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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0054 | En un tablero cuadriculado de tamaño $19 \times 19$, una ficha llamada *dragón* da saltos de la siguiente manera: se desplaza $4$ casillas en una dirección paralela a uno de los lados del tablero y $1$ casilla en dirección perpendicular a la anterior.
Desde $D$, el dragón puede saltar a una ... | [] | Argentina | XXII Olimpiada Iberoamericana de Matemática | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof only | null | |
09bk | $x, y \in \mathbb{Z}$ ба $2010x^2 + y^2 + 1$ болон $x^2 + 2010y^2 - 1$ тоонууд бүтэн квадрат бол $(x, y)$ хосыг “квадратлаг хос” гэе.
a. $(2009^2, 2010^2)$ квадратлаг хос мөн үү?
b. Бүх квадратлаг хосыг ол. | [
"$$\n\\begin{cases}\nx^2 + 2010y^2 - 1 = m^2 \\\\\n2010x^2 + y^2 + 1 = n^2\n\\end{cases}\n$$\n\nЯмар ч тооны квадратыг 4-өөр жишихэд 0, 1 гэсэн хоёр үлдэгдэл л өгөх боломжтой. Иймд\n\na. $\\begin{cases} x^2 = 1(4) \\\\ y^2 = 1(4) \\end{cases}$ бол $x^2 + 2010y^2 - 1 = 1 + 2 \\cdot 1 - 1 = 2(4) \\neq m^2(4)$ болж бү... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Mongolian | proof and answer | a: No. b: None. | |
01yq | Three circles $\omega_1$, $\omega_2$ and $\omega_3$ with non-colinear centres $O_1$, $O_2$ and $O_3$ are drawn such that $\omega_1$ externally touches $\omega_2$ and $\omega_3$ at the points $P$ and $Q$ respectively. An arbitrary point $C$ is chosen on $\omega_1$. The line $CP$ intersects $\omega_2$ for the second time... | [
"Let's first consider the case when the radii of $\\omega_2$ and $\\omega_3$ are different (without loss of generality assume that the radius of $\\omega_3$ is greater than the radius of $\\omega_2$). Consider three homotheties: $l_1$ centered at $Q$ and mapping $\\omega_3$ to $\\omega_1$; $l_2$ centered $P$ and ma... | Belarus | Belarus2022 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circl... | English | proof only | null | |
0kz3 | Suppose $z$ is a complex number with positive imaginary part, with real part greater than $1$, and with $|z| = 2$. In the complex plane, the four values $0$, $z$, $z^2$, and $z^3$ are the vertices of a quadrilateral with area $15$. What is the imaginary part of $z$?
(A) $\frac{3}{4}$ (B) $1$ (C) $\frac{4}{3}$ (D) $\fra... | [
"Let $\\theta$ be the argument of $z$. Because $|z| = 2$ and the real part of $z$ is greater than $1$, it follows that $\\theta$ is less than $60^\\circ$. This ensures that the imaginary parts of $z^2$ and $z^3$ are positive and all the vertices of the quadrilateral other than $0$ lie in the upper half-plane. Thus ... | United States | 2024 AMC 12 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | D | |
00v9 | Prove that for every integer $n$, the number $n^4 - 12n^2 + 144$ is not a perfect cube of an integer. | [
"Suppose otherwise and let $m \\in \\mathbb{Z}$ be such that $n^4 - 12n^2 + 144 = m^3$. Firstly, $m$ is clearly positive and we can assume that $n$ is a positive integer (since $n = 0$ clearly doesn't work). Note that the polynomial $x^4 - 12x^2 + 144$ may be factored as\n$$\nx^4 - 12x^2 + 144 = (x^2 + 12)^2 - (6x)... | Balkan Mathematical Olympiad | 41st Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Diophantine Equations > Techniques: m... | English | proof only | null | |
0krg | Problem:
For a real number $x$, let $[x]$ be $x$ rounded to the nearest integer and $\langle x\rangle$ be $x$ rounded to the nearest tenth. Real numbers $a$ and $b$ satisfy $\langle a\rangle+[b]=98.6$ and $[a]+\langle b\rangle=99.3$. Compute the minimum possible value of $[10(a+b)]$.
(Here, any number equally between... | [
"Solution:\n\nWithout loss of generality, let $a$ and $b$ have the same integer part or integer parts that differ by at most 1, as we can always repeatedly subtract 1 from the larger number and add 1 to the smaller to get another solution.\n\nNext, we note that the decimal part of $a$ must round to .6 and the decim... | United States | HMMT February | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 988 | |
02iv | Problem:
Luíza, Maria, Antônio e Júlio são irmãos. Dois deles têm a mesma altura. Sabe-se que:
- Luíza é maior que Antônio
- Maria é menor que Luíza
- Antônio é maior do que Júlio
- Júlio é menor do que Maria.
Quais deles têm a mesma altura?
A) Maria e Júlio
B) Júlio e Luíza
C) Antônio e Luíza
D) Antônio e Júlio
E) An... | [
"Solution:\n\nUsaremos a notação $a < b$ que significa que $a$ é menor do que $b$, ou equivalentemente, $b$ é maior do que $a$. Assim, $a < b < c$ significa que $a$ é menor do que $b$ e $b$ é menor do que $c$.\n\nPara simplificar, vamos denotar a altura de cada um dos irmãos pela letra inicial de seu nome.\n\nDo en... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | MCQ | E | |
0dgl | Find all integer numbers $m$ and $n$ such that
$$
(5 + 3\sqrt{2})^m = (3 + 5\sqrt{2})^n.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (m, n) = (0, 0) | |
0lb1 | Given a convex pentagon $ABCDE$, of which the length of each edge and of the diagonals $AC$, $AD$ does not exceed $\sqrt{3}$. Choose 2001 arbitrary distinct points in the interior of that pentagon. Show that there exists a unit disk with center lying on the edges of the pentagon, which contains at least 403 of the chos... | [
"To verify the claim, we will show that it is possible to cover the pentagon $ABCDE$ by 5 unit discs with center lying on the edges of the pentagon.\n\nWe have following remark:\n**Remark:** It is possible to cover a triangle $XYZ$ with edges of length not exceeding $\\sqrt{3}$ by 3 unit discs with centers at the v... | Vietnam | Vijetnam 2011 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0hj5 | Problem:
If $x$ and $y$ are integers such that
$$
x^{2} y^{2} = x^{2} + y^{2},
$$
prove that $x = y = 0$. | [
"Solution:\nWe move all the terms to the left side of the equation and add $1$.\n$$\n\\begin{array}{r}\nx^{2} y^{2} - x^{2} - y^{2} + 1 = 1 \\\\\n\\left(x^{2} - 1\\right)\\left(y^{2} - 1\\right) = 1\n\\end{array}\n$$\nSince the factors are integers, they must be both $1$ or both $-1$. If both are $1$, we get $x^{2}... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
084s | Problem:
Rosa e Savino fanno il seguente gioco con le carte napoletane (40 carte numerate da 1 a 10 di 4 semi diversi): inizialmente si dividono le 40 carte (20 per ciascuno), poi a turno appoggiano sul tavolo una carta. Quando alcune delle carte presenti sul tavolo hanno dei valori la cui somma fa esattamente 15, que... | [
"Solution:\n\nRosa ha un 8.\n\nLa somma dei valori di tutte le carte del gioco è $\\frac{10 \\cdot 11}{2} \\cdot 4 = 220$.\nLa somma di quelle eliminate è un multiplo di 15 (vengono tolte a gruppi con somma pari a 15).\nIndicando con $x$ il valore della carta di Rosa si ha:\n$$\n220 = 15k + 5 + 3 + 9 + x\n$$\nquind... | Italy | XXII OLIMPIADE ITALIANA DI MATEMATICA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof and answer | 8 | |
0ahv | Let $a$, $b$ and $c$ be positive real numbers such that $abc=1$. Prove that the following inequality holds
$$
\frac{1}{2}(\sqrt{a} + \sqrt{b} + \sqrt{c}) + \frac{1}{1+a} + \frac{1}{1+b} + \frac{1}{1+c} \ge 3
$$
When does equality hold? | [
"Since $(1-\\sqrt{bc})^2 \\ge 0$ it follows that $1+bc \\ge 2\\sqrt{bc}$, i.e. $\\frac{1}{2\\sqrt{bc}} \\ge \\frac{1}{1+bc}$. We get that\n$$\n\\frac{\\sqrt{a}}{2} + \\frac{1}{1+a} = \\frac{1}{2\\sqrt{bc}} + \\frac{1}{1+a} \\ge \\frac{1}{1+bc} + \\frac{1}{1+a} = \\frac{1}{1+\\frac{1}{a}} + \\frac{1}{1+a} = 1\\ldots... | North Macedonia | Junior Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | Equality holds if and only if a = b = c = 1. | |
0fnn | Para cada entero positivo $n$, se define $s(n)$ como la suma de los dígitos de $n$. Determine el menor entero positivo $k$ tal que
$$
s(k) = s(2k) = s(3k) = \dots = s(2013k) = s(2014k)
$$ | [] | Spain | XXIX Olimpiada Iberoamericana de Matemáticas | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Spanish | proof and answer | 9999 | |
0598 | There are some distinct positive integers written on a blackboard. If we erase the smallest number written on the blackboard, then the ratio of the sum and the product of the remaining numbers will be 4 times greater than the ratio of the sum and the product of the numbers initially on the blackboard. Find all possibil... | [
"*Answer:* $\\{5, 20\\}, \\{5, 6, 14\\}, \\{5, 7, 13\\}, \\{5, 8, 12\\}, \\{5, 9, 11\\}, \\{6, 12\\}$.\n\nClearly there must be at least 2 numbers. Let $n$ be the smallest number, $s$ the sum and $k$ the product of the remaining numbers. We get the equation $4 \\cdot \\frac{n+s}{nk} = \\frac{s}{k}$. Multiplying by ... | Estonia | Estonian Math Competitions | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | {5, 20}; {5, 6, 14}; {5, 7, 13}; {5, 8, 12}; {5, 9, 11}; {6, 12} | |
0d9p | Let $\{x_{n}\}$ be a sequence defined by $x_{1}=2$ and
$$
x_{n+1}=x_{n}^{2}-x_{n}+1
$$
for $n \geq 1$. Prove that
$$
1-\frac{1}{2^{2^{n-1}}}<\frac{1}{x_{1}}+\frac{1}{x_{2}}+\ldots+\frac{1}{x_{n}}<1-\frac{1}{2^{2^{n}}}
$$
for all $n$. | [
"The sequence is increasing since $x_{n+1}-x_{n}=(x_{n}-1)^{2} \\geq 0$. We also have $x_{n+1}-1=x_{n}(x_{n}-1)$ which gives us\n$$\n\\frac{1}{x_{n+1}-1}=\\frac{1}{x_{n}(x_{n}-1)}=\\frac{1}{x_{n}-1}-\\frac{1}{x_{n}}\n$$\nThis identity is valid because given $x_{1}=2$ we have $x_{n} \\geq 2$ and therefore $x_{n}-1 \... | Saudi Arabia | Team selection tests for GMO 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
07vy | The points $A$, $B$, $P$, $Q$ are collinear, and $ABCD$ is a parallelogram. The lines $PD$ and $BC$ meet at $E$. The lines $QC$ and $AD$ meet at $F$. The lines $PF$ and $BC$ meet at $H$. The lines $QE$ and $AD$ meet at $G$.
