned}
$$
Compute the value of $\frac{49 b^{2}-33 b c+9 c^{2}}{a^{2}}$. | [
"Solution:\nConsider a triangle $A B C$ with Fermat point $P$ such that $A P=a$, $B P=b$, $C P=c$. Then\n$$\nA B^{2}=A P^{2}+B P^{2}-2 A P \\cdot B P \\cos \\left(120^{\\circ}\\right)\n$$\nby the Law of Cosines, which becomes\n$$\nA B^{2}=a^{2}+a b+b^{2}\n$$\nand hence $A B=3$. Similarly, $B C=\\sqrt{52}$ and $A C=... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Advanced Configurations > Napoleon and Fermat points",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 52 | |
096e | Problem:
O sferă trece prin toate vârfurile unei fețe a cubului și este tangentă la toate muchiile feței opuse a cubului. Găsiți raportul dintre volumul sferei și volumul cubului. | [
"Solution:\n\nFie sfera trece prin vârfurile $A, B, C, D$. Introducem sistemul de coordonate astfel încât originea coordonatelor să coincidă cu punctul $A$, axa $Ox$ este dreapta $(AD)$, axa $Oy$ este dreapta $(AB)$, axa $Oz$ este dreapta $(AA_1)$.\nFie muchia cubului este egală cu $a$. Atunci în coordonate\n$$\nA=... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 41π√41/384 | |
04nm | Let $n$ be a positive integer. Prove that for all $x_1, x_2, \dots, x_n \in [0, 1]$ the following holds:
$$
(x_1 + x_2 + \dots + x_n + 1)^2 \ge 4(x_1^2 + x_2^2 + \dots + x_n^2). \quad (\text{ASU 1979})
$$ | [
"Since $x_i \\in [0, 1]$, we conclude that $x_i \\ge x_i^2$, i.e.\n$$\n4(x_1 + x_2 + \\dots + x_n) \\ge 4(x_1^2 + x_2^2 + \\dots + x_n^2).\n$$\nDenote $S = x_1 + x_2 + \\dots + x_n$. Notice that, because of the previous bound, it suffices to show that\n$$\n(S + 1)^2 \\ge 4S.\n$$\nThis inequality obviously holds sin... | Croatia | Croatia_2018 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
08i8 | Problem:
Let $ABC$ be an acute-angled triangle such that $m(\angle ACB) \neq 45^\circ$. The points $M$ and $N$ are the feet of the altitudes, drawn from the vertices $A$ and $B$ respectively. The points $P$ and $Q$ lie on the half-straight lines $MA$ and $NB$ respectively so that $MP = MB$ and $NQ = NA$. Prove that th... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0hdt | Given a prime number $p > 2$. There are $N$ people in a circle, each one came up with a positive integer and then wrote down the remainder when divided by $p$ on a piece of paper. Then everyone showed their piece of paper to the neighbor on the right, computed the product of their number on a piece of paper and their n... | [
"Clearly, there cannot be more than $p$ different numbers. It follows from the condition of the problem, that $0$ could not be someone's number on the first step, since then that person will have it on the second step as well. Also, no one can have $1$ as a first number, since then the neighbor on the left will hav... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | p-2 | |
06wp | Alice is given a rational number $r>1$ and a line with two points $B \neq R$, where point $R$ contains a red bead and point $B$ contains a blue bead. Alice plays a solitaire game by performing a sequence of moves. In every move, she chooses a (not necessarily positive) integer $k$, and a bead to move. If that bead is p... | [
"Answer: All $r=(b+1) / b$ with $b=1, \\ldots, 1010$.\n\nDenote the red and blue beads by $\\mathcal{R}$ and $\\mathcal{B}$, respectively. Introduce coordinates on the line and identify the points with their coordinates so that $R=0$ and $B=1$. Then, during the game, the coordinate of $\\mathcal{R}$ is always small... | IMO | IMO 2021 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions ... | null | proof and answer | All r = (b+1)/b for integers b from 1 to 1010 | |
0e3w | Three equal circles are inscribed into the equilateral triangle with the sides of length $1$ (see figure). What is the length of the radii of the circles?
(A) $\frac{1}{3}$
(B) $\frac{1}{4}$
(C) $\frac{\sqrt{3}}{4}$
(D) $\frac{\sqrt{3}+1}{4}$
(E) $\frac{\sqrt{3}-1}{4}$ | [
"If we denote the radii by $r$, we have $r\\sqrt{3} + 2r + r\\sqrt{3} = 1$. It follows that $r = \\frac{1}{2(1+\\sqrt{3})} = \\frac{\\sqrt{3}-1}{4}$."
] | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | E | |
0kr5 | Problem:
A random binary string of length $1000$ is chosen. Let $L$ be the expected length of its longest (contiguous) palindromic substring. Estimate $L$.
An estimate of $E$ will receive $\left\lfloor 20 \min \left(\frac{E}{L}, \frac{L}{E}\right)^{10}\right\rfloor$ points. | [
"Solution:\nThe probability that there exists a palindromic substring of length $2n+1$ is approximately $2^{-n} \\cdot 1000$. Thus, we can expect to often see a length $21$ palindrome, and sometimes longer ones. This leads to a guess a bit above $21$.\n$L$ was approximated with $10^{7}$ simulations (the answer is g... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | final answer only | null | |
06uq | Given any set $S$ of positive integers, show that at least one of the following two assertions holds:
(1) There exist distinct finite subsets $F$ and $G$ of $S$ such that $\sum_{x \in F} 1 / x = \sum_{x \in G} 1 / x$;
(2) There exists a positive rational number $r < 1$ such that $\sum_{x \in F} 1 / x \neq r$ for all fi... | [
"Solution 1. Argue indirectly. Agree, as usual, that the empty sum is $0$ to consider rationals in $[0,1)$; adjoining $0$ causes no harm, since $\\sum_{x \\in F} 1 / x = 0$ for no nonempty finite subset $F$ of $S$. For every rational $r$ in $[0,1)$, let $F_{r}$ be the unique finite subset of $S$ such that $\\sum_{x... | IMO | IMO Shortlisted Problems | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
0k2x | Problem:
An integer $c$ is square-friendly if it has the following property: For every integer $m$, the number $m^{2}+18 m+c$ is a perfect square. (A perfect square is a number of the form $n^{2}$, where $n$ is an integer. For example, $49=7^{2}$ is a perfect square while $46$ is not a perfect square. Further, as an e... | [
"Solution:\n\na. $c=81$ is square-friendly, since $m^{2}+18 m+81=(m+9)^{2}$ which is an integer whenever $m$ is an integer.\n\nb. Suppose there are two different square-friendly integers $c$ and $c^{\\prime}$. In that case, for every integer $m$, we have that $m^{2}+18 m+c$ and $m^{2}+18 m+c^{\\prime}$ are perfect ... | United States | Bay Area Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 81 | |
04nb | Determine all real numbers $x$ such that
$$
(x - 1009)^3 + (2x - 1009)^3 + (2018 - 3x)^3 = 0.
