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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ldu | For two positive integers $n$ and $d$, let $S_n(d)$ be the set of all ordered $d$-tuples $(x_1, x_2, \dots, x_d)$ that satisfy all the following conditions:
i) $x_i \in \{1, 2, \dots, n\}$ for every $i \in \{1, 2, \dots, d\}$;
ii) $x_i \neq x_{i+1}$ for every $i \in \{1, 2, \dots, d-1\}$;
iii) There do not exist indic... | [
"a) To calculate $S_3(5)$, we need to count the number of $(a, b, c, d, e)$ such that $a, b, c, d, e \\in \\{1, 2, 3\\}$ and satisfy the conditions ii), iii). We investigate these two cases.\n\n* If the first 3 terms are distinct, we consider the case where $(a, b, c) = (1, 2, 3)$ represents. For $(1, 2, 3, d, e)$ ... | Vietnam | VMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 12
b) |S_n(d)| > 0 if and only if d ≤ 2n − 1 | |
066q | Let $ABC$ be a triangle. At the extensions of the side $AB$ we consider the points $A_1$, $B_1$ such that $AB_1 = BA_1 = x$ ($B_1$ closer to $A$). At the extensions of the side $BC$ we consider the points $B_4$, $C_4$ such that $CB_4 = BC_4 = y$ ($B_4$ closer to $C$). At the extensions of the side $AC$ we consider the ... | [] | Greece | Selection Examination B | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors... | English | proof only | null | |
0c0k | Consider $n$ weights, $n \ge 2$, of masses $m_1, m_2, \dots, m_n$, where $m_k$ are positive integers such that $1 \le m_k \le k$ for all $k \in \{1, 2, \dots, n\}$. Prove that we can place the weights on the two pans of a balance such that the pans stay in equilibrium if and only if the number $m_1 + m_2 + \dots + m_n$... | [
"Obviously, if the two pans can stay in equilibrium, the sum of the masses must be even (namely twice the sum of the masses of the weights placed on one of the pans of the balance).\n\nConversely, we prove by \"strong\" induction after $n$ that, if the sum of the masses is even, then one can place the weights conve... | Romania | 69th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
035w | Problem:
Let $k$ be the circumcircle of $\triangle ABC$ with $\angle ACB > 90^\circ$, and $BD$ be the diameter of $k$ through $B$. The circle $k_1$ with center $D$ and radius $DC$ meets $k$ at point $E$ and $AB$ at point $G$. If $F$ is the intersection point of $GE$ and $BD$, prove that $\angle DCG = \angle EFD$. | [
"Solution:\nSince $\\angle ACB > 90^\\circ$, the points $G$ and $E$ lie on different sides to the diameter $BD$. If $\\angle DCG = \\angle EFD$, then $\\angle DCG + \\angle DFG = 180^\\circ$, i.e. it is enough to prove that the quadrilateral $CDFG$ is cyclic. On the other hand, $CE \\perp BD$ and $D$ is the center ... | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08zd | How many permutations $(p_1, p_2, \dots, p_{1000})$ of $\{1, 2, \dots, 1000\}$ are there such that $p_i$ is a multiple of $i$ for any $1 \le i \le 999$? | [
"Let $S = \\{1, 2, \\dots, 1000\\}$. We want to find the number of bijective functions $f: S \\to S$ such that $f(i)$ is a multiple of $i$ for any $1 \\le i \\le 999$. Denote $f^k(i) = \\underbrace{f(f(\\dots f(i)\\dots))}_{k \\text{ times}}$.\n\nConsider the sequence $\\{1000, f(1000), f^2(1000), \\dots\\}$. Since... | Japan | Japan 2022 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 504 | |
08gd | Problem:
Vittorio calcola la somma dei numeri da $1$ a $2024$, ma per sbaglio ne salta uno! Sapendo che ottiene un multiplo di $5$, quale dei seguenti può essere il numero saltato?
(A) $1$
(B) $100$
(C) $222$
(D) $2023$
(E) $2024$ | [
"Solution:\n\nLa risposta è (B). Innanzitutto, notiamo che la somma dei numeri da $1$ a $2024$ è un multiplo di $5$. Ciò si può osservare direttamente sapendo che la somma dei numeri da $1$ a $n$ è $\\frac{n(n+1)}{2}$; alternativamente, si può notare che ciascuno dei seguenti numeri è multiplo di $5$: $1+2024$, $2+... | Italy | Italian Mathematical Olympiad - February Round | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | B | |
0f48 | Problem:
Given 6 points inside a $3 \times 4$ rectangle, show that we can find two points whose distance does not exceed $\sqrt{5}$. | [] | Soviet Union | 15th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06p4 | Let $a > b > 1$ be relatively prime positive integers. Define the weight of an integer $c$, denoted by $w(c)$, to be the minimal possible value of $|x| + |y|$ taken over all pairs of integers $x$ and $y$ such that
$$
a x + b y = c.
$$
An integer $c$ is called a local champion if $w(c) \geq w(c \pm a)$ and $w(c) \geq w(... | [
"Call the pair of integers $(x, y)$ a representation of $c$ if $a x + b y = c$ and $|x| + |y|$ has the smallest possible value, i.e. $|x| + |y| = w(c)$.\nWe characterise the local champions by the following three observations.\n\nLemma 1. If $(x, y)$ a representation of a local champion $c$ then $x y < 0$.\n\nProof... | IMO | IMO 2006 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | All local champions are exactly the integers of the form ±(a x − b y) with positive integers x, y satisfying x < b and x + y = floor((a + b) / 2). The total number of local champions is b − 1 if both a and b are odd, and 2(b − 1) otherwise. | |
05ze | Problem:
Soient $ABC$ un triangle non-isocèle et $I$ le centre de son cercle inscrit. Soit $D$ un point sur $[BC]$ (autre que $B$ et $C$). Le cercle circonscrit à $DIB$ coupe $(AB)$ une deuxième fois en $E \neq B$, et le cercle circonscrit à $DIC$ coupe $(AC)$ une deuxième fois en $F \neq C$. Le cercle circonscrit à $... | [
"Solution:\n\n\n\nSur la figure, il semble que $I$ est le centre du cercle $(DEF)$, et que les droites $(MF)$, $(EN)$, $(PQ)$ sont toutes perpendiculaires à $(AI)$ : c'est donc la stratégie qu'on va utiliser pour résoudre l'exercice.\n\n- **Lemme 1:** Le point $I$ est le centre du cercle ci... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Mis... | null | proof only | null | |
01c3 | Let $p^n$ be a prime power. Find the number of quadruples $(a_1, a_2, a_3, a_4)$ with $a_i \in \{0, 1, \dots, p^n - 1\}$ for $i = 1, 2, 3, 4$, such that
$$
p^n \mid (a_1a_2 + a_3a_4 + 1).