Prove that $GH$ is parallel to $AB$. | [
"The Intercept Theorem applied to $AG \\parallel BH$ and lines intersecting at $P$ gives\n$$\n\\frac{|PA|}{|PB|} = \\frac{|AF|}{|BH|} \\quad \\text{and} \\quad \\frac{|PA|}{|PB|} = \\frac{|AD|}{|BE|}, \\text{ hence } |AF| \\cdot |BE| = |BH| \\cdot |AD|.\n$$\n\n\n\nConsidering lines that int... | Ireland | IRL_ABooklet_2023 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0e8l | Problem:
Izračunaj koordinate točke, ki leži na premici skozi točki $A(-5,-2)$ in $B(-3,-1)$ ter je enako oddaljena od točk $C(-1,0)$ in $D(4,-1)$. | [] | Slovenia | Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | final answer only | (17/9, 13/9) | |
08xl | Find the value of
$$
\sum \frac{1}{d + \sqrt{10!}}
$$
where the sum is taken over all the positive factors $d$ of the number $10!$. | [
"$$\n\\boxed{\\frac{3}{16\\sqrt{7}}}\n$$\nFrom $10! = 2^8 \\cdot 3^4 \\cdot 5^2 \\cdot 7$ it follows that there are $(8+1) \\cdot (4+1) \\cdot (2+1) \\cdot (1+1) = 270$ positive factors of $10!$. Suppose we enumerate them as $d_1, d_2, \\dots, d_{270}$ in increasing order of magnitude, starting with the smallest on... | Japan | Japan Mathematical Olympiad Initial Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | English | proof and answer | 3/(16√7) | |
0eda | Prove that for all positive integers $n \ge 2$ we have
$$
\frac{1}{2} + \sqrt{\frac{1}{2}} + \sqrt[3]{\frac{2}{3}} + \dots + \sqrt[n]{\frac{n-1}{n}} < \frac{n^2}{n+1}.
$$ | [
"We use proof by induction.\nFor $n = 2$ we have\n$$\n\\frac{1}{2} + \\sqrt{\\frac{1}{2}} < \\frac{4}{3} \\Leftrightarrow \\sqrt{\\frac{1}{2}} < \\frac{5}{6} \\Leftrightarrow \\frac{1}{2} < \\frac{25}{36},\n$$\nwhich is true. For $n = 2$ the inequality holds.\n\nNow assume that the inequality holds for $n$ and let ... | Slovenia | Slovenija 2016 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0dto | Let $p$ be a prime number. Determine the largest possible $n$ such that the following holds. It is possible to fill an $n \times n$ table with integers $a_{ik}$ in the $i$-th row and $k$-th column, for $1 \le i, k \le n$, such that for any quadruple $i, j, k, l$ with $1 \le i < j \le n$ and $1 \le k < l \le n$, the num... | [
"The answer is $n = p + 1$. We first show that $n \\le p + 1$. Since we are interested only in divisibility by $p$, we assume that $0 \\le a_{ik} \\le p - 1$. Note that each row and each column can have at most one zero. Also notice that we can scale each row or column by scalar not divisible by $p$ without affecti... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | n = p + 1 | |
0dkk | Find all pairs $(b, c)$ of positive integers, such that the sequence defined by $a_1 = b$, $a_2 = c$ and
$$
a_{n+2} = |4a_{n+1} - 3a_n|, \forall n \ge 1
$$
has only finite number of composite terms. | [
"Suppose firstly that there exists a $k \\ge 2$ such that $a_{k+1} > a_k$. Then by induction we have $a_{n+1} > a_n$ for all $n \\ge k$ (when taking out the modulus the expression does not change sign) and so $a_{n+2} = 4a_{n+1} - 3a_n$ for all $n \\ge k$. Solving the characteristic equation $t^2 - 4t + 3 = 0$ impl... | Saudi Arabia | Saudi Booklet | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | {(p, p), (3p, p), (11, 9), (13, 9)} where p is any prime | |
0b9e | Consider two equilateral triangles $ABC$ and $MNP$ with $AB \parallel MN$, $BC \parallel NP$ and $CA \parallel PM$, intersecting over a convex hexagon. The distances between the pairs of parallel sides do not exceed $1$. Show that at least one of the triangles has the side length less than or equal to $\sqrt{3}$. | [
"Let $P$ be an interior point to the hexagon, therefore also interior to the triangles. Denote by $a$, respectively $b$, the lengths of the sides of the two triangles. The sum of the distances from $P$ to the sides of an equilateral triangle is equal to the altitude of the triangle, hence the sum of the distances f... | Romania | NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0k6x | Problem:
Let $T$ be a triangle with area $1$. We let $T_{1}$ be the medial triangle of $T$, i.e. the triangle whose vertices are the midpoints of sides of $T$. We then let $T_{2}$ be the medial triangle of $T_{1}$, $T_{3}$ the medial triangle of $T_{2}$, and so on. What is the sum of the areas of $T_{1}, T_{2}, T_{3}, ... | [
"Solution:\nIn general, the medial triangle has side length half the original triangle, hence $\\frac{1}{4}$ the area. Thus, $T_{1}$ has area $\\left(\\frac{1}{4}\\right)$; then $T_{2}$ has area $\\left(\\frac{1}{4}\\right)^{2}$, and so on. Thus the answer is\n$$\n\\frac{1}{4}+\\left(\\frac{1}{4}\\right)^{2}+\\left... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | 1/3 | |
0ak8 | An integer $a \ge 1$ is called Aegean, if none of the numbers $a^{n+2} + 3a^n + 1$ with $n \ge 1$ is prime. Prove that there are at least 500 Aegean integers in the set $\{1, 2, ..., 2018\}$. | [] | North Macedonia | Mediterranean Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
020n | Problem:
Let $A B C$ be a triangle with orthocentre $H$, and let $D, E$, and $F$ denote the respective midpoints of line segments $A B, A C$, and $A H$. The reflections of $B$ and $C$ in $F$ are $P$ and $Q$, respectively.
a. Show that lines $P E$ and $Q D$ intersect on the circumcircle of triangle $A B C$.
b. Prove ... | [
"Solution:\n\na.\n\nSolution 1. Since $F$ is the midpoint of $[A H]$ and $[B P]$, $B H P A$ is a parallelogram. Similarly, $C A Q H$ is a parallelogram, too. Let $O$ denote the circumcentre of triangle $A B C$ and let $R$ be the reflection of $A$ in $O$, so that $R$ lies on the circumcircle of $A B C$. Since $[A R]... | Benelux Mathematical Olympiad | 10th Benelux Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformation... | null | proof only | null | |
0fhl | Problem:
Dibujado el triángulo de vértices $A, B, C$, se pide determinar gráficamente el punto $P$ tal que
$$
\widehat{PAB} = \widehat{PBC} = \widehat{PCA}
$$
Expresar una función trigonométrica de este ángulo $\widehat{PAB}$ en función de las funciones trigonométricas de los ángulos $A, B$ y $C$. | [
"Solution:\n\n\n\nPongamos $\\varphi = \\widehat{PAB} = \\widehat{PBC} = \\widehat{PCA}$.\nLa determinación de $P$ se hace en los siguientes pasos:\n\n1) Se traza la circunferencia pasando por $A$ y tangente en $B$ al lado $BC$.\n\n2) La paralela por $B$ al lado $AC$ corta en $M$ a la circu... | Spain | OME 28 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | cot φ = cot A + cot B + cot C | |
0820 | Problem:
In una mappa di una certa regione ci sono dieci città ai vertici di un decagono regolare, e i dieci lati del decagono rappresentano altrettante strade. Nella regione ci sono dei lavori, per cui ogni strada è aperta con una probabilità $\frac{1}{2}$ indipendentemente dalle altre.