$$ | [] | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 1009/2, 1009, 2018/3 | |
0idu | Problem:
I have chosen five of the numbers $\{1,2,3,4,5,6,7\}$. If I told you what their product was, that would not be enough information for you to figure out whether their sum was even or odd. What is their product? | [
"Solution: 420\nGiving you the product of the five numbers is equivalent to telling you the product of the two numbers I didn't choose. The only possible products that are achieved by more than one pair of numbers are $12(\\{3,4\\}$ and $\\{2,6\\})$ and $6(\\{1,6\\}$ and $\\{2,3\\})$. But in the second case, you at... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 420 | |
056n | A sequence of positive real numbers $a_1, a_2, a_3, \dots$ satisfies $a_n = a_{n-1} + a_{n-2}$ for all $n \ge 3$. A sequence $b_1, b_2, b_3, \dots$ is defined by equations $b_1 = a_1$, $b_n = a_n + (b_1 + b_2 + \dots + b_{n-1})$ for even $n > 1$, $b_n = a_n + (b_2 + b_4 + \dots + b_{n-1})$ for odd $n > 1$. Prove that i... | [
"The definition of sequence $(b_n)$ indicates that $b_n - b_{n-2} = a_n - a_{n-2} + b_{n-1}$ for all $n \\ge 3$. Therefore $b_n - b_{n-2} = a_{n-1} + b_{n-1}$, i.e., $b_n = a_{n-1} + b_{n-1} + b_{n-2}$.\n\nNotice that the left-hand inequality $\\frac{1}{3} < \\frac{b_n}{n \\cdot a_n}$ holds for $n=2$ and $n=3$ as $... | Estonia | IMO Team Selection Contest | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
07iz | We say a set $\{a_1, \dots, a_n\}$ of real numbers is *interesting* if for all $1 < i < n$ we have $0 < a_{i+1} - a_i < a_i - a_{i-1}$. For each positive integer $m$ find the smallest positive integer $k$ such that there is an interesting set $A$ with $k$ elements such that $1$ can be written in at least $m$ different ... | [
"**First Solution.** Defining $b_i = a_{i+1} - a_i$, $i = 1, \\dots, n-1$. Thus, $(b_i)$ is a decreasing sequence of positive real numbers. Then, the difference of any two elements of $A$ can be written as the sum of some of the $b_i$'s. Assume\n$$\nb_{i_1} + b_{i_2} + \\dots + b_{i_j} = b_{i_2} + \\dots + b_{i_j} ... | Iran | 41th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | k = 2m | |
0cge | Let $a, b \in \mathbb{R}$, $a > 1$, $b > 0$. Determine the minimum value of the real number $\alpha$ such that:
$$
(a + b)^x \geq a^x + b, \quad \forall x \geq \alpha.
$$ | [
"The given relation is equivalent to:\n$$\n\\left(1 + \\frac{b}{a}\\right)^x - b\\left(\\frac{1}{a}\\right)^x \\geq 1, \\quad \\forall x \\geq \\alpha.\n$$\nConsidering the function $f : \\mathbb{R} \\to \\mathbb{R}$, $f(x) = (1 + \\frac{b}{a})^x - b(\\frac{1}{a})^x$, we can notice that the function is increasing, ... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | 1 | |
0809 | Problem:
Diciamo che tre numeri dispari $a < b < c$ sono consecutivi se $c - b = b - a = 2$. Chiamiamo "speciali" quei numeri interi che hanno tutte le cifre uguali e che si possono scrivere come somma dei quadrati di tre numeri dispari consecutivi.
a) Determinare tutti i numeri speciali di 4 cifre.
b) Esistono nume... | [
"Solution:\n\nSupponiamo che un certo numero $N$ sia speciale e abbia 4 cifre. Allora esisterà un numero dispari $d$ tale che\n$$\nN = (d-2)^2 + d^2 + (d+2)^2 = 3d^2 + 8\n$$\nPoiché $d$ è dispari, anche $N$ è dispari e dunque può essere solo $1111$, $3333$, $5555$, $7777$, $9999$. Ora però $N$ non può essere $3333$... | Italy | Cesenatico | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) 5555; b) No | |
0a7c | Problem:
Determine the last two digits of the number
$$
2^{5}+2^{5^{2}}+2^{5^{3}}+\cdots+2^{5^{1991}}
$$
written in decimal notation. | [
"Solution:\nWe first show that all numbers $2^{5^{k}}$ are of the form $100p + 32$. This can be shown by induction. The case $k = 1$ is clear ($2^{5} = 32$). Assume $2^{5^{k}} = 100p + 32$. Then, by the binomial formula,\n$$\n2^{5^{k+1}} = \\left(2^{5^{k}}\\right)^{5} = (100p + 32)^{5} = 100q + 32^{5}\n$$\nand\n$$\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 5 | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 12 | |
0bcj | Problem:
Fie $A$ o matrice neinversabilă de ordin $n$, $n>1$, cu elemente în mulţimea numerelor complexe, toate elementele având modulul egal cu $1$.
a) Arătaţi că pentru $n=3$, două dintre liniile sau două dintre coloanele matricei $A$ sunt proporţionale.