$$ | [
"$$\np^{3n} - p^{3n-2}.\n$$\nWe have $p^n - p^{n-1}$ choices for $a_1$ such that $p \\nmid a_1$. In this case, for any of the $p^n \\cdot p^n$ choices of $a_3$ and $a_4$, there is a unique choice of $a_2$. Namely\n$$\na_2 \\equiv a_1^{-1}(-1 - a_3a_4) \\mod p^n.\n$$\nThis gives $p^{2n}(p^n - p^{n-1})$ quadruples.\n... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | p^{3n} - p^{3n-2} | |
08ks | Problem:
Let $A$ be a set of positive integers containing the number $1$ and at least one more element. Given that for any two different elements $m, n$ of $A$ the number $\frac{m+1}{(m+1, n+1)}$ is also an element of $A$, prove that $A$ coincides with the set of positive integers. | [
"Solution:\n\nLet $a > 1$ be the lowest number in $A \\setminus \\{1\\}$. For $m = a$, $n = 1$ one gets $y = \\frac{a+1}{(2, a+1)} \\in A$. Since $(2, a+1)$ is either $1$ or $2$, then $y = a+1$ or $y = \\frac{a+1}{2}$.\n\nBut $1 < \\frac{a+1}{2} < a$, hence $y = a+1$. Applying the given property for $m = a+1$, $n =... | JBMO | 2007 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0jbv | For integer $n \ge 2$, let $x_1, x_2, \dots, x_n$ be real numbers satisfying
$$
x_1 + x_2 + \dots + x_n = 0, \quad \text{and} \quad x_1^2 + x_2^2 + \dots + x_n^2 = 1.
$$
For each subset $A \subseteq \{1, 2, \dots, n\}$, define
$$
S_A = \sum_{i \in A} x_i.
$$
(If $A$ is the empty set, then $S_A = 0$.)
Prove that for any... | [
"This problem is a form of Chebyshev's inequality for random variables. For each subset $A \\subseteq \\{1, 2, \\dots, n\\}$, define\n$$\n\\Delta_A = 2S_A = \\sum_{i \\in A} x_i - \\sum_{i \\in \\{1, 2, \\dots, n\\} \\setminus A} x_i = \\sum_{i=1}^n \\epsilon_A(i)x_i,\n$$\nwhere $\\epsilon_A(i) = 1$ if $i \\in A$ a... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds precisely when exactly two of the numbers are nonzero, one positive and one negative of equal magnitude, with all others zero, and the threshold equals that magnitude. Concretely, for some distinct indices j and k, take x_k = sqrt(2)/2, x_j = -sqrt(2)/2, and all other x_i = 0, with lambda = sqrt(2)/2. In... | |
0a6i | Problem:
Find all pairs of positive integers $m$ and $n$ such that the centres of the unit squares in a $m$ by $n$ grid of unit squares can be paired up so that the distance between the centres of each pair is exactly $2$.
(A unit square has side length $1$.) | [
"Solution:\nWe will show that the answer is all pairs $(a,b)$ where either $a$ or $b$ (or both) is a multiple of $4$. First, partition the grid into four subgrids $A$, $B$, $C$, and $D$\n\n| A | B | A | B | A | B |\n|---|---|---|---|---|---|\n| C | D | C | D | C | D |\n| A | B | A | B | A | B |\n| C | D | C | D | C... | New Zealand | NZMO Round Two | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | All positive integer pairs where at least one of the two numbers is a multiple of four. | |
0l48 | Let $M$ be the greatest integer such that both $M + 1213$ and $M + 3773$ are perfect squares. What is the units digit of $M$?
(A) 1 (B) 2 (C) 3 (D) 6 (E) 8 | [
"Suppose $M + 1213 = j^2$ and $M + 3773 = k^2$ for nonnegative integers $j$ and $k$. Then $(k + j)(k - j) = k^2 - j^2 = 3773 - 1213 = 2560 = 5 \\cdot 2^9$.\nBecause $k + j$ and $k - j$ have the same parity and their product is even, they must both be even, and it follows that one of them is $5 \\cdot 2^i$ and the o... | United States | AMC 12 A | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
026x | Problem:
Júlia está treinando para olimpíadas de matemática. Um dia ela decide dividir $2014$ por cada um dos divisores inteiros positivos de $2015$. Para cada divisão, ela escreve o quociente no seu caderno e o resto em uma lousa.
a. Vamos ajudar Júlia. Escreva os oito divisores inteiros positivos de $2015$.
b. Par... | [
"Solution:\n\na. A fatoração de $2015$ em primos é $5 \\cdot 13 \\cdot 31$. Então seus oito divisores inteiros positivos são:\n$$\n\\{1, 5, 13, 31, 65, 155, 403, 2015\\}\n$$\n\nb. Escrevendo as divisões na forma $n = m \\cdot q + r$, onde $q$ é o quociente e $r$ é o resto, teremos\n$$\n\\begin{aligned}\n& 2014 = 1 ... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) Divisors of 2015: 1, 5, 13, 31, 65, 155, 403, 2015.