Qual è la probabilità che da o... | [
"Solution:\n\nLa risposta è (C). Se tutte le strade sono aperte, evidentemente è possibile spostarsi liberamente da una città all'altra. Se una sola strada è chiusa, le nove strade aperte costituiscono una linea spezzata continua attraverso cui è ancora possibile spostarsi da una città all'altra. Se ci sono almeno ... | Italy | Progetto Olimpiadi di Matematica | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | MCQ | C | |
0jvf | Problem:
Find all twice differentiable functions $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfying
$$
f(x)^2 - f(y)^2 = f(x+y) f(x-y)
$$
for all real numbers $x$ and $y$. | [
"Solution:\nThe answer is $f(x) = kx$, $f(x) = a \\sin(c x)$, $f(x) = a \\sinh(c x)$, where $a, c \\in \\mathbb{R}$. The given functional equation is\n$$\nf(x)^2 - f(y)^2 = f(x+y) f(x-y).\n$$\nObserve that $x = y = 0$ gives $f(0) = 0$.\nSince $f$ is smooth, we may differentiate with respect to $x$ and obtain\n$$\n2... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | All twice differentiable solutions are f(x) = k x, f(x) = a sin(c x), or f(x) = a sinh(c x), for real parameters a, c, k. | |
0g9g | 設三角形 $ABC$ 的內切圓為 $\omega$, $\omega$ 切 $BC$ 邊於 $D$ 點。設 $AD$ 與 $\omega$ 的另一個交點為 $L$。令三角形 $ABC$ 在角 $A$ 內的旁心為 $K$。設 $M$ 為 $BC$ 的中點, 而 $N$ 為 $KM$ 的中點。證明: $B$, $C$, $N$, $L$ 共圓。 | [
"\n\n設 $A$ 對圓 $\\omega$ 的極線與 $BC$ 交於 $X$。因 $AD$ 是 $X$ 對圓 $\\omega$ 的極線, 所以 $XL$ 與 $\\omega$ 相切, 且 $(X, D; B, C) = -1$。\n\n由於 $XL = XD$ 以及 $N'M = N'D$, 所以 $\\angle XLD = \\angle LDX = \\angle N'DM = \\angle DMN'$, 得 $L$, $X$, $M$, $N'$ 共圓。\n\n因此有 $DL \\cdot DN' = DX \\cdot DM = DB \\cdot DN$... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a26 | Let $\triangle ABC$ be a triangle. Let $P$ be a point on the segment $BC$, such that the circle with diameter $BP$ passes through the incentre of $\triangle ABC$. Prove that
$$
\frac{|BP|}{|PC|} = \frac{c}{s-c},
$$
where $c$ is the length of the segment $AB$, and $s$ is half the perimeter of $\triangle ABC$. | [
"Let $\\ell$ be the second tangent through $P$ to the circle aside from $BC$. Note that\n$$\n\\begin{aligned}\n\\angle ABP &= 2\\angle IBP = 2(90^\\circ - \\angle BPI) = 180^\\circ - 2\\angle BPI \\\\\n&= 180^\\circ - \\angle (BP, \\ell) = \\angle (CP, \\ell),\n\\end{aligned}\n$$\nwhere we use the fact that $BI$ is... | Netherlands | IMO Team Selection Test 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0713 | Problem:
Let $X$ be a set with $n$ elements. Given $k > 2$ subsets of $X$, each with at least $r$ elements, show that we can always find two of them whose intersection has at least $r - \dfrac{n k}{4k - 4}$ elements. | [] | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0j39 | Problem:
Point $P$ lies inside a convex pentagon $A F Q D C$ such that $F P D Q$ is a parallelogram. Given that $\angle F A Q=\angle P A C=10^{\circ}$, and $\angle P F A=\angle P D C=15^{\circ}$. What is $\angle A Q C$? | [
"Solution:\n\nAnswer: $\\frac{\\pi}{12}$\n\nLet $C'$ be the point such that there is a spiral similarity between $\\triangle A F P$ and $\\triangle A Q C'$. In other words, one triangle can be formed from the other by dilating and rotating about one of the triangle's vertices (in this case, $A$). We will show that ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | π/12 | |
0joi | Problem:
Define a sequence $a_{i, j}$ of integers such that $a_{1, n} = n^{n}$ for $n \geq 1$ and $a_{i, j} = a_{i-1, j} + a_{i-1, j+1}$ for all $i, j \geq 1$. Find the last (decimal) digit of $a_{128,1}$. | [
"Solution:\n\nBy applying the recursion multiple times, we find that $a_{1,1} = 1$, $a_{2, n} = n^{n} + (n+1)^{n+1}$, and $a_{3, n} = n^{n} + 2(n+1)^{n+1} + (n+2)^{n+2}$. At this point, we can conjecture and prove by induction that\n$$\na_{m, n} = \\sum_{k=0}^{m-1} \\binom{m-1}{k} (n+k)^{n+k} = \\sum_{k \\geq 0} \\... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | 4 | |
06z8 | Problem:
The incircle of the triangle $ABC$ touches $AC$ at $M$ and $BC$ at $N$ and has center $O$. $AO$ meets $MN$ at $P$ and $BO$ meets $MN$ at $Q$. Show that $MP \cdot OA = BC \cdot OQ$. | [
"Solution:\n\n\n\nThe key to getting started is to notice that angle $AQB = 90^\\circ$.\nAngle $BAQ = 90^\\circ - B/2$, so angle $OAQ = 90^\\circ - B/2 - A/2 = C/2$. So $OQ = AO \\sin (C/2)$. Thus we have to show that $MP = BC \\sin (C/2)$.\n\nLet the incircle touch $AB$ at $L$ and let $Y$ ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
03rm | Draw a tangent line of parabola $y = x^2$ at the point $A(1, 1)$. Suppose the line intersects the $x$-axis and $y$-axis at $D$ and $B$ respectively. Let point $C$ be on the parabola and point $E$ on $AC$ such that $\frac{AE}{EC} = \lambda_1$. Let point $F$ be on $BC$ such that $\frac{BF}{FC} = \lambda_2$ and $\lambda_1... | [
"The slope of the tangent line passing through $A$ is $y' = 2x|_{x=1} = 2$. So the equation of the tangent line $AB$ is $y = 2x - 1$. Hence the coordinates of $B$ and $D$ are $B(0, -1)$, $D(\\frac{1}{2}, 0)$. Thus $D$ is the midpoint of line segment $AB$.\n\nConsider $P(x, y)$, $C(x_0, x_0^2)$, $E(x_1, y_1)$, $F(x_... | China | China Mathematical Competition (Jiangxi) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | y = (1/3)(3x - 1)^2, with x ≠ 2/3 | |
08s2 | Find one of the polynomials $f(x, y, z)$ whose degree is $3$, with real coefficients, that satisfy the following conditions.