b) Rămâne adevărată concluzia de la punctul anterior pentru ... | [] | Romania | Olimpiada Naţională de Matematică | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | No | |
0e4i | Let $P$ be a point inside the triangle $ABC$, such that the angles $\angle CBP$ and $\angle PAC$ are equal. Denote the intersection of the line $AP$ and the segment $BC$ by $D$, and the intersection of the line $BP$ with the segment $AC$ by $E$. The circumcircles of the triangles $ADC$ and $BEC$ meet at $C$ and $F$. Sh... | [
"Using the equalities between the inscribed angles in the quadrilaterals $AFDC$ and $BCEF$ as well as the given equality $\\angle CBP = \\angle PAC$ we find that\n$$\n\\begin{aligned}\n\\angle CFE &= \\angle CBE = \\angle CBP = \\angle PAC \\\\\n&= \\angle DAC = \\angle DFC.\n\\end{aligned}\n$$\n\n![](attached_imag... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0d18 | Given a triangle $ABC$, the internal bisector of $\hat{A}$ intersects $BC$ at $D$ and the external bisector of $\hat{A}$ intersects $BC$ at $E$. Let $C_1$ be the circumcircle of $ADE$. $AC$ intersects $C_1$ at $F$. Let $C_2$ be the circumcircle of $ABF$ and the tangent of $C_2$ at $A$ intersects $C_1$ at $G$. Prove tha... | [
"Let $\\widehat{ABC} = 2\\beta$, $\\widehat{BAC} = 2\\alpha$, $\\widehat{ACB} = 2\\gamma$. Thus, the law of sines on triangle $AEF$ gives us\n\n$$\n\\frac{AE}{AF} = \\frac{\\sin(\\alpha + 2\\beta)}{\\sin(\\alpha - \\beta + \\gamma)}.\n$$\n\n\nOn the other hand, using the law of sines on $AB... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof only | null | |
0a4h | Problem:
Vind alle functies $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ zo dat voor alle positieve gehele getallen $m, n$ geldt dat
$$(f(m))^2 + 2mf(n) + f(n^2)$$
het kwadraat van een geheel getal is. | [
"Solution:\n\nVul in $m = n = 1$, dan moet $f(1)^2 + 3f(1)$ een kwadraat zijn. Aangezien\n$$(f(1) + 1)^2 \\le f(1)^2 + 3f(1) < (f(1) + 2)^2,$$\nmoet er links gelijkheid gelden, en dus is $f(1) = 1$. Laat nu $p = 2k+1$ een oneven priemgetal en vul in $m = k = (p-1)/2$ en $n = 1$. Dan zien we dat $f(k)^2 + p$ een kwa... | Netherlands | IMO-selectietoets II | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order anal... | null | proof and answer | f(m) = m for all positive integers m | |
0isy | Problem:
In a triangle $ABC$, take point $D$ on $BC$ such that $DB = 14$, $DA = 13$, $DC = 4$, and the circumcircle of $ADB$ is congruent to the circumcircle of $ADC$. What is the area of triangle $ABC$? | [] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 108 | |
0ezd | Problem:
The product of three positive numbers is $1$, their sum is greater than the sum of their inverses. Prove that just one of the numbers is greater than $1$. | [
"Solution:\nThe product of the numbers is $1$, so they cannot all be greater than $1$ or all less than $1$. If all equalled $1$, then the sum would not be greater than the sum of the inverses. So we must have either one or two greater than $1$. Thus it is sufficient to show that we cannot have two of the numbers gr... | Soviet Union | 4th ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0467 | Let $a$, $b$, and $d$ be integers satisfying $d \ge 0$, $|a| \ge 2$, and $b \ge (|a| + 1)^{d+1}$. Suppose $f(x)$ is a real-coefficient polynomial of degree $d$, and for each positive integer $n$, let $r_n$ denote the remainder of $\lfloor f(n)a^n \rfloor$ modulo $b$.
*Proof:* If the sequence $r_n$ is ultimately period... | [
"**Proof. Lemma** Let integers $a$ and real number $z$ satisfy $|a| \\ge 2$. For any non-negative integer $n$, if\n$$\n||a^n z|| < \\frac{1}{|a| + 1},\n$$\nthen $z$ is an integer. Here, $||x||$ is defined as the distance between the real number $x$ and the nearest integer, that is, $||x|| = \\min(\\{x\\}, 1 - \\{x\... | China | China-TST-2023B | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0554 | Solve the system $a^3 + b = 4c$, $a + b^3 = c$, $ab = -1$. | [
"From the third equation we get $b = -\\frac{1}{a}$. By substituting this in the first and second equations we obtain a new system: $a^3 - \\frac{1}{a} = 4c$, $a - \\frac{1}{a^3} = c$.\n\nIf $c = 0$, we have $a^3 = \\frac{1}{a}$ and $a^4 = 1$, whence $a = 1$ or $a = -1$, since $a = 0$ is not possible. We have respe... | Estonia | Open Contests | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | (1, -1, 0), (-1, 1, 0), (2, -1/2, 15/8), (-2, 1/2, -15/8) | |
07ti | For each integer $n \ge 100$ we define $T(n)$ to be the number obtained from $n$ by moving the two leading digits to the end. For example, $T(12345) = 34512$ and $T(100) = 10$. Find all integers $n \ge 100$ for which:
$$
n + T(n) = 10n.
$$ | [
"The general solution is\n$$\nn = \\frac{10}{91} \\times (10^{6k} - 1) = (1 + 10^6 + 10^{12} + \\dots + 10^{6(k-1)}) \\times 109890\n$$\nfor any positive integer $k$. In other words, the number $n$ consists of $k$ replicates of the digits 109890.\n\nLet $m$ be the number of digits of $n$ so $m \\ge 3$ when $n \\ge ... | Ireland | IRL_ABooklet | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All n of the form n = 10(10^{6k} − 1)/91 for integers k ≥ 1; equivalently, n is the k-fold concatenation of the six-digit block 109890. | |
0cqq | A grid square $n \times n$ is given. Initially its cells are colored in white and black in a chessboard manner so that at least one of the corner cells is black. By one move it is allowed to recolor all 4 cells in some square $2 \times 2$ following the rule: each white cell is recolored to black, each black cell is rec... | [
"All positive integer $n$ divisible by $3$.\n\nSuppose we managed to recolor the cells as required by the problem. Let us call cells of the first type those that were originally white, and cells of the second type those that were originally black. Note that if a cell is recolored three times, it ends up with its or... | Russia | Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | All positive integers n divisible by 3 | |
08gp | Problem:
Dato il polinomio $p(x) = x(x+1)(x-2)^2$, consideriamo il polinomio $q(x) = p(p(p(\ldots(p(x)) \ldots)))$ dato dalla composizione di $p(x)$ con se stesso 2024 volte. Quanti sono gli interi $k$ tali che si abbia $q(k) = 0$?