b) Quotients when dividing 2014 by these divisors: 2014, 402, 154, 64, 30, 12, 4, 0. Remainders: 0, 4, 12, 30, 64, 154, 402, 2014.
c) In general, for any n, if one divides n−1 by each positive divisor of n, the multiset of quotients equals the multiset of remainder... | |
08bu | Problem:
Andrea incolla 27 normali dadi a 6 facce tra loro in modo da formare un grande cubo. I dadi sono orientati in modo che le somme dei valori leggibili su ciascuna faccia del cubo siano, in un qualche ordine, $14$, $22$, $30$, $38$, $46$, $54$. Quanto vale la somma di tutte le facce dei dadi che, essendo state i... | [
"Solution:\n\nLa risposta è $(E)$. La somma dei valori sulle facce di un dado è $1+2+3+4+5+6=21$. La somma su tutte le facce, visibili e non, è quindi $27 \\cdot 21=567$. Per ottenere la somma sulle facce nascoste possiamo sottrarre a questo numero la somma dei numeri scritti su quelle visibili: la risposta è quind... | Italy | Gara di Febbraio | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | E | |
0gqx | Let $P$ be a point inside the triangle $ABC$ satisfying $\angle PAC = \angle PCB$ and $D$ be the midpoint of the line segment $PC$. Let $AP$ intersect $BC$ at $E$ and $ED$ intersect $BP$ at $Q$ such that $P$ is on the line segment $BQ$. Prove that $\angle BAP + \angle BCQ = 180^\circ$. | [
"\n\nLet the line passing through $B$ and parallel to $PC$ intersect the line $AE$ at $F$ and $CQ$ at $F'$. Let $QD$ and $BF$ intersect at $K$. Since the triangles $PEC$ and $FEB$ are similar, $PD = DC$ and $D, E, K$ are collinear, we obtain that $K$ is the midpoint of the line segment $BF$... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08bp | Problem:
Sia $k$ un intero positivo. Poniamo $a_{1}=1$ e, per ogni $n \geq 1$, $a_{n+1}=k\left(a_{1}+\cdots+a_{n}\right)+1$ (quindi per esempio $a_{2}=k a_{1}+1$ e $a_{3}=k\left(a_{1}+a_{2}\right)+1$ ). Qual è il più piccolo valore di $k$ tale che $a_{2016}$ sia multiplo di $3^{4031} \cdot 7^{4027}$ ?
(A) 1271
(B) 13... | [
"Solution:\n\nLa risposta è (B). Dimostriamo innanzitutto che $a_{n}=(k+1)^{n-1}$ : abbiamo $a_{1}=1=(k+1)^{1-1}$ e $a_{n+1}=k\\left(a_{n}+\\ldots+a_{1}\\right)+1=k a_{n}+k\\left(a_{n-1}+\\ldots+a_{1}\\right)+1=k a_{n}+a_{n}=(k+1) a_{n}$; per ricorrenza, si ottiene la formula voluta.\n\nOra consideriamo $a_{2016}=(... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization"
] | null | MCQ | B | |
0bc9 | Prove that, if $x, y > 0$, then
$$
0 < \frac{x}{x+y^4} + \frac{y}{y+x^4} - \frac{1}{x^2y^2+1} < 1.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0c1j | Let $f : [a, b] \to \mathbb{R}$ be an integrable function, and let $(a_n)_{n \ge 1}$ be a sequence of positive real numbers such that $\lim_{n \to \infty} a_n = 0$.
a) If $A = \{m \cdot a_n \mid m, n \in \mathbb{N}^*\}$, prove that any open interval of positive real numbers has elements from the set $A$.
b) If, for a... | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Calculus > Integral Calculus > Applications"
] | null | proof only | null | |
02hv | Problem:
Uma professora tem 237 balas para dar a seus 31 alunos. Qual é o número mínimo de balas a mais que ela precisa conseguir para que todos os alunos recebam a mesma quantidade de balas, sem sobrar nenhuma?
A) 11
B) 20
C) 21
D) 31
E) 41 | [
"Solution:\n\nO algoritmo de divisão de Euclides nos dá $237 = 7 \\times 31 + 20$; logo 237 não é divisível por 31. Isso quer dizer que a professora realmente vai ter que comprar mais balas para que todos os alunos recebam o mesmo número de balas. De acordo com o enunciado, devemos então adicionar à expressão $7 \\... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0cxa | Find all real numbers $x$ that can be written as
$$
x = \frac{a_{0}}{a_{1} a_{2} \ldots a_{n}} + \frac{a_{1}}{a_{2} a_{3} \ldots a_{n}} + \frac{a_{2}}{a_{3} a_{4} \ldots a_{n}} + \ldots + \frac{a_{n-2}}{a_{n-1} a_{n}} + \frac{a_{n-1}}{a_{n}},
$$
where $n, a_{1}, a_{2}, \ldots, a_{n}$ are positive integers and
$$
1 = a_... | [