• $f(x, y, z) + x$ is divisible by $y + z$
• $f(x, y, z) + y$ is divisible by $z + x$
• $f(x, y, z) + z$ is divisible by $x + y$
A polynomial $P(x, y, z)$ is divisible by a polynomial $Q(x, y, z)... | [
"$f(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z$, ($k \\neq 0$) satisfies the conditions (in fact, only this form is a solution).\n\nPut $g(x, y, z) = f(x, y, z) + x + y + z$. Then the conditions are equivalent to the condition that $g(x, y, z)$ is divisible by $x + y$, $y + z$, $z + x$. And the condition that th... | Japan | Japan 2007 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | f(x, y, z) = k(x + y)(y + z)(z + x) - x - y - z, where k is any nonzero real constant | |
0kts | Problem:
Let $ABCDEF$ be a regular hexagon, and let $P$ be a point inside quadrilateral $ABCD$. If the area of triangle $PBC$ is $20$, and the area of triangle $PAD$ is $23$, compute the area of hexagon $ABCDEF$. | [
"Solution:\n\nIf $s$ is the side length of the hexagon, $h_1$ is the length of the height from $P$ to $BC$, and $h_2$ is the length of the height from $P$ to $AD$, we have $[PBC] = \\frac{1}{2} s \\cdot h_1$ and $[PAD] = \\frac{1}{2}(2s) \\cdot h_2$. We also have $h_1 + h_2 = \\frac{\\sqrt{3}}{2} s$. Therefore,\n$$... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 189 | |
01pa | Find the smallest positive integer $n$ such that the number $2013n$ can be presented as the difference of two cubes of positive integer numbers. | [
"Answer: $n = 39$.\n(Solution of A. Semchankau, A. Zhuk.) Let\n$$\n2013n = a^3 - b^3. \\qquad (1)\n$$\nThen\n$$\n(1) \\Leftrightarrow 61 \\cdot 11 \\cdot 3n = 2013n = (a-b)^3 + 3ab(a-b) \\Rightarrow (a-b) \\vdash 3,\n$$\ni.e. $(a-b) = 3k, k \\in \\mathbb{N}$. So $61 \\cdot 11 \\cdot 3n = 3^3 k^3 + 3^2 abk$, whence ... | Belarus | BelarusMO 2013_s | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 39 | |
02m3 | Let $\triangle ABC$ be a triangle and $O$ its circumcenter. Lines $AB$ and $AC$ meet the circumcircle of $OBC$ again in $B_1 \neq B$ and $C_1 \neq C$, respectively, lines $BA$ and $BC$ meet the circumcircle of $OAC$ again in $A_2 \neq A$ and $C_2 \neq C$, respectively, and lines $CA$ and $CB$ meet the circumcircle of $... | [
"One can guess, by drawing a good diagram, that the common point is $O$ and that the lines $A_2A_3$, $B_1B_3$ and $C_1C_2$ are the perpendicular bisectors of the triangle $ABC$. So it is sufficient to prove that $A_2A_3$ is the perpendicular bisector of $BC$. For the sake of simplicity, let $\\angle A = \\angle BAC... | Brazil | XXXI Brazilian Math 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 > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
0f10 | Problem:
The triangle $ABC$ has area $1$. $D$, $E$, $F$ are the midpoints of the sides $BC$, $CA$, $AB$. $P$ lies in the segment $BF$, $Q$ lies in the segment $CD$, $R$ lies in the segment $AE$. What is the smallest possible area for the intersection of triangles $DEF$ and $PQR$? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | 1/8 | |
0auk | Problem:
Rationalize the denominator of $\frac{6}{\sqrt[3]{4}+\sqrt[3]{16}+\sqrt[3]{64}}$ and simplify. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 2 - \sqrt[3]{2} | |
0fsp | Problem:
Die reellen Zahlen $a_{1}, a_{2}, \ldots, a_{16}$ erfüllen die beiden Bedingungen
$$
\sum_{i=1}^{16} a_{i}=100 \quad \text{ und } \quad \sum_{i=1}^{16} a_{i}^{2}=1000
$$
Was ist der grösstmögliche Wert, den $a_{16}$ annehmen kann? | [
"Solution:\n\nSetze $S=a_{1}+\\ldots+a_{15}$ und $Q=a_{1}^{2}+\\ldots+a_{15}^{2}$. Nach AM-QM gilt $S^{2} \\leq 15 Q$. Die Nebenbedingungen lauten $100-a_{16}=S$ und $1000-a_{16}^{2}=Q$. Quadriert man die erste und subtrahiert 15 mal die zweiten, dann folgt\n$$\n16 a_{16}^{2}-200 a_{16}-5000=S^{2}-15 Q \\leq 0\n$$\... | Switzerland | IMO - Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 25 | |
0617 | Problem:
Wir betrachten - mit 1 beginnend - alle positiven Teiler einer natürlichen Zahl $n$ der Größe nach geordnet: $1 = d_{1} < d_{2} < d_{3} < \ldots < n$.
Man bestimme alle natürlichen Zahlen $n$ mit den Eigenschaften:
(1) $n = d_{13} + d_{14} + d_{15}$
und
$$
\left(d_{5} + 1\right)^{3} = d_{15} + 1
$$ | [] | Germany | Auswahlwettbewerb zur IMO 2000 | [
"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, inequalities"
] | null | proof and answer | 1998 | |
0j15 | Problem:
While Travis is having fun on cubes, Sherry is hopping in the same manner on an octahedron. An octahedron has six vertices and eight regular triangular faces. After five minutes, how likely is Sherry to be one edge away from where she started? | [
"Solution:\n\nAnswer: $\\frac{11}{16}$\n\nLet the starting vertex be the 'bottom' one. Then there is a 'top' vertex, and 4 'middle' ones. If $p(n)$ is the probability that Sherry is on a middle vertex after $n$ minutes, $p(0)=0$,\n$p(n+1) = (1 - p(n)) + p(n) \\cdot \\frac{1}{2}$.\n\nThis recurrence gives us the fol... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 11/16 | |
06jh | During a school year $44$ competitions were held. Exactly $7$ students won in each of the competitions. For any two competitions, there exists exactly $1$ student who won in both competitions. Is it true that there exists a student who won all of the competitions? | [
"Yes. Consider any competition $C_1$. Since every other competition shares a common winner with $C_1$, there must be a winner of $C_1$ who won at least $\\lfloor \\frac{43}{7} \\rfloor = 6$ more competitions by the pigeonhole principle. WLOG assume $A$ won the competitions $C_1, C_2, \\dots, C_8$.\n\nConsider any o... | Hong Kong | Year 2016 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
047g | We say a prime number $p$ is “good”, if there exists a bijection $f$ from the set $\{0, 1, \dots, p-1\}$ to itself satisfying the following condition: for any pair of elements $a, b \in \{0, 1, \dots, p-1\}$, if $p \mid a^2 - b$, then $|f(a) - f(b)| \le 2024$. If no such bijection $f$ exists, we say that the prime $p$ ... | [
"**Proof.** First, we show that there are infinitely many good primes. It is well-known that there are infinitely many primes $p \\equiv 3 \\pmod 4$. We prove that if $p \\equiv 3 \\pmod 4$, then $p$ is a good prime. Let $f(0) = 0$. Consider all quadratic residues of $p$, denoted as $r_1, r_2, \\dots, r_{(p-1)/2}$.... | China | 2024 CMO | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
05wt | Problem:
Soient $ABC$ un triangle, $O$ le centre de son cercle circonscrit. On suppose que $\widehat{CBA}=60^{\circ}$ et $\widehat{\mathrm{CBO}}=45^{\circ}$. Soit $D$ le point d'intersection des droites $(\mathrm{AC})$ et $(BO)$. Montrer que $\mathrm{AD}=\mathrm{DO}$. | [
"Solution:\n\n\n\nPour montrer que $AD=DO$ (c'est-à-dire que $ADO$ est isocèle en $D$), nous allons montrer que $\\widehat{DOA}=\\widehat{OAD}$. Introduisons $P$ le point d'intersection (autre que $B$) de $(BO)$ avec le cercle circonscrit de $ABC$.\n\nD'une part, $\\widehat{DOA}=\\widehat{P... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"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 | |
0hvg | Problem:
Six students taking a test sit in a row of seats with aisles only on the two sides of the row. If they finish the test at random times, what is the probability that some student will have to pass by another student to get to an aisle? | [
"Solution:\n\nThe probability $p$ that no student will have to pass by another student to get to an aisle is the probability that the first student to leave is one of the students on the end, the next student to leave is on one of the ends of the remaining students, etc.:\n\n$$\np = \\frac{2}{6} \\cdot \\frac{2}{5}... | United States | null | [
"Statistics > Probability > Counting Methods > Permutations"
] | null | proof and answer | 43/45 | |
0bmp | $$
\lim_{n \to \infty} \frac{1}{n} \sum_{k=0}^{n-1} k \int_{\frac{k}{n}}^{\frac{k+1}{n}} \sin(\pi x^2) dx.
$$ | [
"Let $f: [0, 1] \\to \\mathbb{R}$, $f(x) = \\sin(\\pi x^2)$, and let $F: [0, 1] \\to \\mathbb{R}$, $F(x) = \\int_0^x \\sin(\\pi t^2) dt$. Write\n$$\n\\begin{aligned}\n\\frac{1}{n} \\sum_{k=0}^{n-1} k \\int_{\\frac{k}{n}}^{\\frac{k+1}{n}} \\sin(\\pi x^2) dx &= \\frac{1}{n} \\sum_{k=0}^{n-1} \\left( kF\\left(\\frac{k... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof and answer | 1/π | |
0fg3 | Problem:
Diremos que una matriz cuadrada es de suma constante si la suma de los elementos de cada fila, de cada columna, y de cada diagonal, son valores iguales. Análogamente, una matriz cuadrada es de producto constante si son iguales los productos de los elementos de cada fila, de cada columna y de cada diagonal. De... | [
"Solution:\n\nLas condiciones para que la matriz\n$$\n\\left(\\begin{array}{lll}\na & d & g \\\\\nb & e & h \\\\\nc & f & i\n\\end{array}\\right)\n$$\nsea de suma constante son\n$a+b+c=a+d+g=a+e+i=d+e+f=b+e+h=c+f+i=g+h+i=c+e+g$.\nDespejando convenientemente se obtiene\n$$\n\\begin{aligned}\na & =-i+2e \\\\\nb & =2e... | Spain | OME 21 | [
"Algebra > Linear Algebra > Matrices"
] | null | proof and answer | All 3×3 real matrices that are simultaneously constant-sum and constant-product are exactly the following families:
1) \(\begin{pmatrix}0 & h & -h\\ -h & 0 & h\\ h & -h & 0\end{pmatrix}\), with \(h \in \mathbb{R}\).