(A) 1
(B) 3
(C) 4
(D) 5
(E) 6 | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Chiamiamo $q_{k}(x) = p(p(\\ldots(p(x))))$ il polinomio ottenuto componendo $k$ volte il polinomio $p$ con se stesso, così che $q(x) = q_{2024}(x)$. È chiaro che $q_{k}(x)$ è un polinomio a coefficienti interi. Poiché per $k > 1$ si ha $q_{k}(x) = p\\left(q_{k-1}(x)\\rig... | Italy | Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
03as | The convex quadrilateral $ABCD$ is inscribed in the circle with diameter $AC$. The point $A$ lies on the smaller arc $BD$. Prove the inequality $|AB - AD| \ge |CB - CD|$. | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0go1 | Some of $n \ge 4$ students in a class are friends. Any $n-1$ students in this class can form a circle so that any two students next to each other on the circle are friends, but all $n$ students cannot form a similar circle. Find the smallest possible value of $n$. | [
"We want to show that if $G - v$ has a Hamiltonian cycle for each vertex $v$ in a graph $G$, but $G$ does not have a Hamiltonian cycle, then $n = |G| \\ge 10$, and produce such a graph $G$ with $n = 10$.\n\nSince $v$ cannot be adjacent with two consecutive vertices in a Hamiltonian cycle in $G - v$, we have $\\deg ... | Turkey | Team Selection Test for IMO 2009 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 10 | |
0kfy | Let $ABC$ be a scalene triangle with incenter $I$. The incircle of $ABC$ touches $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $P$ be the foot of the altitude from $D$ to $EF$, and let $M$ be the midpoint of $BC$. The rays $AP$ and $IP$ intersect the circumcircle of triangle $ABC$ again at points $G$ and ... | [
"Refer to the figure below.\n\n\n**Claim** — The point $Q$ is the Miquel point of $BFEC$. Also, $\\overline{QD}$ bisects $\\angle BQC$.\n*Proof*. Inversion around the incircle maps line $EF$ to $(AIEF)$ and the nine-point circle of $\\triangle DEF$ to the circumcircle of $\\triangle ABC$ (a... | United States | USA TSTST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configur... | English | proof only | null | |
0aet | **Даден е триаголникот $ABC$, ($BC < AB$).** Низ точката $C$ е повлечена права $l$, нормална на симетралата $BE$ на аголот $\angle B$. Правата $l$ ја сече $BE$ во точка $F$, а тежишната линија $BD$ во точка $G$. Да се докаже дека отсечката $DF$ ја преполовува отсечката $EG$. | [
"Нека $CF \\cap AB = \\{K\\}$ и $DF \\cap BC = \\{M\\}$. Бидејќи $BF \\perp KC$ и $BF$ е симетрала на $\\angle KBC$ следува дека $\\angle KBC$ е рамнокрак, т.е. $\\overline{BK} = \\overline{BC}$ и уште $F$ е средина на $a > b > 0$.\n\n\n\nСпоред тоа, $DF$ е средна линија за $B$, односно $\\... | North Macedonia | Републички натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Macedonian, English | proof only | null | |
0705 | Problem:
Find all solutions in real numbers $x_{1}, x_{2}, \ldots, x_{n+1}$ all at least $1$ such that:
(1) $x_{1}^{1/2} + x_{2}^{1/3} + x_{3}^{1/4} + \ldots + x_{n}^{1/(n+1)} = n x_{n+1}^{1/2}$;
and
(2) $\dfrac{x_{1} + x_{2} + \ldots + x_{n}}{n} = x_{n+1}$. | [
"Solution:\n\nBy Cauchy-Schwarz, $\\left(\\sum x_{i}^{1/2}\\right)^{2} \\leq \\left(\\sum 1\\right)\\left(\\sum x_{i}\\right)$, with equality if and only if all $x_{i}$ are equal. In other words, if we put $x_{n+1} = \\dfrac{x_{1} + x_{2} + \\ldots + x_{n}}{n}$, then $\\sum x_{i}^{1/2} \\leq n x_{n+1}^{1/2}$.\n\nBu... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | x1 = x2 = ... = x_{n+1} = 1 | |
0hos | Problem:
If the sum of digits in a decimal representation of a natural number $n$ is equal to 2006, prove that $n$ can't be a perfect square of an integer. | [
"Solution:\n\nThe remainder of $n$ upon division by $3$ is equal to the sum of its digits, i.e. $2006$. Hence number $n$ has a remainder $2$ upon division by $3$ and no square can have that remainder."
] | United States | Berkeley Math Circle Monthly Contest 2 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof only | null | |
0coz | Given a triangle $ABC$ with $\angle BAC = 60^\circ$. Let $BB_1$ and $CC_1$ be the angle bisectors in this triangle. Prove that the point symmetrical to $A$ with respect to $B_1C_1$ lies on side $BC$.
В треугольнике $ABC$ угол $A$ равен $60^\circ$. Пусть $BB_1$ и $CC_1$ — биссектрисы этого треугольника. Докажите, что т... | [
"Пусть $I$ — точка пересечения биссектрис треугольника $ABC$. Тогда $\\angle B_1IC_1 = \\angle BIC = 180^\\circ - \\angle IBC - \\angle ICB = 180^\\circ - (\\angle ABC + \\angle ACB)/2 = 180^\\circ - 60^\\circ = 120^\\circ = 180^\\circ - \\angle B_1AC_1$. Значит, четырёхугольник $AB_1IC_1$ вписан в окружность. Отсю... | Russia | Regional round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0ejl | Problem:
V pokončno prizmo, katere osnovna ploskev je paralelogram z enim notranjim kotom $\alpha$, je včrtana sfera z radijem $r$, ki se dotika vseh mejnih ploskev prizme. Izrazi prostornino prizme s pomočjo $r$ in $\alpha$. | [
"\n\nKer se sfera dotika spodnje in zgornje ploskve prizme, je višina prizme enaka premeru sfere, torej $h=2r$.\n\nNaj bo $\\Sigma$ ravnina, ki vsebuje središče sfere in je vzporedna osnovni ploskvi prizme. Označimo s $\\mathcal{P}$ presečišče prizme in ravnine $\\Sigma$ ter s $\\mathcal{K}... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 8 r^3 / sin α | |
0dd4 | Let $ABC$ be an acute, non-isosceles triangle with $AD$, $BE$, $CF$ as altitudes and $d$ is the tangent line of the circumcircle of triangle $ABC$ at $A$. The line through $H$ and parallel to $EF$ cuts $DE$, $DF$ at $Q$, $P$ respectively. Prove that $d$ is tangent to the ex-circle with respect to vertex $D$ of triangle... | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard ... | null | proof only | null | |
0fjt | Problem:
Sean $a$ y $b$ enteros. Demostrar que la ecuación
$$
(x-a)(x-b)(x-3)+1=0
$$
admite a lo sumo una solución entera. | [
"Solution:\nSea el entero $p$ una raíz, entonces: $(x-a)(x-b)(x-3)+1$ se anula para $x=p$, es decir\n$$\n(p-a)(p-b)(p-3) = -1\n$$\nDistingamos varios casos\n$$\n\\text{1.- } (p-3) = 1 \\Rightarrow p = 4\n$$\nentonces para los otros factores tenemos dos posibilidades:\n$$\n\\begin{aligned}\n& \\left.\\begin{array}{c... | Spain | Spanish National Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0853 | Problem:
Consideriamo le quattro affermazioni seguenti:
Manuela ha un cane e un gatto.