"It is clear that $x \\in \\mathbb{Q}$ and $x > 0$.\n\nOn the other hand, if $x$ has such a representation, then from the inequalities\n$$\n\\begin{aligned}\n& \\frac{1}{a_{1} a_{2} \\ldots a_{n}} \\leq \\frac{1}{a_{2} a_{3} \\ldots a_{n}} \\\\\n& \\frac{a_{1}}{a_{2} a_{3} \\ldots a_{n}} \\leq \\frac{a_{2} - 1}{a_{... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | All rational numbers in the interval (0, 1], i.e., Q ∩ (0, 1]. | |
036h | Problem:
Given a triangle $ABC$ with $\angle ACB = 60^\circ$, define the sequence of points $A_0, A_1, \ldots, A_{2006}$ in the following way: $A_0 = A$, $A_1$ is the orthogonal projection of $A_0$ on $BC$, $A_2$ is the orthogonal projection of $A_1$ on $AC$, $\ldots$, $A_{2005}$ is the orthogonal projection of $A_{20... | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | proof only | null | |
0d3y | Given $x \geq 0$, prove that
$$
\frac{\left(x^{2}+1\right)^{6}}{2^{7}}+\frac{1}{2} \geq x^{5}-x^{3}+x
$$ | [
"Because $x^{2}+1 \\geq 2x$, we have\n$$\n\\frac{\\left(x^{2}+1\\right)^{7}}{2^{7}} \\geq \\frac{(2x)^{7}}{2^{7}} = x^{7}, \\quad \\text{and} \\quad \\frac{x^{2}+1}{2} \\geq x,\n$$\nand deduce\n$\\left(x^{2}+1\\right) \\cdot \\left(\\frac{\\left(x^{2}+1\\right)^{6}}{2^{7}}+\\frac{1}{2}\\right) = \\frac{\\left(x^{2}... | Saudi Arabia | SAMC | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English, Arabic | proof only | null | |
0jhv | Problem:
A single-elimination ping-pong tournament has $2^{2013}$ players, seeded in order of ability. If the player with seed $x$ plays the player with seed $y$, then it is possible for $x$ to win if and only if $x \leq y+3$. For how many players $P$ is it possible for $P$ to win? (In each round of a single eliminati... | [
"Solution:\n\n6038\n\nWe calculate the highest seed $n$ that can win. Below, we say that a player $x$ vicariously defeats a player $y$ if $x$ defeats $y$ directly or indirectly through some chain (i.e. $x$ defeats $x_{1}$, who defeated $x_{2}, \\ldots$, who defeated $x_{n}$, who defeated $y$ for some players $x_{1}... | United States | HMMT | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6038 | |
0gla | Let $M$ be a positive integer, and $A = \{1, 2, \dots, M+1\}$. Show that if $f$ is a bijection from $A$ to $A$ then
$$
\sum_{n=1}^{M} \frac{1}{f(n) + f(n+1)} > \frac{M}{M+3}.
$$ | [
"From the AM-HM inequality\n$$\n\\sum_{n=1}^{M} \\frac{1}{f(n) + f(n+1)} > \\frac{M^2}{\\sum_{n=1}^{M} (f(n) + f(n+1))} = \\frac{M^2}{2 \\sum_{n=1}^{M+1} f(n) - f(1) - f(M+1)}.\n$$\n\nSince $f$ is a bijection we have\n$$\n\\sum_{n=1}^{M+1} f(n) = \\sum_{n=1}^{M+1} n = \\frac{(M+1)(M+2)}{2}.\n$$\n\nNote that\n$$\nf(... | Thailand | The 13th Thailand Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
090q | For an integer $n \ge 3$, a sequence of integers $a_1, a_2, \dots, a_n$ is called a **beautiful sequence** if the following conditions are all satisfied.
* $0 = a_1 < a_2 < \dots < a_n$.
* There exists an integer $i$ such that $1 \le i \le n$ and $a_i = 2025$.
* For all integers $i, j, k$ with $1 \le i < j < k \le n$, ... | [
"Let $a_1, a_2, \\dots, a_n$ be a beautiful sequence of length $n$. Take an integer $m$ such that $a_m = 2025$. Then $m \\ge 2$ since $a_1 = 0$.\nWe first prove $n \\le 13$. Let $t$ be an integer with $1 \\le t \\le n - 2$. By applying the third condition to $(i, j, k) = (t, t+1, n)$, we obtain\n$$\n\\frac{a_t + a_... | Japan | The 35th Japanese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | N = 13; minimum a_N = 2057 | |
0d8l | Find all pairs of positive integers ($p$, $q$) such that both equations $x^{2} - p x + q = 0$ and $x^{2} - q x + p = 0$ have integral solutions. | [
"Since $x^{2} - p x + q = 0$ has integral solution, then we may assume that\n$$\n\\begin{cases}\np = a + b \\\\\nq = a b\n\\end{cases} \\text{ with } a, b \\in \\mathbb{Z}^{+} .\n$$\nNote that $q - p + 1 = a b - (a + b) + 1 = (a - 1)(b - 1) \\geq 0$ then $q + 1 \\geq p$.\nBy the same way on the second equation $x^{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (4, 4), (5, 6), (6, 5) | |
09vq | By replacing each * in the expression $1 * 2 * 3 * 4 * 5 * \cdots * 2019 * 2020$ by a $+$ or a $-$ sign, we get a long calculation. Put the $+$ and $-$ signs in such a way that the outcome is a positive number (greater than $0$) which is as small as possible.