2) \(\begin{pmatrix}h & -h & 0\\ -h & 0 & h\\ 0 & h & -h\end{pmatrix}\), with \(h \neq 0\).
3) \(\begin{... | |
01w2 | A circle $\omega$ of radius $1$ is given. A collection $T$ of triangles is called good, if the following conditions hold:
(i) each triangle from $T$ is inscribed in $\omega$;
(ii) no two triangles from $T$ have a common interior point.
Determine all positive real numbers $t$ such that, for each positive integer $n$ the... | [
"1. See IMO-2018 Shortlist, Problem G3."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geomet... | English | proof and answer | 0 < t < 4 | |
0k42 | Problem:
Let $\omega_{1}, \omega_{2}, \ldots, \omega_{100}$ be the roots of $\frac{x^{101}-1}{x-1}$ (in some order). Consider the set
$$
S=\left\{\omega_{1}^{1}, \omega_{2}^{2}, \omega_{3}^{3}, \ldots, \omega_{100}^{100}\right\}
$$
Let $M$ be the maximum possible number of unique values in $S$, and let $N$ be the minim... | [
"Solution:\nThroughout this solution, assume we're working modulo $101$.\n\nFirst, $N=1$. Let $\\omega$ be a primitive $101$st root of unity. We then let $\\omega_{n}=\\omega^{1 / n}$, which we can do because $101$ is prime, so $1 / n$ exists for all nonzero $n$ and $1 / n=1 / m \\Longrightarrow m=n$. Thus the set ... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 98 | |
037t | Problem:
Let $p$ be a prime number such that $p^2$ divides $2^{p-1}-1$. Prove that for any positive integer $n$ the integer $(p-1)\left(p!+2^{n}\right)$ has at least three distinct prime divisors. | [
"Solution:\nSince $p-1$ is a divisor of $p!$ the greatest common divisor of $p-1$ and $p!+2^{n}$ is a power of two. We shall show that both numbers $p-1$ and $p!+2^{n}$ have at least one odd divisor.\n\nSuppose that $p-1=2^{k}$, i.e. $p=2^{k}+1$. If $s \\geq 3$ is an odd divisor of $k$ then\n$p=2^{s t}+1=\\left(2^{... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0b55 | Problem:
Rezolvaţi în mulţimea numerelor prime ecuaţia
$$
x^{y} - y^{x} = x y^{2} - 19
$$ | [] | Romania | BMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (x,y) = (2,3) and (2,7) | |
00q3 | Let $ABC$ be an acute triangle with $AB < AC < BC$ inscribed in a circle $(c)$ and let $E$ be an arbitrary point on its altitude $CD$. The circle $(c_1)$ with diameter $EC$, intersects the circle $(c)$ at points $K$ (different than $C$), the line $AC$ at point $L$ and the line $BC$ at point $M$. Finally the line $KE$ i... | [
"From the orthogonal triangle $ACD$ we have: $\\angle ECL = 90^\\circ - \\hat{A}$. From the inscribed quadrilateral $ELCM$ we have: $\\angle EML = \\angle ECL = 90^\\circ - \\hat{A}$. Also $\\hat{EMC} = 90^\\circ$ (since $EC$ is a diameter of the circle $(c_1)$).\n\nHence we have:\n$$\n\\an... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0g07 | Problem:
Wir haben ein $8 \times 8$ Brett. Eine innere Kante ist eine Kante zwischen zwei $1 \times 1$ Feldern. Wir zerschneiden das Brett in $1 \times 2$ Dominosteine. Für eine innere Kante $k$ bezeichnet $N(k)$ die Anzahl Möglichkeiten, das Brett so zu zerschneiden, dass entlang der Kante $k$ geschnitten wird. Berec... | [
"Solution:\n\nZuerst berechnen wir, entlang wie vielen inneren Kanten bei einer Zerlegung geschnitten wird. Insgesamt gibt es je $7 \\cdot 8$ horizontale und vertikale innere Kanten. Bei einer Zerlegung wird das Brett entlang allen inneren Kanten, ausser den 32 innerhalb eines Dominosteines, geschnitten. Somit wird... | Switzerland | SMO - Finalrunde | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 0 | |
0ja5 | Problem:
Find the number of ordered triples of divisors $\left(d_{1}, d_{2}, d_{3}\right)$ of $360$ such that $d_{1} d_{2} d_{3}$ is also a divisor of $360$. | [
"Solution:\nAnswer: $800$\n\nSince $360=2^{3} \\cdot 3^{2} \\cdot 5$, the only possible prime divisors of $d_{i}$ are $2$, $3$, and $5$, so we can write $d_{i}=2^{a_{i}} \\cdot 3^{b_{i}} \\cdot 5^{c_{i}}$, for nonnegative integers $a_{i}, b_{i}$, and $c_{i}$. Then, $d_{1} d_{2} d_{3} \\mid 360$ if and only if the f... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 800 | |
01p9 | Find all triples $(x, y, z)$ of nonnegative integers $x$, $y$, $z$ such that $7^x = 3^z - 2^y$. | [
"$\\{(x, y, z)\\} = \\{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\\}$.\nWe rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.\nConsider three cases: $y = 1$, $y = 2$ and $y \\ge 3$.... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | [[0, 1, 1], [1, 1, 2], [0, 3, 2], [2, 5, 4]] | |
06a5 | We write $99$ circles in a line and in their interior we write the numbers from $1$ to $99$, as follows:
We color each circle with one of the two colors: red (R) and green (G). We say that a coloring is «good», if the following happens:
*The number of red circles in the first part of numbers f... | [
"(a) Each circle can be colored with two different colors independently of the coloring of the other circles. Therefore, according to the multiplicative principle, the different colorings are\n$$\n\\underbrace{2 \\cdot 2 \\cdot 2 \\dotsm 2}_{99} = 2^{99}.\n$$\n\n(b) We consider the coloring $X: 1\\ 2\\ 3\\ \\dots\\... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | a) 2^99; b) 2^98 | |
06rj | Let $ABC$ be a triangle with $AB = AC$, and let $D$ be the midpoint of $AC$. The angle bisector of $\angle BAC$ intersects the circle through $D$, $B$, and $C$ in a point $E$ inside the triangle $ABC$. The line $BD$ intersects the circle through $A$, $E$, and $B$ in two points $B$ and $F$. The lines $AF$ and $BE$ meet ... | [
"Let $D'$ be the midpoint of the segment $AB$, and let $M$ be the midpoint of $BC$. By symmetry at line $AM$, the point $D'$ has to lie on the circle $BCD$. Since the $\\operatorname{arcs} D'E$ and $ED$ of that circle are equal, we have $\\angle ABI = \\angle D'BE = \\angle EBD = IBK$, so $I$ lies on the angle bise... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Q... | null | proof only | null | |
01r4 | Let $n$ points be given inside a rectangle $R$ such that no two of them lie on a line parallel to one of the sides of $R$. The rectangle $R$ is to be dissected into smaller rectangles with sides parallel to the sides of $R$ in such a way that none of these rectangles contains any of the given points on its interior.
Pr... | [] | Belarus | SELECTION and TRAINING SESSION | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
03qa | In an acute triangle $ABC$, point $H$ is the intersection point of altitude $CE$ to $AB$ and altitude $BD$ to $AC$. A circle with $DE$ as its diameter intersects $AB$ and $AC$ at points $F$ and $G$, respectively. $FG$ and $AH$ intersect at point $K$. If $BC = 25$, $BD = 20$, and $BE = 7$, find the length of $AK$. | [
"We know that $\\angle ADB = \\angle AEC = 90^\\circ$, therefore\n\n$$\n\\triangle ADB \\sim \\triangle AEC,\n$$\nand\n$$\n\\frac{AD}{AE} = \\frac{BD}{CE} = \\frac{AB}{AC} \\quad (1)\n$$\nBut $BC = 25$, $BD = 20$, and $BE = 7$, so $CD = 15$, and $CE = 24$. From (1), we obtain\n\nThus, point $D$ is the midpoint of t... | China | China Mathematical Competition (Extra Test) | [
"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",
"Geometry > Plane Geometry > Miscellaneou... | English | proof and answer | 216/25 | |
01df | Consider triangles where each corner has integer coordinates. Such a triangle can be legally *transformed* by moving one corner parallel to the opposite side to a different point with integer coordinates. Show that if two triangles with integer coordinates have the same area, then there exists a series of legal transfo... | [
"We will first show that any such triangle can be transformed to a *special* triangle whose corners are at $(0,0)$, $(0,1)$ and $(n,0)$. Since every transformation preserves the triangle's area, triangles with the same area will have the same value for $n$.\n\nDefine *y-span* of a triangle to be the difference betw... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
05ak | The altitudes of an acute-angled triangle $ABC$ intersect at point $H$. The tangent at point $A$ to the circumcircle of triangle $AHB$ intersects the line $CH$ at point $K$. The tangent at point $A$ to the circumcircle of triangle $AHC$ intersects the line $BH$ at point $L$. Prove that the points $B$, $C$, $K$, $L$ lie... | [
"Denote $\\angle CAB = \\alpha$, $\\angle ABC = \\beta$, and $\\angle BCA = \\gamma$. By tangency, $\\angle KAH = \\angle ABH = 90^\\circ - \\alpha$ (Fig. 41). Since $\\angle ACH = 90^\\circ - \\alpha$ as well, triangles $AHC$ and $KHA$ are similar. Consequently, the corresponding third angles are equal, i.e., $\\a... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
01y6 | Call a polygon on a Cartesian plane to be *integer* if all its vertices are *integer*. A convex integer 14-gon is cut into integer parallelograms with areas not greater than $C$.