Manuela non ha né un cane né un gatto.
Se Manuela ha un cane, allora ha anche un gatto.
Manuela non ha un cane, ma ha un gatto.
Quante di esse, al massimo, possono essere false contemporaneamente?
(A) 0
(B) 1
(C) 2
(D) 3
(E) 4 . | [
"Solution:\nLa risposta è (E). Le affermazioni \"Manuela ha un cane e un gatto\", \"Manuela non ha né un cane né un gatto\" e \"Manuela non ha un cane, ma ha un gatto\" sono tra loro incompatibili. L'affermazione \"Se Manuela ha un cane, allora ha anche un gatto\" è vera esattamente quando una (ed una sola) delle a... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0bod | Let $(x_n)_{n \ge 1}$ be a sequence of real numbers from $[1, \infty)$. It is known that the sequences $(y_n^{(k)})_{n \ge 1}$, defined by $y_n^{(k)} = \lfloor x_n^k \rfloor$, $n \ge 1$, are convergent for every $k \in \mathbb{N}^*$. Prove that the sequence $(x_n)_{n \ge 1}$ is convergent.
Mihai Piticari, Vlad Cerbu | [
"For $k \\in \\mathbb{N}^*$, the sequence $(y_n^{(k)})_{n \\ge 1}$ is convergent and its terms are integers. Then there exist $n_k, a_k \\in \\mathbb{N}^*$ so that $y_n^{(k)} = a_k, \\forall n \\ge n_k$. Consequently, $x_n^k \\in [a_k, a_k + 1), \\forall n \\ge n_k$. In particular, $x_n \\in [a_1, a_1 + 1), \\foral... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0frk | Problem:
Determinar todas las funciones $f$ tales que
$$
f(x f(y)+y)=f(x y)+f(y)
$$
para cualesquiera números reales $x, y$. | [
"Solution:\n\nLlamamos (E) a la ecuación del enunciado. Para cada $k \\in \\mathbb{R}$, la función $f(x)=k$ es solución si y sólo si $k=k+k$, es decir $k=0$. Así, la función $f(x)=0$ verifica (E), y es la única solución constante.\n\nPor otra parte, la función identidad $f(x)=x$ también verifica (E), ya que los tér... | Spain | Spain | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; and f(x) = x for all real x | |
0ju1 | Problem:
For a positive integer $n$, denote by $\tau(n)$ the number of positive integer divisors of $n$, and denote by $\phi(n)$ the number of positive integers that are less than or equal to $n$ and relatively prime to $n$. Call a positive integer $n$ good if $\varphi(n)+4 \tau(n)=n$. For example, the number $44$ is ... | [
"Solution:\n\nWe claim that $44, 56, 72$ are the only good numbers. It is easy to check that these numbers work.\n\nNow we prove none others work. First, remark that as $n=1,2$ fail so we have $\\varphi(n)$ is even, thus $n$ is even. This gives us $\\varphi(n) \\leq n / 2$. Now remark that $\\tau(n)<2 \\sqrt{n}$, s... | United States | HMMT February | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, ineq... | null | proof and answer | 172 | |
01xx | The circles $\omega_1$ and $\omega_2$ centered at $O_1$ and $O_2$, respectively, intersect at two different points $E$ and $F$. The line $O_1E$ intersects for the second time $\omega_1$ at the point $A$, and $\omega_2$ at the point $C$. The line $O_2E$ intersects for the second time $\omega_2$ at the point $B$, and $\o... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fz0 | Problem:
Zeige, dass es mehr als $10^{13}$ Möglichkeiten gibt, 81 Könige so auf einem $18 \times 18$ Schachbrett zu platzieren, dass sich keine zwei Könige attackieren. | [
"Solution:\n\nSei $A_{i, j}$ das $2 \\times 2$ Quadrat mit den Feldern $(2i-1, 2j-1), (2i-1, 2j), (2i, 2j-1), (2i, 2j)$. Dadurch wird das $18 \\times 18$ Brett paarweise disjunkt in $81$ $2 \\times 2$ Quadrate aufgeteilt. Falls 2 Könige auf demselben $2 \\times 2$ Quadrat stehen, müssen sie sich notwendigerweise an... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
02v7 | Problem:
Considere um triângulo $ABC$ e quadrados $ABDE$ e $ACFG$ construídos exteriormente sobre seus lados. Os segmentos $BG$ e $EC$ se intersectam em $P$.
a) Prove que os triângulos $BAG$ e $EAC$ são congruentes.