What is this outcome? | [
"2"
] | Netherlands | First Round | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2 | |
0is0 | Problem:
Let $n>6$ be a positive integer. Determine the number of ways to walk from $(0,0)$ to $(n, 3)$ using only up and right unit steps such that the path does not meet the lines $y=x$ or $y=x-n+3$ except at the start and at the end. | [
"Solution:\nAnswer: $\\quad \\frac{1}{6}(n-6)(n-1)(n+1)$\n\nConsider the first point of the path that lies on $x=3$. There are two possibilities for this point: $(3,0)$ and $(3,1)$, and there is exactly one valid way of getting to each point from the origin. Similarly, consider the last point of the path that lies ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (n-6)(n-1)(n+1)/6 | |
0276 | Problem:
Sejam $a$ e $b$ números inteiros positivos tais que $a > b$. O professor Fernando disse ao aluno Raul que se ele calculasse o número $A = a^{2} + 4b + 1$, o resultado seria um quadrado perfeito. Raul, por engano, trocou os números $a$ e $b$ e calculou o número $B = b^{2} + 4a + 1$ que, por acaso, também é um ... | [
"Solution:\n\na) Note que\n$$\na^{2} < A = a^{2} + 4b + 1 < a^{2} + 4a + 1 < (a+2)^{2}\n$$\nAssim, $A$ é um quadrado perfeito entre os quadrados perfeitos $a^{2}$ e $(a+2)^{2}$, e então $A = (a+1)^{2}$.\n\nb) Pelo item anterior,\n$$\na^{2} + 4b + 1 = A = (a+1)^{2} = a^{2} + 2a + 1\n$$\ne então $a = 2b$. Substituind... | Brazil | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | A = (a+1)^2; a = 8, b = 4, A = 81, B = 49 | |
0fsx | Problem:
Jede Ecke eines regelmässigen $2n$-Ecks $(n \geq 3)$ soll mit einer Zahl aus der Menge $\{1,2, \ldots, 2n\}$ beschriftet werden, und zwar so, dass die Summe der Zahlen benachbarter Ecken stets gleich ist wie die Summe der Zahlen in den beiden diametral gegenüberliegenden Ecken. Zudem müssen die in den $2n$ Ec... | [
"Solution:\n\nDie Zahlen in den Ecken seien $a_{1}, \\ldots, a_{2n}$ in dieser Reihenfolge, die Indizes werden wir modulo $2n$ betrachten. Wir können oBdA annehmen, dass $a_{1} > a_{n+1}$ ist. Da $a_{1} + a_{2} = a_{n+1} + a_{n+2}$ gilt, muss $a_{2} < a_{n+2}$ sein. Mit der gleichen Argumentation folgt induktiv $a_... | Switzerland | IMO - Selektion | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n is odd | |
0i4k | Problem:
Divide an $m$-by-$n$ rectangle into $mn$ nonoverlapping $1$-by-$1$ squares. A polyomino of this rectangle is a subset of these unit squares such that for any two unit squares $S, T$ in the polyomino, either
(1) $S$ and $T$ share an edge or
(2) there exists a positive integer $n$ such that the polyomino cont... | [
"Solution:\n\nTo span an $a \\times b$ rectangle, we need at least $a + b - 1$ squares. Indeed, consider a square of the polyomino bordering the left edge of the rectangle and one bordering the right edge. There exists a path connecting these squares; suppose it runs through $c$ different rows. Then the path requir... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 470 | |
05zc | Problem:
Soit $ABC$ un triangle, et soit $E$ le milieu de $[BC]$. Soit $m$ la médiatrice de $[AC]$, et soit $D$ un point de $m$ tel que le cercle circonscrit à $ABD$ soit tangent à $m$. Enfin, soit $K$ le point d'intersection, autre que $A$, entre le cercle circonscrit à $ABD$ et la droite $(AC)$, et soit $F$ le milie... | [
"Solution:\n\n\n\nUne première difficulté est de tracer la figure : comment tracer un point tangent à une droite et passant par deux points donnés? Une stratégie usuelle, en pareil cas, est d'inverser le tracé : on commence par placer les points $A, B$ et $D$, puis on en déduit la tangente ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05s9 | Problem:
Soit $ABC$ un triangle, soit $I$ le centre du cercle inscrit dans $ABC$. Soit $A'$, $B'$ et $C'$ trois points respectivement situés sur $(AI)$, $(BI)$ et $(CI)$. On suppose que $A'$, $B'$ et $C'$ sont distincts de $A$, $B$, $C$ et $I$, et qu'ils sont alignés. Enfin, soit $P_A$ le point d'intersection des médi... | [
"Solution:\n\nOn commence par tracer une figure comme ci-dessous, sans oublier la droite $\\ell$ qui passe par $A'$, $B'$ et $C'$ ni les cercles $\\Omega$ et $\\Gamma$, respectivement circonscrits à $ABC$ et à $P_A P_B P_C$.\n\n\n\nAu vu de l'énoncé, une approche raisonnable est de trouver ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate... | null | proof only | null | |
02ns | Problem:
Uma companhia de eletricidade instalou um poste num terreno plano. Para fixar bem o poste, foram presos cabos no poste, a uma altura de $1{,}4$ metros do solo e a $2$ metros de distância do poste, sendo que um dos cabos mede $2{,}5$ metros, conforme a figura.
Um professor de Matemática, após analisar estas... | [] | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | Yes. The professor is right; the pole is not perpendicular to the ground. | |
05oe | Problem:
Soit $ABC$ un triangle. $H$ son orthocentre et $P$, $Q$ et $R$ les pieds des hauteurs issues de $A$, $B$ et $C$.
Montrer que $H$ est le centre du cercle inscrit à $PQR$.
 | [
"Solution:\n\nOn sait que les points $A$, $B$, $P$ et $Q$ sont cocycliques sur le cercle de diamètre $[AB]$, donc :\n$$\n\\widehat{HPQ} = \\widehat{APQ} = \\widehat{ABQ} = 90 - \\widehat{BAQ} = 90 - \\widehat{BAC}\n$$\nDe même, $A$, $C$, $P$ et $R$ sont cocycliques sur le cercle de diamètre $[AC]$ donc :\n$$\n\\wid... | France | Olympiades Françaises de Mathématiques | [