Find the minimal possible $C$.
(A. Yuran) | [
"Answer: minimal possible $C$ equals $5$.\n\nFirst we prove two lemmas.\n\n**Lemma 1:** the given 14-gon has $7$ pairs of opposite parallel sides.\nSince the 14-gon is convex, it suffices to prove that each side has a side parallel to it. Consider an arbitrary side $AB$ of the given 14-gon. Draw the line perpendicu... | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof and answer | 5 | |
0cqi | A convex pentagon $P$ is given. Peter wrote down the five values of sines of the angles of $P$, while Basil wrote down the five values of cosines of the angles of $P$. It appears that among five Peter's numbers, no four are pairwise distinct. Determine whether the five Basil's numbers could be pairwise distinct.
Дан в... | [
"No, they could not.\n\nSuppose the contrary; then all angles of the pentagon are distinct numbers from the interval $(0, \\pi)$. Note immediately that then Peter cannot have three equal numbers, since in this interval there are not three distinct angles with equal sines.\n\nTherefore, Peter must have two pairs of ... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof and answer | No | |
0dah | Let $n \geq 2$ be a positive integer. A subset of positive integers $S$ is called comprehensive if for every integer $0 \leq x < n$, there is a subset of $S$ whose sum of elements has remainder $x$ when divided by $n$. Note that the empty set has sum $0$. Show that if a set $S$ is comprehensive then there is some (not ... | [
"We will show that if $|S| \\geq n$, we can remove one element from $S$ and still have a comprehensive set. Doing this repeatedly will always allow us to find a comprehensive subset of size at most $n-1$.\n\nDenote $S = \\{s_1, s_2, \\ldots, s_k\\}$ for some $k \\geq n$. Now start with the empty set and add in the ... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
07tm | Determine the largest integer which is the side length of a square tile which can be used to completely tile a rectangle that is inscribed in a circle of radius $75$. | [
"Let $x$ be the side length of the square tile. If the rectangle is completely tiled with such square tiles, there exist integers $a, b$ such that the side lengths of the rectangle are $ax$ and $bx$. We let $r = 75$ denote the radius of the circle.\n\n\n\nThe diagonals of the rectangle are ... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Alge... | null | proof and answer | 30 | |
0j34 | Problem:
Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$. | [
"Solution:\nWe have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\\frac{\\sqrt{2 b-1} \\pm 1}{2}$. Then\n\\[\n\\begin{aligned}\nx^{20}+y^{20} & =\\left(\\frac{\\sqrt{2 b-1}+1}{2}\\right)^{20}+\\left(\\frac{\\sqrt{... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | -90 | |
056x | The diagonals of a tangential quadrilateral $ABCD$ intersect at point $P$. The side $AB$ is longer than any other side of $ABCD$. Prove that the angle $APB$ is obtuse. | [
"Let $a, b, c, d$ be the lengths of the tangent line segments at vertices $A, B, C, D$, respectively, and $\\alpha = \\angle APB$ (Fig. 25). We have $a + b > b + c$ and $a + b > a + d$ as $AB$ is the longest side, implying $a > c$ and $b > d$. By the law of cosines:\n$$\n(a+b)^2 = PA^2 + PB^2 - 2 \\cdot PA \\cdot P... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
09mg | How many seven-digit integers can be formed using only the digits $0$, $1$, and $2$, with the condition that there are at most four $1$'s and at most three $2$'s? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | final answer only | 1066 | |
0hp0 | Problem:
A circle is inscribed in a sector that is one sixth of a circle of radius $6$. (That is, the circle is tangent to both segments and the arc forming the sector.) Find, with proof, the radius of the small circle.
 | [
"Solution:\nBecause the circles are tangent, we can draw the line $AT$, which passes through the center $O$ of the other circle. Note that triangles $ADO$ and $AEO$ are symmetric (this is HL congruence: $AO$ is shared, radii $OD$ and $OE$ are equal, and angles $ADO$ and $AEO$ are right). Therefore, since $\\angle B... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 | |
0cta | A point $X$ is chosen inside a convex 100-gon ($X$ does not lie on sides or diagonals of the 100-gon). Initially, the vertices of the polygon are not marked. Pete and Bazil mark the vertices in turn. Pete starts and marks two vertices by his first move; after that, by any move a player marks one vertex. If after someon... | [
"Paint the sides of the polygon alternately in black and white. Consider some convex quadrilateral $ABCD$ where $AB$ and $CD$ are sides of the same color; let $K$ be the meeting point of its diagonals. If $X$ lies in the triangle $KBC$ (see Fig. 17), then Pete can mark $B$ and $C$ by his first move and win (no succ... | Russia | Russian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
0i2o | Problem:
Given a line segment $A B$, construct a segment half as long as $A B$ using only a compass. Construct a segment one-third as long as $A B$ using only a compass. | [
"Solution:\nFirst, we provide (part of) an algorithm for circular inversion. Suppose we are given point $O$ and a circle centered at $O$ of some radius $r$. If $P$ is a point outside the circle, we wish to construct point $Q$ on ray $O P$, satisfying $O P \\cdot O Q = r^{2}$. Let the circle centered at $P$, with ra... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eyq | Problem:
The difference between the longest and shortest diagonals of the regular $n$-gon equals its side. Find all possible $n$. | [
"Solution:\nAnswer: $n = 9$\n\nFor $n < 6$, there is at most one length of diagonal. For $n = 6$, $7$ the longest and shortest, and a side of the $n$-gon form a triangle, so the difference between the longest and shortest is less than the side.\n\nFor $n > 7$ the side has length $2R \\sin \\dfrac{\\pi}{n}$, the sho... | Soviet Union | 2nd ASU | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | n = 9 | |
06bn | Two circles $C_1, C_2$ with different radii are given in the plane. They touch each other externally at $T$. Consider any points $A \in C_1$ and $B \in C_2$, both different from $T$, such that $\angle ATB = 90^\circ$.
a. Show that all such lines $AB$ are concurrent.
b. Find the locus of midpoints of all such segments... | [
"a.\nLet $O_1$ and $O_2$ be the centres of $C_1$ and $C_2$ respectively. Let $O_1O_2$ meet $C_1$ and $C_2$ again at $C$ and $D$ respectively. Note that $CT$ and $DT$ are diameters of the two circles. Therefore, we have\n$$\n\\angle CAT = \\angle BTA = \\angle TBD = 90^\\circ.\n$$\nThis implies $AC \\parallel BT$ an... | Hong Kong | 1997-2023 IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a. All such lines AB are concurrent at the external center of homothety of the two circles.
b. The locus of the midpoints is the circle with diameter O1O2. | |
0arb | Problem:
Find the probability of obtaining two numbers $x$ and $y$ in the interval $[0,1]$ such that $x^{2}-3 x y+2 y^{2}>0$. | [
"Solution:\nLet $x, y \\in [0,1]$. We are to find the probability that $x^{2} - 3 x y + 2 y^{2} > 0$.\n\nWe can factor the quadratic:\n$$\nx^{2} - 3 x y + 2 y^{2} = (x - y)(x - 2y)\n$$\nSo, $(x - y)(x - 2y) > 0$.\n\nThis inequality holds if both factors are positive or both are negative:\n\nCase 1: $x - y > 0$ and ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3/4 | |
0gbl | 平面上有 $\triangle ABC$ 及一點 $O$, $\Gamma$ 為 $\triangle ABC$ 的外接圓。設直線 $CO$ 與直線 $AB$ 交於點 $D$, 直線 $BO$ 與直線 $CA$ 交於點 $E$。設直線 $AO$ 與 $\Gamma$ 再交於點 $F$。令點 $I$ 為 $\Gamma$ 和 $\triangle ADE$ 外接圓的另一個交點, 點 $Y$ 為直線 $BE$ 與 $\triangle CEI$ 外接圓的另一個交點, 而點 $Z$ 為直線 $CD$ 與 $\triangle BDI$ 外接圓的另一個交點。在 $\Gamma$ 上分別作以 $B$, $C$ 為切點的兩條切線, 設它們相交於... | [
"\n\n解：設 $BC$ 與 $ED$ 交於點 $X$。因為 $I$ 是 $BCED$ 的密克點，故 $X$ 在 $\\odot(BDI)$，$\\odot(CEI)$ 上。因為 $BFCU$ 為調和四邊形，故\n$$\nA(B, C; F, U) = -1 = A(B, C; O, X),\n$$\n從而 $X \\in AU$。\n\n因\n$$\n\\angle BYX = \\angle ACB = \\angle XUB = \\angle BGX,\n$$\n故 $X \\in \\odot(BGY)$。\n\n同理可得 $X$ 在 $\\odot(CGZ)$ ... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 模擬競賽(一) | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03b3 | Let $a$, $b$ and $c$ be the lengths of sides of a triangle. Prove that
$$
\left| \sqrt{\frac{a}{b}} - \sqrt{\frac{b}{a}} + \sqrt{\frac{b}{c}} - \sqrt{\frac{c}{b}} + \sqrt{\frac{c}{a}} - \sqrt{\frac{a}{c}} \right| < \frac{1}{10}.