b) Como esses dois triângulos são congruentes e possuem o ponto $A$ em comum, ... | [
"Solution:\nConsidere a figura a seguir em que marcamos os lados iguais dos dois quadrados.\n\n\na) Podemos afirmar que os triângulos $BAG$ e $EAC$ são congruentes pelo caso $LAL$, já que $BA=EA$, $\\angle BAG=\\angle EAC=90^\\circ+\\angle BAC$ e $AG=AC$.\n\nb) Como o ângulo formado entre $... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | Rotation angle = 90°, and ∠BPC = 90° | |
0e2h | Problem:
Zapiši definicijsko območje funkcije $f(x) = \log_{\frac{1}{2}} \left( \frac{x}{x+2} - \frac{1}{x} \right) - \frac{1}{2}$. | [
"Solution:\n\nUgotovimo, da mora biti logaritmand $\\frac{x}{x+2} - \\frac{1}{x}$ večji od $0$. Zapišemo neenačbo $\\frac{x}{x+2} - \\frac{1}{x} > 0$. Neenačbo uredimo $\\frac{x^2 - x - 2}{x(x+2)} > 0$. Določimo ničle $x_1 = 2$ in $x_2 = -1$ in pole $x_1 = 0$ in $x_2 = -2$. Na številski premici določimo predznake i... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | (-∞, -2) ∪ (-1, 0) ∪ (2, ∞) | |
0lfd | Let $n \ge 3$ be an integer and $S$ be a set of $n$ elements. Determine the largest integer $k_n$ such that: for each selection of $k_n$ 3-subsets of $S$, there exists a way to color elements of $S$ with two colors such that none of the chosen 3-subset is monochromatic. | [
"The answer to the problem depends on the values of $n$. We have the following cases\n\n* If $n = 3$ then it is obvious that $k_n = 1$.\n\n* If $n = 4$ then $k_n = 4$ because $A_4$ has 4 subsets with 3 elements. For example, color (1, 2) blue and (3, 4) red. Each set contains one of the two numbers above, so it has... | Vietnam | Team selection tests | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | k3 = 1, k4 = 4, k5 = 9, k6 = 9, and k_n = 6 for all n ≥ 7 | |
05tx | Problem:
Soit $a$, $b$, $c$ et $d$ quatre nombres réels. On suppose qu'il existe une permutation $(x, y, z, t)$ des nombres $a, b, c$ et $d$ telle que
$$
x \leqslant 2 a-b, \quad y \leqslant 2 b-c, \quad z \leqslant 2 c-d \text{ et } t \leqslant 2 d-a.
$$
Démontrer que $a = b = c = d$. | [
"Solution:\n\nPour mieux exploiter la première inégalité, que l'on réécrit comme $x + b \\leqslant 2a$, l'idéal serait que $x$ et $b$ soient maximaux, ou bien que $a$ soit minimal. Évidemment, il se pourrait très bien que ni $a$, ni $b$, ni $x$ ne soit extrémal. Néanmoins, puisque les variables jouent des rôles cyc... | France | Préparation Olympique Française de Mathématiques - Test du 14 et du 21 Février 2021 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0dp2 | The set $f_1, f_2, \dots, f_n$ of polynomials with real coefficients is called *special*, if for all distinct $i, j, k \in \{1, 2, \dots, n\}$ the polynomial $\frac{2}{3}f_i + f_j + f_k$ has no real roots, but for all distinct $p, q, r, s \in \{1, 2, \dots, n\}$ the polynomial $f_p + f_q + f_r + f_s$ has a real root.
1... | [
"1) The polynomials $x^2 + x + 100$, $x^2 + x + 100$, $-x^2 + x - 200$, $1$ give a needed example.\n\n2) To the contrary, assume that there exists a special set of polynomials $f_1, f_2, f_3, f_4, f_5$ with real coefficients. Consider a complete graph with a set of vertices $\\{1, 2, 3, 4\\}$. An edge $(i, j)$ is c... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | English | proof and answer | 1) x^2 + x + 100, x^2 + x + 100, -x^2 + x - 200, 1. 2) No; a special set of five polynomials does not exist. | |
0963 | Problem:
Fie $ABC$ un triunghi isoscel cu $AC = BC$. Fie $M$ mijlocul laturii $AB$, $N$ - piciorul perpendicularei duse din $M$ pe $AC$, iar $P$ - mijlocul segmentului $MN$. Demonstraţi că dreptele $BN$ şi $CP$ sunt reciproc perpendiculare. | [
"Solution:\n\nFie $Q$ - mijlocul segmentului $AN$, atunci $MQ$ este linia mijlocie a triunghiului $ABN \\Rightarrow MQ \\parallel BN$. Vom demonstra că $CP \\perp MQ$.\n\nFie $CP \\cap MQ = \\{K\\}$. Fie $m(\\angle A) = \\alpha$. Unghiurile $CMN$ şi $MAN$, fiind cu laturile respectiv perpendiculare, sunt congruente... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0fbf | Problem:
Se sabe que un dodecaedro regular es un poliedro regular con 12 caras pentagonales iguales y concurriendo 3 aristas en cada vértice. Se pide calcular, razonadamente,
a) el número de vértices,
b) el número de aristas,
c) el número de diagonales de todas las caras,
d) el número de segmentos rectilíneos determin... | [
"Solution:\n\n1. Número de vértices: $\\frac{12 \\cdot 5}{3} = 20$.\n\n2. Número de aristas: $\\frac{12 \\cdot 5}{2} = 30$.\n\n3. Número de diagonales de todas las caras\n$$\n12 \\left( \\binom{5}{2} - 5 \\right ) = 60\n$$\n\n4. Número de segmentos rectilíneos determinados por cada dos vértices\n$$\n\\binom{20}{2} ... | Spain | OME 11 | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 20, b) 30, c) 60, d) 190, e) 100 | |
08ds | Problem:
Su una circonferenza consideriamo nell'ordine cinque punti $A, B, C, D, E$. Supponiamo che le rette $BC$ e $DE$ si intersechino in un punto $F$, che $F$ e $A$ siano da parti opposte rispetto alla retta $BE$, e che la circonferenza circoscritta al triangolo $BFE$ sia tangente (in $E$) alla retta $AE$.
a. Dimo... | [] | Italy | XXXVI Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
054k | Positive integer $b$ is obtained by reordering the digits in a positive integer $a$. Which of the following claims are definitely true?
a) The sums of the digits of numbers $2a$ and $2b$ are equal.
b) The sums of the digits of numbers $3a$ and $3b$ are equal.