"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"
] | null | proof only | null | |
0k9b | Problem:
Scalene triangle $ABC$ satisfies $\angle A = 60^{\circ}$. Let the circumcenter of $ABC$ be $O$, the orthocenter be $H$, and the incenter be $I$. Let $D, T$ be the points where line $BC$ intersects the internal and external angle bisectors of $\angle A$, respectively. Choose point $X$ on the circumcircle of $\... | [
"Solution:\n\nLet $I_A$ denote the $A$-excenter. Because $\\angle A = 60^{\\circ}$, $AI$ is the perpendicular bisector of $OH$ and $B, H, O, C$ all lie on the circle with diameter $II_A$. We are given that $X$ is on this circle as well, and since $HI = OI$, $XIOI_A$ is also an isosceles trapezoid. But $II_A$ is a d... | United States | HMMT February 2019 Team Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
0arp | Problem:
In $\triangle A B C$, let $X$ and $Y$ be the midpoints of $A B$ and $A C$, respectively. On segment $B C$, there is a point $D$, different from its midpoint, such that $\angle X D Y = \angle B A C$. Prove that $A D$ is perpendicular to $B C$. | [
"Solution:\n\nLet $Z$ be the midpoint of $B C$. Without loss of generality, we assume that $D$ is between $B$ and $Z$. Then $X Y \\parallel B C$, $X Z \\parallel A C$, and $Y Z \\parallel A B$, and so $\\angle A B C = \\angle X Y Z$ and $\\angle X D Y = \\angle B A C = \\angle X Z Y$. It follows that quadrilateral ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0248 | Problem:
Prove que se um polígono possui uma circunferência tangente a todos os seus lados, então é possível encontrar três lados do polígono que formam um triângulo. | [
"Solution:\n\nSeja $DE$ o maior lado do polígono e considere os outros dois lados adjacentes a ele, denotados por $CD$ e $EF$ na figura a seguir.\n\nAlém disso, sejam $K$, em $CD$, e $I$, em $DE$, e $J$, em $EF$ os pontos de tangências desses lados com a circunferência mencionada no enuncia... | Brazil | null | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
01ns | Find all triples $(a; b; c)$ of real numbers for which there exists a non-zero function $f$, $f: \mathbb{R} \to \mathbb{R}$, such that
$$
af(xy + f(z)) + bf(yz + f(x)) + cf(zx + f(y)) = 0
$$
for all real $x, y, z$. | [
"First, note that if $a + b + c = 0$, then the function $f \\equiv \\lambda \\neq 0$ satisfies the condition.\n\nNow suppose that $a + b + c \\neq 0$. In particular, at least one of $a, b, c$ is not zero. Say, $a \\neq 0$. Let $(*)$ denote the given functional equation.\n\nSet $x = y = 0$ in $(*)$, then $af(f(z)) =... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All real triples with a + b + c = 0 | |
0fci | Problem:
Dado un triángulo de vértices $A$, $B$ y $C$, y con lados de longitud $a = BC$, $b = AC$ y $c = AB$, llamemos $D$ al punto de intersección del lado $AB$ con la bisectriz del ángulo $C$.
Demuestra que:
$$
CD = \frac{2ab \cos \frac{C}{2}}{a + b}
$$ | [
"Solution:\n\n- A partir del vértice $B$ trazamos una paralela a la bisectriz $CD$ y prolongamos el lado $AC$ hasta obtener el punto $E$.\nY, también, $CF$ perpendicular a $BE$.\n- Así, $CB = CE = a$\n\n\n\n- Por ángulos alternos-internos, en el triángulo $BCF$ tenemos: $\\cos \\frac{C}{2} ... | Spain | Fase Local | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ca2 | Problem:
Fie $p, q$ numere naturale nenule. Pentru fiecare $a, b \in \mathbb{R}$ definim mulțimile
$$
P(a)=\left\{\left.a_{n}=a+n \cdot \frac{1}{p} \right| n \in \mathbb{N}\right\}, \quad Q(b)=\left\{\left.b_{n}=b+n \cdot \frac{1}{q} \right| n \in \mathbb{N}\right\}
$$
Numim distanța de la $P(a)$ la $Q(b)$ valoarea mi... | [] | Romania | Al treilea test de selecţie pentru OBMJ Bucureşti | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 1/(2·lcm(p,q)) | |
0h84 | Numbers $a$, $b$, $c$ satisfy the conditions:
$$
a^2 + 2 = b^4, \quad b^2 + 2 = c^4, \quad c^2 + 2 = a^4.
$$
What values can take expression
$(a^2 - 1)(b^2 - 1)(c^2 - 1)$? | [
"Let us subtract $1$ from the left and right parts of equality and obtain:\n$$\na^2 + 1 = b^4 - 1 = (b^2 - 1)(b^2 + 1),\n$$\n\nand analogously from other $2$ equalities. Next multiply the resulting equalities:\n\n$$\n(a^2 + 1)(b^2 + 1)(c^2 + 1) = (a^2 - 1)(a^2 + 1)(b^2 - 1)(b^2 + 1)(c^2 - 1)(c^2 + 1)\n$$\n$$\n(a^2 ... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 1 | |
0a8z | Problem:
Let $B$ and $C$ be points on two fixed rays emanating from a point $A$ such that $A B + A C$ is constant. Prove that there exists a point $D \neq A$ such that the circumcircles of the triangles $A B C$ pass through $D$ for every choice of $B$ and $C$. | [
"Solution:\n\n(See Figure 16.) Let $E$ and $F$ be the points on rays $A B$ and $A C$, respectively, such that $A E = A F = A B + A C$. Let the perpendicular bisectors of the segments $A E$ and $A F$ intersect at $D$. It is easy to see, for instance from the right triangles with $A D$ as the common hypothenuse and t... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 20 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
073u | Let $r$, $t$ and $a$ be three positive integers such that $r \le t$ and let
$$
A = \{1, 2, 3, \dots, r\}, \quad B = \{a+r+1, a+r+2, a+r+3, \dots, a+r+t\}.
$$
For any set $P$ of integers and any integer $x$, we define $P + x = \{p + x : p \in P\}$, called the translate of $P$ by $x$.