$$ | [] | Bulgaria | Selection test for 27. Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0dib | Let $ABC$ be a triangle with $AB \neq BC$; and let $BD$ be the internal bisector of $\angle ABC$, $(D \in AC)$. Denote by $M$ the midpoint of the arc $AC$ which contains point $B$ in the circumscribed circle of the triangle $ABC$. The circumscribed circle of the triangle $\triangle BDM$ intersects the segment $AB$ at p... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0a44 | Problem:
Beschouw de rij $y_0, y_1, \ldots$ met $y_0 = -\frac{1}{4}$ en $y_1 = 0$ en die verder voldoet aan
$$y_{n+1} + y_{n-1} = 4y_n + 1$$
voor alle $n \ge 1$. Bewijs dat voor alle $n \ge 0$ de uitdrukking $2y_{2n} + \frac{3}{2}$
a) een positief geheel getal is en
b) het kwadraat van een geheel getal is. | [
"Solution:\nWe maken de substitutie $x_n = 4y_n + 2$. Dan wordt de vergelijking homogeen:\n$$x_{n+1} + x_{n-1} = 4y_{n+1} + 2 + 4y_{n-1} + 2 = 4(4y_n + 1) + 4 = 16y_n + 8 = 4x_n,$$\nmet beginvoorwaarden $x_0 = 4(-\\frac{1}{4}) + 2 = 1$ en $x_1 = 4 \\cdot 0 + 2 = 2$. Alle getallen in de rij $(x_i)$ zijn dus geheel e... | Netherlands | IMO-selectietoets I | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ffc | Problem:
En un plano vertical se consideran los puntos $A$ y $B$ situados sobre una recta horizontal, y la semicircunferencia de extremos $A, B$ situada en el semiplano inferior. Un segmento de longitud $a$, igual al diámetro de la semicircunferencia, se mueve de manera que contiene siempre el punto $A$, y que uno de ... | [
"Solution:\n\nSea $C$ el extremo del segmento que recorre la semicircunferencia dada, $M$ el punto medio del mismo y $\\alpha$ el ángulo que forma $AC$ con la horizontal $AB$, $\\left(0<\\alpha<\\frac{\\pi}{2}\\right)$.\nSea $N$ la proyección ortogonal del punto medio $M$ sobre el segmento $AB$. Ponemos $y=MN$. Se ... | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | (1+sqrt(33))/8 | |
0c1s | The infinite grid of lines of the form $\mathbb{R} \times \{m\}$ and $\{m\} \times \mathbb{R}$, where $m$ runs through all integers, subdivide the Euclidean plane $\mathbb{R} \times \mathbb{R}$ into $1 \times 1$ cells. Let $S$ be the set-theoretic union of a finite number of such cells, and let $a$ be a positive real n... | [
"Let $n = \\lfloor a^{-1/2} \\rfloor$ and notice that $n \\ge 2$, since $a \\le 1/4$. Choose a large enough integer $k$ to cover $S$ by an $n^k \\times n^k$ array $Q$ so that the ratio of the area of $S$ to the area of $Q$ is at most $a$. Subdivide $Q$ into $n^2$ congruent square subarrays $Q'$, and notice that the... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
07he | Given triangle $ABC$, variable points $X$ and $Y$ are chosen on segments $AB$ and $AC$, respectively. Let $Z$ be a point on the line $BC$ such that $ZX = ZY$. The circumcircle of $XYZ$ intersects the line $BC$ at $T$, for the second time. Point $P$ is chosen on line $XY$ such that $\angle PTZ = 90^\circ$. Let $Q$ be a ... | [
"Let $F'$ be the intersection of $CX$ with the circumcircle of $ABC$ and $G$ be the intersection of $XY, BC$. Note that $TZ$ is the external angle bisector of $\\angle XTY$ and $PT$ is perpendicular to $TZ$, so $PT$ is the angle bisector of $\\angle XTZ$. Thus $(GP, XY) = -1$ and $BY, CX, AP$ are concurrent. Let $R... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
01cw | Find all real numbers $a$ for which there exists a non-constant function $f: \mathbb{R} \to \mathbb{R}$ satisfying the equations
$$
1) \quad f(ax) = a^2 f(x)
$$
$$
2) \quad f(f(x)) = a f(x).
$$ | [
"Examining $f(f(f(x)))$ we can write\n$$\n\\begin{align*}\na^2 f(x) &\\stackrel{(2)}{=} a f(f(x)) &\\stackrel{(2)}{=} f(f(f(x))) \\\\\n&\\stackrel{(2)}{=} f(a f(x)) &\\stackrel{(1)}{=} a^2 f(f(x)) &\\stackrel{(2)}{=} a^3 f(x)\n\\end{align*}\n$$\nwhich implies $a \\in \\{0, 1\\}$ or $f(x) = 0$.\n\nIf $a = 1$, then f... | Baltic Way | Baltic Way 2016 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | a = 0 or a = 1 | |
05lx | Problem:
Trouver le nombre de suites $\left(u_{n}\right)_{n \geqslant 1}$ d'entiers relatifs telles que $u_{n} \neq -1$ pour tout entier $n \geqslant 1$ et telles que
$$
u_{n+2}=\frac{2014+u_{n}}{1+u_{n+1}}$$
pour tout entier $n \geqslant 1$. | [
"Solution:\n\nLa relation de l'énoncé impose que $\\left(u_{n+1}-u_{n-1}\\right)\\left(u_{n}+1\\right)=u_{n}-u_{n-2}$ pour $n \\geqslant 3$. Par récurrence, il vient\n$$\n\\mathbf{u}_{3}-\\mathbf{u}_{1}=\\prod_{i=3}^{n}\\left(u_{i}+1\\right)\\left(u_{n+1}-u_{n-1}\\right)\n$$\nSupposons par l'absurde que $u_{3} \\ne... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 14 | |
0cm8 | Problem:
Find all pairs of integers $(c, d)$, both greater than $1$, such that the following holds:
For any monic polynomial $Q$ of degree $d$ with integer coefficients and for any prime $p > c(2c+1)$, there exists a set $S$ of at most $\left(\frac{2c-1}{2c+1}\right)p$ integers, such that
$$
\bigcup_{s \in S} \{s, Q(s... | [] | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Other"
] | null | proof and answer | All integer pairs (c, d) with c ≥ d and both greater than 1. | |
0hgn | When dividing with remainder some four consecutive positive integers by some three-digit integer it turned out, that the sum of these four remainders is equal to $983$. Find the remainder under the division of the smallest of these four numbers by $109$. | [
"Denote these four consecutive integers by $n$, $n+1$, $n+2$ and $n+3$. Denote the three-digit number that we were dividing by as $b$. Let $n = bq + r$. Consider possible values of $r$.\n\nIf $r \\le b - 4$, then these remainders are $r$, $r+1$, $r+2$ and $r+3 \\Rightarrow r + r+1 + r+2 + r+3 = 983 \\Rightarrow 4r ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 108 | |
0483 | Find the smallest real number $M$ such that there exist complex numbers $a, b, c, d$ with $|a| = |b| = |c| = |d| = 1$ satisfying: for any complex number $z$ with $|z| = 1$,
$$
|az^3 + bz^2 + cz + d| \le M.
$$ | [
"Let $L$ denote the maximum squared modulus of the polynomial $f(z) = az^3 + bz^2 + cz + d$ on the unit circle.\nBy choosing a unit complex number $s$ with $s^3 = d/a$ and considering $f_1(z) = d^{-1}f(sz)$, we may assume without loss of generality that $a = d = 1$. Thus we only need to consider polynomials of the ... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity"
] | English | proof and answer | 2√5 - 2 | |
0hcj | <table><tr><td>$a_1$</td><td>$a_2$</td><td>$a_3$</td><td>$a_4$</td></tr><tr><td>$a_5$</td><td>$a_6$</td><td>$a_7$</td><td>$a_8$</td></tr><tr><td>$a_9$</td><td>$a_{10}$</td><td>$a_{11}$</td><td>$a_{12}$</td></tr><tr><td>$a_{13}$</td><td>$a_{14}$</td><td>$a_{15}$</td><td>$a_{16}$</td></tr></table>
**Fig. 7**
A $4 \times... | [
"We will start with the following lemma.\n\n**Lemma 1.** There exists a diagonal that has two equal numbers for any $2 \\times 2$ square.\n**Proof.** Let $X$ be the greatest number in $2 \\times 2$ square. Let his neighbors in this $2 \\times 2$ square be $A$ and $B$. Clearly, they are located on a diagonal, so if ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof only | null | |
06u5 | Let $a_{1}, a_{2}, \ldots, a_{n}, k$, and $M$ be positive integers such that
$$
\frac{1}{a_{1}}+\frac{1}{a_{2}}+\cdots+\frac{1}{a_{n}}=k \quad \text{ and } \quad a_{1} a_{2} \ldots a_{n}=M .