c) The sums of the digits of numbers $5a$ and $5b$ are equa... | [
"Call digits $0, 1, 2, 3, 4$ small and digits $5, 6, 7, 8, 9$ large. Denote the digits of $k$-digit number $n$ from right to left by $(n)_0, (n)_1, \\dots, (n)_{k-1}$. Denote by $\\Sigma(n)$ the sum of all digits of $n$ and by $l(n)$ the number of large digits of $n$.\n\na) If $(a)_i$ is small then $(2a)_i = 2(a)_i... | Estonia | National Olympiad Final Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | a and c | |
0fbb | Problem:
Sean $C$ y $C'$ dos circunferencias concéntricas de radios $r$ y $r'$ respectivamente. Determinar cuánto ha de valer el cociente $r'/r$ para que en la corona limitada por $C$ y $C'$ existan ocho circunferencias $C_i$, $i=1, \ldots, 8$, que sean tangentes a $C$ y a $C'$, y también que $C_i$ sea tangente a $C_{... | [
"Solution:\n\nEn la figura vemos que\n$$\n\\sen \\frac{\\pi}{8} = \\frac{\\frac{r' - r}{2}}{\\frac{r + r'}{2}} = \\frac{\\frac{r'}{r} - 1}{1 + \\frac{r'}{r}}\n$$\nde donde se tiene que\n$$\n\\frac{r'}{r} = \\frac{1 + \\sen \\frac{\\pi}{8}}{1 - \\sen \\frac{\\pi}{8}}\n$$"
] | Spain | OME 10 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | (1 + sin(pi/8)) / (1 - sin(pi/8)) | |
0gl9 | A Pattano coin is a coin which has a blue side and a yellow side. A positive integer not exceeding $100$ is written on each side of every coin (the sides may have different integers).
Two Pattano coins are *identical* if the number on the blue side of both coins are equal and the number on the yellow side of both coin... | [
"We represent each Pattano coin by an ordered pair $(i, j)$, where $i$ and $j$ are the numbers written on the blue and yellow side respectively. Let $C$ be the set of non-identical $2559$ Pattano coins. Let $b_i$ denote the number of Pattano coins in $C$ with $i$ on its blue side, and $y_j$ denote the number of Pat... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0jya | Problem:
Let $S$ be a set of 2017 distinct points in the plane. Let $R$ be the radius of the smallest circle containing all points in $S$ on either the interior or boundary. Also, let $D$ be the longest distance between two of the points in $S$. Let $a, b$ be real numbers such that $a \leq \frac{D}{R} \leq b$ for all ... | [
"Solution:\n\nIt is easy to verify that the smallest circle enclosing all the points will either have some 2 points in $S$ as its diameter, or will be the circumcircle of some 3 points in $S$ who form an acute triangle.\n\nNow, clearly $\\frac{D}{R} \\leq 2$. Indeed consider the two farthest pair of points $S_{1}, ... | United States | February 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (sqrt(3), 2) | |
0h0l | Let $h$ be the altitude of triangle $ABC$ passing through the vertex $A$ and $\alpha = \angle BAC$. Prove that the following inequality holds:
$$
AB + AC \ge BC \cdot \cos \alpha + 2h \sin \alpha.
$$
In what triangles does equality hold? | [
"If $\\alpha \\ge 90^\\circ$, then $AB + AC > 2h \\ge 2h \\sin \\alpha \\ge BC \\cdot \\cos \\alpha + 2h \\sin \\alpha$, because $\\cos \\alpha \\le 0$, that is, the inequality is strict.\n\nLet now $\\alpha < 90^\\circ$, denote by $H_1, H_2, H_3$ the feet of altitudes from vertices $A, B, C$ respectively, $K_2$ be... | Ukraine | The Problems of Ukrainian Authors | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | Equilateral triangles | |
0htu | Problem:
An apple is in the shape of a ball of radius $31~\mathrm{mm}$. A worm gets into the apple and digs a tunnel of total length $61~\mathrm{mm}$, and then leaves the apple. (The tunnel need not be a straight line.) Prove that one can cut the apple with a straight slice through the center so that one of the two ha... | [
"Solution:\n\nLet $A$ and $B$ be the endpoints of the tunnel, and let $C$ be the point diametrically opposite to $A$. We claim that the plane $\\pi$ which perpendicularly bisects $BC$ gives the desired cut. Indeed, this plane contains the center of the apple since it is equidistant from $B$ and $C$; now assume for ... | United States | Berkeley Math Circle Take-Home Contest #1 | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
08pa | Problem:
Let $ABC$ be an acute angled triangle, let $O$ be its circumcentre, and let $D, E, F$ be points on the sides $BC, AC, AB$, respectively. The circle $(c_1)$ of radius $FA$, centred at $F$, crosses the segment $(OA)$ at $A'$, and the circumcircle $(c)$ of the triangle $ABC$ again at $K$. Similarly, the circle $(... | [
"Solution:\nWe will prove that the quadrilateral $BKFA'$ is cyclic and its circumcircle passes through the center $O$ of the circle $(c)$.\n\nThe triangle $AFK$ is isosceles, so $m(\\widehat{KFB}) = 2 m(\\widehat{KAB}) = m(\\widehat{KOB})$. It follows that the quadrilateral $BKFO$ is cyclic.\n\nThe triangles $OFK$ ... | JBMO | Junior Balkan Mathematics Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
04wj | Given an integer $n \ge 16$, consider the set
$$
G = \{(x, y) : x, y \in \{1, 2, \dots, n\}\}
$$
consisting of $n^2$ points in the plane. Let $A$ be any subset of $G$ containing at least $4n\sqrt{n}$ points. Prove that there are at least $n^2$ convex quadrangles with all their vertices in $A$ such that their diagonals ... | [] | Czech-Polish-Slovak Mathematical Match | 9-th Czech-Slovak-Polish Match | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
0jv1 | Problem:
Define the sequence $a_{1}, a_{2} \ldots$ as follows: $a_{1}=1$ and for every $n \geq 2$,
$$
a_{n}= \begin{cases}n-2 & \text{ if } a_{n-1}=0 \\ a_{n-1}-1 & \text{ if } a_{n-1} \neq 0\end{cases}
$$
A non-negative integer $d$ is said to be jet-lagged if there are non-negative integers $r, s$ and a positive inte... | [
"Solution:\n\nLet $N=n+r$, and $M=n$. Then $r=N-M$, and $s=a_{N}-a_{M}$, and $d=r+s=\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right)$. So we are trying to find the number of possible values of $\\left(a_{N}+N\\right)-\\left(a_{M}+M\\right)$, subject to $N \\geq M$ and $a_{N} \\geq a_{M}$.\nDivide the $a_{i}$ into the ... | United States | HMMT February 2016 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 51 | |
01bz | Prove that for any positive integer $d$ it is possible to find two distinct positive integers $n_1$ and $n_2$ such that
* $n_2$ can be obtained from $n_1$ by permuting its digits,
* both $n_1$ and $n_2$ are divisible by $d$,
* none of them starts with "0".