a. If $r = t$, prove that $Z$ (the ... | [
"First observe that $A$ and $B$ are separated by $a$ numbers and as such $A \\cup B$ is not a set of consecutive integers.\n\na. ($\\Rightarrow$) If $r = t$ and $a = kr$ for some $k \\in \\mathbb{N}$, then it is clear that $Z$ is covered by the mutually disjoint sets $(A \\cup B) + s$, where $s$ ranges over the set... | India | Indija TS 2008 | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
0g7a | 設點 $H$ 為 $\triangle ABC$ 的垂心，且點 $P$ 是 $\triangle ABC$ 外接圓上異於頂點 $A$, $B$, $C$ 的任意點。令點 $E$, $F$ 分別是 $P$ 對 $BC$, $AB$ 邊的垂足。試證：直線 $EF$ 平分線段 $PH$。 | [] | Taiwan | 二0一三數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > ... | null | proof only | null | |
0j4l | Problem:
Let $ABC$ be a non-isosceles, non-right triangle, let $\omega$ be its circumcircle, and let $O$ be its circumcenter. Let $M$ be the midpoint of segment $BC$. Let the circumcircle of triangle $AOM$ intersect $\omega$ again at $D$. If $H$ is the orthocenter of triangle $ABC$, prove that $\angle DAH = \angle MAO$... | [
"Solution:\n\nLet $X$ be the intersection of the line tangent to $\\omega$ at $B$ with the line tangent to $\\omega$ at $C$. Note that $\\triangle OMC \\sim \\triangle OCX$ since $\\angle OMC = \\angle OCX = \\frac{\\pi}{2}$. Hence $\\frac{OM}{OC} = \\frac{OC}{OX}$, or, equivalently, $\\fra... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis ... | null | proof only | null | |
0gl2 | Find all polynomials $P(x)$ with integral coefficients such that
$$
P(n) \mid 2557^n + (213 \times 2014)
$$
for all positive integers $n$. | [
"First, we note that the constant polynomials $P(x) \\equiv 1$ and $P(x) \\equiv -1$ satisfy the above divisibility condition. We show that these two polynomials are the only ones satisfying the divisibility condition.\n\nSuppose that the polynomial $P(x)$ with the integral coefficients satisfying the above divisib... | Thailand | Tajland 2014 | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | P(x) ≡ 1 or P(x) ≡ −1 | |
0hut | Problem:
Let $ABCD$ be a trapezoid with $AB \parallel CD$. Let $M$ and $N$ be the respective midpoints of $AB$ and $CD$. Diagonals $AC$ and $BD$ meet at $P$, and lines $AD$ and $BC$ meet at $Q$. Prove that $M$, $N$, $P$, and $Q$ are collinear. | [
"Solution:\n\nNote that $\\triangle QAB \\sim \\triangle QDC$ (by corresponding angles). Since $QM$ and $QN$ are corresponding medians, we have $\\angle AQ M = \\angle DQ N$ and hence $Q$, $M$, and $N$ are collinear.\n\n\n\nSimilarly, note that $\\triangle PAB \\sim \\triangle PCD$. Since $... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k2b | Problem:
How many graphs are there on $10$ vertices labeled $1,2, \ldots, 10$ such that there are exactly $23$ edges and no triangles? | [
"Solution:\n\nNote that the sum of the degrees of the graph is $23 \\cdot 2 = 46$, so at least one vertex has degree $5$ or more. We casework on the maximal degree $n$.\n\nCase 1: $n \\geq 7$, then none of the $n$ neighbors can have an edge between each other, for $\\binom{n}{2}$ edges unusable, and the vertex with... | United States | HMMT February 2018 | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 42840 | |
0k91 | Problem:
Isosceles triangle $A B C$ with $A B = A C$ is inscribed in a unit circle $\Omega$ with center $O$. Point $D$ is the reflection of $C$ across $A B$. Given that $D O = \sqrt{3}$, find the area of triangle $A B C$. | [
"Solution:\n\nObserve that\n$$\n\\angle D B O = \\angle D B A + \\angle A B O = \\angle C B A + \\angle B A O = \\frac{1}{2}(\\angle C B A + \\angle B C A) + \\frac{1}{2}(\\angle B A C) = \\frac{1}{2}\\left(180^{\\circ}\\right) = 90^{\\circ}.\n$$\nThus $B C = B D = \\sqrt{2}$ by the Pythagorean Theorem on $\\triang... | United States | HMMT February 2019 February 16, 2019 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (sqrt(2) ± 1)/2 | |
0bll | Let $ABC$ be an acute triangle such that $AB \neq AC$. Let $M$ be the midpoint of $[BC]$, $H$ be the orthocenter of $ABC$, $O_1$ be the midpoint of $[AH]$ and $O_2$ be the circumcenter of $BCH$. Prove that $O_1AMO_2$ is a parallelogram.
JBMO ShortList 2014 | [
"We use the following well-known facts:\n(1) The reflection of $H$ in line $BC$ lies on the circumcircle of triangle $ABC$.\n(2) $AH = 2MO$, where $O$ is the circumcenter of $ABC$.\nFrom (1) it follows that the reflection of the circumcenter of $ABC$ is the circumcenter of $HBC$, hence $O_2$ is the reflection of $O... | Romania | 66th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gu8 | Let $\mathcal{S}$ be a family of $n$ sets, each has size $k$, such that $A \Delta B \in \mathcal{S}$ whenever $A, B \in \mathcal{S}, A \neq B$.
a) Is it possible that $n = 2023$? If it is, find all possible values of $k$.
b) Is it possible that $n = 1023$? If it is, find all possible values of $k$.
Notation: $A \Del... | [
"**a.** We show that $n = 2023$ is impossible. Let $\\mathcal{S} = \\{S_1, S_2, \\ldots, S_{2023}\\}$ and $T = \\{1, 2, \\ldots, 2023\\}$. We call a non-empty set $I \\subseteq T$ as *nice* if every non-empty subset $J \\subseteq I$ satisfies $\\Delta_{j \\in J} S_j \\neq \\emptyset$. Note that any set of size one ... | Turkey | Team Selection Test for IMO 2023 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) Not possible. b) Possible, and the permissible values of k are exactly the positive integers divisible by 512. | |
02ny | Define the polynomials $\binom{x}{j} = \frac{x(x-1)(x-2)\cdots(x-j+1)}{j!}$ for $j$ positive integer and $\binom{x}{0} = 1$.
a. Prove that all non-nil polynomials can be written uniquely as a linear combination of such polynomials $\binom{x}{n}$.
b. Let $c(n,k)$ be the coefficient of $\binom{x}{k}$ in $x^n$ (as descr... | [
"a. Induct on the degree $n$ of the polynomial $P(x) = \\sum_{k=0}^{n} a_k x^k$. It is immediate if $n=0$. Since the degree of $(x^k)$, $k < n$, is $k$, the coefficient on $(x^n)$ on $P(x)$ is $a_n \\cdot n!$. Now consider the polynomial $P(x) - a_n \\cdot n! (x^n)$. This is a polynomial with degree less than $n$, ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | 1/(k+1) | |
048v | Let $\overline{AB}$ be a chord of a circle, parallel to its diameter $\overline{MN}$. Let $t$ be the tangent of that circle in $M$ and let $C, D$ be the intersections of the lines $NA, NB$ with $t$, respectively. Prove
$$
|MC| \cdot |MD| = |MN|^2.