$$
If $M>1$, prove that the polynomial
$$
P(x)=M(x+1)^{k}-\left(x+a_{1}\right)\left(x+a_{2}\right) \cdots\left(x+a_{n}\right)
$$
... | [
"We first prove that, for $x>0$,\n$$\n\\begin{equation*}\na_{i}(x+1)^{1 / a_{i}} \\leqslant x+a_{i}, \\tag{1}\n\\end{equation*}\n$$\nwith equality if and only if $a_{i}=1$. It is clear that equality occurs if $a_{i}=1$.\nIf $a_{i}>1$, the AM-GM inequality applied to a single copy of $x+1$ and $a_{i}-1$ copies of 1 ... | IMO | International Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
01e7 | Let $n$ be a positive integer. Elfie the Elf lives in a three dimensional space $\mathbb{Z}^3$. She starts at the origin: $(0,0,0)$. In each turn she can teleport into any point in $\mathbb{Z}^3$ which lies at the distance $\sqrt{n}$ from her current location. However, teleportation is a complicated procedure. Elfie st... | [
"Answer: there are no such $n$.\nWe colour all the points in $\\mathbb{Z}^3$ white and black: The point $(x, y, z)$ is colored white if $x+y+z \\equiv_2 0$ and black if $x + y + z \\equiv_2 1$.\nAfter the first move Elfie is at a point $(a, b, c)$ where $a^2 + b^2 + c^2 = n$. Thus, $a + b + c \\equiv_2 n$\nNow, if ... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | There are no such n. | |
06a7 | Let $AB\Gamma$ be an isosceles triangle and a point $\Delta$ in its interior such that $\Delta\hat{B}\Gamma = 30^\circ$, $\Delta\hat{B}A = 50^\circ$, $B\hat{\Gamma}\Delta = 55^\circ$.
a. Prove that $\hat{B} = \hat{\Gamma} = 80^\circ$
b. Find the measure of the angle $\Delta\hat{A}\Gamma$. | [
"a.\nWe have $\\hat{B} = 50^\\circ + 30^\\circ = 80^\\circ$.\nSuppose that $\\hat{A} = \\hat{B} = 80^\\circ$. Then\n$$\n\\hat{A} + \\hat{B} + \\hat{\\Gamma} = 80^\\circ + 80^\\circ + \\hat{\\Gamma} > 160^\\circ + 55^\\circ = 215^\\circ,\n$$\nabsurd.\nIf $A\\Gamma = \\frac{180^\\circ - 80^\\circ}{2} - 50^\\circ$, th... | Greece | 39th Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | ∠B = ∠Γ = 80°, ∠ΔAΓ = 5° | |
0d8n | A positive integer $k>1$ is called nice if for any pair ($m, n$) of positive integers satisfying the condition $k n + m \mid k m + n$ we have $n \mid m$.
1. Prove that $5$ is a nice number;
2. Find all nice numbers. | [
"1) For $k=5$, we need to prove that for all $m, n$ satisfying $5 n + m \\mid 5 m + n$ then $n \\mid m$. Note that $5 n + m \\leq 5 m + n$ or $n \\leq m$, then $1 \\leq \\frac{5 m + n}{5 n + m} < 5$.\n\nThen $A = \\frac{5 m + n}{5 n + m} \\in \\{1, 2, 3, 4\\}$. We consider some cases:\n\n- If $A = 1$ then $m = n$.\... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 2, 3, 5 | |
0fcp | Problem:
Un cuadrado de papel $ABCD$, de lado unidad, se dobla de modo que el vértice $A$ toque en un punto arbitrario $E$ del lado $CD$. Así, se obtienen tres triángulos rectos formados por una sola capa de papel.
Determinar la longitud de sus lados en función de $x=DE$ y demostrar que el perímetro del triángulo mayo... | [
"Solution:\n\nDenominamos, con letras mayúsculas, los puntos característicos que produce el plegado, y con minúsculas, los lados de los triángulos.\n\nLos lados del triángulo $DEF$ se obtienen, resolviendo el sistema:\n$$\n\\begin{gathered}\n\\left.\\left.\\begin{array}{c}\nx^{2}+z^{2}=y^{2} \\\\\nz+y=1\n\\end{arra... | Spain | Fase Local | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Let x be the length from the nearer corner of the base to the touch point. For triangle DEF: sides are x, (1 + x^2)/2, (1 − x^2)/2 and its perimeter is x + 1. For triangle ECI: sides are 1 − x, (1 + x^2)/(1 + x), 2x/(1 + x) and its perimeter is 2. For triangle IHG: sides are x(1 − x)/(1 + x), ((1 + x^2)(1 − x))/(2(1 + ... | |
0cjb | Consider the sets
$$
A = \{(x, y) \mid x, y \in \mathbb{R} \text{ and } x + y + 1 = 0\}
$$
and
$$
B = \{(x, y) \mid x, y \in \mathbb{R} \text{ and } x^3 + y^3 + 1 = 3xy\}.
$$
a) Show that $A \subset B$.
b) Prove that the set $B \setminus A$ has exactly one element. | [
"a. If $(x, y) \\in A$, then $y = -x - 1$. We get\n$$\nx^3 + y^3 + 1 = x^3 + (-x - 1)^3 + 1 = x^3 - (x + 1)^3 + 1 = x^3 - (x^3 + 3x^2 + 3x + 1) + 1 = -3x^2 - 3x = 3x(-x - 1) = 3xy,\n$$\nso $(x, y) \\in B$.\n\nb. If $(x, y) \\in B$, then $x^3 + y^3 - 3xy + 1 = 0$, so\n$$\n(x + y)^3 - 3xy(x + y) - 3xy + 1 = 0,\n$$\no... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0k6e | Problem:
In $\triangle ABC$, the incircle centered at $I$ touches sides $AB$ and $BC$ at $X$ and $Y$, respectively. Additionally, the area of quadrilateral $BXIY$ is $\frac{2}{5}$ of the area of $ABC$. Let $p$ be the smallest possible perimeter of a $\triangle ABC$ that meets these conditions and has integer side leng... | [
"Solution:\n\nNote that $\\angle BXI = \\angle BYI = 90^\\circ$, which means that $AB$ and $BC$ are tangent to the incircle of $ABC$ at $X$ and $Y$ respectively. So $BX = BY = \\frac{AB + BC - AC}{2}$, which means that $\\frac{2}{5} = \\frac{[BXIY]}{[ABC]} = \\frac{AB + BC - AC}{AB + BC + AC}$. The smallest perimet... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2√5 | |
0hnk | Problem:
What is the remainder when $100!$ is divided by $101$? | [
"Solution:\n\nWilson's theorem says that for $p$ a prime, $(p-1)! \\equiv -1 \\pmod{p}$, so the remainder is $100$."
] | United States | null | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 100 | |
0dsc | Let $n \ge 3$ be an integer. Prove that there exist positive integers $x_1, \dots, x_n$ in geometric progression and positive integers $y_1, \dots, y_n$ in arithmetic progression such that $x_1 < y_1 < x_2 < y_2 < \dots < x_n < y_n$. | [
"Let $x_k = n^{2n} \\left( 1 + \\frac{1}{n^2} \\right)^k$ and $y_k = n^{2n} + (k+1)n^{2n-2}$, $k = 1, \\dots, n$. Then\n$x_1 < y_1 < x_2 < y_2 < \\dots < x_n < y_n$.\n\nBy the binomial theorem, we have for $k \\ge 2$ and $a \\le 1/k^2$,\n$$\n\\begin{align*}\n(1+a)^k &= 1+ka+a \\left( \\frac{ak(k-1)}{2!} + \\frac{a^... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0b9k | Call a positive integer *balanced* if the number of its distinct prime factors is equal to the number of its digits in the decimal representation; for example, the number $385 = 5 \cdot 7 \cdot 11$ is balanced, while $275 = 5^2 \cdot 11$ is not. Prove that there exist only a finite number of balanced numbers. | [
"Let $p_1 = 2$, $p_2 = 3$, $p_3 = 5$, \\ldots be the sequence of primes. Any balanced number $a$ with $n$ digits satisfies $a \\ge p_1 p_2 \\cdots p_n$. Since $p_1 p_2 \\cdots p_{11} = 2 \\cdot 3 \\cdot 5 \\cdots 29 \\cdot 31 > 10^{11}$ and $p_k > 10$, for any $k > 11$, it follows that there are no balanced numbers... | Romania | 62nd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null |
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