Denote by $|x|$ the number of digits in $x$. Prove that there e... | [
"One can take $n_1 = \\overline{dd0}$ and $n_2 = \\overline{d0d}$. They consist of the same digits, they are different (the middle digit is nonzero for $n_1$ and zero for $n_2$) and both of them are divisible by $d$.\n\nFor $d=5$ one can easily check, that there are no two-digit numbers $n_1$ and $n_2$ satisfying t... | Baltic Way | Baltic Way | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
02xz | Problem:
Se $n$ é um número inteiro positivo, qual o menor valor que a soma dos algarismos da representação decimal de $3 n^{2}+n+1$ pode assumir? | [
"Solution:\nSe $n=8$, temos que $3 n^{2}+n+1=201$ e soma de seus dígitos é $3$. Verificaremos agora que a soma dos dígitos de $3 n^{2}+n+1$ não pode ser $1$ ou $2$ e concluiremos que o menor valor possível é $3$.\n\nComo $n(n+1)$ é o produto de dois números consecutivos, ele é par e assim $3 n^{2}+n+1=2 n^{2}+n(n+1... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
03yw | Prove for any integer $n \ge 4$, there exists a polynomial of degree $n$,
$$
f(x) = x^n + a_{n-1}x^{n-1} + \cdots + a_1x + a_0
$$
with the following properties.
(1) $a_0, a_1, \dots, a_{n-1}$ are all positive integers;
(2) For any positive integer $m$ and arbitrary $k$ ($k \ge 2$) positive integers $r_1, r_2, \dots, r_... | [
"Let\n$$\nf(x) = (x+1)(x+2)\\cdots(x+n)+2. \\qquad \\textcircled{1}\n$$\nObviously, $f(x)$ is a monic polynomial of degree $n$ with positive integer coefficients. We are going to prove that $f(x)$ has property (2).\nFor any integer $t$, since $n \\ge 4$, we know that there exists definitely a multiple of 4 in any $... | China | China Mathematical Competition (Complementary Test) | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | f(x) = (x+1)(x+2)⋯(x+n) + 2 | |
02a0 | Problem:
Para a festa de aniversário da escola, Ana, Pedro, Miriam e Fábio levaram juntos 90 docinhos. A professora deles observou que:
- se Ana tivesse levado 2 docinhos a mais;
- se Pedro tivesse levado 2 docinhos a menos;
- se Miriam tivesse levado o dobro;
- se Fábio tivesse levado a metade;
os 4 amigos teriam lev... | [
"Solution:\n\nRepresentando o número de docinhos que cada um dos 4 amigos levou pela inicial de seu nome temos:\n$$\n\\left\\{\\begin{array}{l}\nA+P+M+F=90 \\\\\nA+2=P-2=2M=\\frac{F}{2}\n\\end{array}\\right.\n$$\nSegue da segunda equação que:\n$$\nP=A+4 \\quad ; \\quad M=\\frac{A+2}{2} \\quad ; \\quad F=2(A+2)\n$$\... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | Ana 18, Pedro 22, Miriam 10, Fábio 40 | |
08ry | Let $\Gamma$ be the circumcircle of triangle $ABC$. Let $\Gamma_A$ be the circle which is tangent to $AB$, $AC$ and tangent to $\Gamma$ internally, $\Gamma_B$ be the circle which is tangent to $AB$, $BC$ and tangent to $\Gamma$ internally, and $\Gamma_C$ be the circle which is tangent to $AC$, $BC$ and tangent to $\Gam... | [
"Let $M_A, M_B, M_C$ be the midpoint of the arcs $BC$, $CA$, $AB$, not containing $A$, $B$, $C$ respectively, and $l_A, l_B, l_C$ be the tangent of $\\Gamma$ at $M_A, M_B, M_C$. Then, note that $l_A \\parallel BC$, $l_B \\parallel CA$, $l_C \\parallel AB$.\n\nNow, let $D$ be the common point of lines $l_B$ and $l_C... | Japan | Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
03n6 | Points $P$ and $Q$ lie inside parallelogram $ABCD$ and are such that triangles $ABP$ and $BCQ$ are equilateral. Prove that the line through $P$ perpendicular to $DP$ and the line through $Q$ perpendicular to $DQ$ meet on the altitude from $B$ in triangle $ABC$. | [
"Let $\\angle ABC = m$ and let $O$ be the circumcenter of triangle $DPQ$. Since $P$ and $Q$ are in the interior of $ABCD$, it follows that $m = \\angle ABC > 60^\\circ$ and $\\angle DAB = 180^\\circ - m > 60^\\circ$ which together imply that $60^\\circ < m < 120^\\circ$. Now note that $\\angle DAP = \\angle DAB - 6... | Canada | CMO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0dpb | Let $N = 2010! + 1$. Prove that
a) $N$ is not divisible by $4021$;
b) $N$ is not divisible by $2027$, $2029$, $2039$;
c) $N$ has a prime divisor greater than $2050$. | [] | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
05wj | Problem:
Soient $ABC$ un triangle et $H_C$ le pied de la hauteur issue de $C$. Soit $P$ un point sur le segment $[CH_C]$ (autre que $C$), et soient $E, F, G, H$ les milieux respectifs de $[AP]$, $[BP]$, $[BC]$, $[AC]$. Montrer que $EFGH$ est un rectangle. | [
"Solution:\n\n\n\nPour démontrer que $EFGH$ est un rectangle, nous allons montrer que c'est un parallélogramme qui possède un angle droit.\n\nConsidérons le triangle $ACP$. Les points $E$ et $H$ sont les milieux respectifs des côtés $[AP]$ et $[AC]$. Par le théorème de la droite des milieux... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
02os | Do there exist $2011$ positive integers $a_1 < a_2 < \dots < a_{2011}$ such that $\gcd(a_i, a_j) = a_j - a_i$ for any $i, j$ such that $1 \le i < j \le 2011$? | [
"The answer is yes and you can construct an example in several ways. The main observation is that $\\gcd(a_i, a_j) = a_j - a_i \\iff a_j - a_i \\mid a_i$. In fact, if $\\gcd(a_i, a_j) = a_j - a_i$ then $a_j - a_i \\mid a_i$ and, conversely, if $a_j - a_i \\mid a_i$ then $a_j - a_i \\mid a_i + (a_j - a_i) \\iff a_j ... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes |
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