$$ | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0bgl | Problem:
Fie $n$ un număr natural nenul şi $a_{1}, a_{2}, \ldots, a_{n}$ numere reale astfel încât $a_{1}+a_{2}+\ldots+a_{k} \leq k$, oricare ar fi $k \in \{1,2, \ldots, n\}$. Arătaţi că
$$
\frac{a_{1}}{1}+\frac{a_{2}}{2}+\ldots+\frac{a_{n}}{n} \leq \frac{1}{1}+\frac{1}{2}+\ldots+\frac{1}{n}
$$ | [] | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jwk | Problem:
Yannick has a bicycle lock with a 4-digit passcode whose digits are between $0$ and $9$ inclusive. (Leading zeroes are allowed.) The dials on the lock are currently set at $0000$. To unlock the lock, every second he picks a contiguous set of dials, and increases or decreases all of them by one, until the dial... | [
"Solution:\n\nTo simplify the solution, we instead consider the equivalent problem of reducing a passcode to $0000$ using the given move.\n\nGiven a passcode $a_{1} a_{2} a_{3} a_{4}$, define a differential of the passcode to be a quintuple $(d_{1}, d_{2}, d_{3}, d_{4}, d_{5})$ such that $d_{i} \\equiv a_{i}-a_{i-1... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | (12, 2) | |
0cv4 | $n > 3$ distinct positive integers are written on a board, all these numbers are smaller than $(n-1)! = 1 \cdot 2 \cdot \dots \cdot (n-1)$. For each pair $a > b$ of these numbers, Sergey found a partial quotient of $a$ by $b$ and wrote up the result into his notebook (e.g., if $a = 100$ and $b = 7$, then $100 = 14 \cdo... | [
"Enumerate the numbers in increasing order: $a_1 < a_2 < \\dots < a_n$.\n\nSuppose, for contradiction, that all the partial quotients $q_{i,j}$ (for $a_i > a_j$) are distinct.\n\nConsider the sequence of ratios:\n$$\n\\frac{a_n}{a_{n-1}} \\cdot \\frac{a_{n-1}}{a_{n-2}} \\cdots \\frac{a_2}{a_1} = \\frac{a_n}{a_1}.\n... | Russia | XLIII Russian mathematical olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English; Russian | proof only | null | |
069s | Let $\alpha$ and $\beta$ be two positive integers. Prove that the integer
$$
\alpha^2 + \left\lfloor \frac{4\alpha^2}{\beta} \right\rfloor
$$
is not a square of an integer. | [
"We suppose that:\n$$\n\\alpha^2 + \\left\\lfloor \\frac{4\\alpha^2}{\\beta} \\right\\rfloor = (\\alpha + \\kappa)^2, \\kappa \\in \\mathbb{N}^* \\Leftrightarrow \\left\\lfloor \\frac{4\\alpha^2}{\\beta} \\right\\rfloor = (2\\alpha + \\kappa)\\kappa, \\kappa \\in \\mathbb{N}^*.\n$$\nBy putting $2\\alpha = x$, we ha... | Greece | SELECTION EXAMINATION 2019 | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
03iv | Problem:
Find all positive integers $w, x, y$ and $z$ which satisfy $w! = x! + y! + z!$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (w, x, y, z) = (3, 2, 2, 2) | |
02vw | Problem:
Existem $2m$ miçangas de $m$ cores distintas, sendo duas de cada cor. Essas miçangas são distribuídas em $m$ caixas, com duas em cada caixa, de modo que é possível escolher uma miçanga em cada uma delas e obter um conjunto de $m$ miçangas de cores distintas. Prove que o número de maneiras de fazermos esse tip... | [
"Solution:\n\nAntes de descrevermos a solução para o caso geral, considere o caso particular com $m=5$, em que as cores das miçangas serão denotadas por $A, B, C, D$ e $E$, com as seguintes distribuições de caixas:\n$$\n\\begin{array}{ccccc}\nC_{1} & C_{2} & C_{3} & C_{4} & C_{5} \\\\\n\\hline A, B & B, C & C, A & ... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
06iv | Find the maximum number of sets which simultaneously satisfy the following 3 conditions:
(i) any of the sets consists of 4 elements;
(ii) any two different sets have exactly 2 common elements;
(iii) no two elements are common to all the sets. | [
"The maximum number of sets is $7$.\nWe first assume there are at least $8$ sets. WLOG assume $1$ and $2$ are common elements of at least two sets.\n\n* If there are $4$ sets containing both $1$ and $2$, say $\\{1, 2, 3, 4\\}$, $\\{1, 2, 5, 6\\}$, $\\{1, 2, 7, 8\\}$, $\\{1, 2, 9, 10\\}$, then all other sets must co... | Hong Kong | 1997-2023 IMO HK TST | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 7 | |
0fau | Problem:
Find all integers $k > 1$ such that for some distinct positive integers $a$, $b$, the number $k^{a} + 1$ can be obtained from $k^{b} + 1$ by reversing the order of its (decimal) digits. | [
"Solution:\n$k = 3$, $3^{3} + 1 = 28$, $3^{4} + 1 = 82$.\n\n$k = 10$ does not work because $k^{a} + 1$ is a palindrome. If $k > 10$, then for $a < b$ we have $k^{b} + 1 \\geq k^{a + 1} + 1 \\geq 11k^{a} + 1 \\geq 10(k^{a} + 1) + k^{a} - 9 > 10(k^{a} + 1)$. So $k^{b} + 1$ has more digits than $k^{a} + 1$. So we only... | Soviet Union | 1st CIS | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | k = 3 | |
0bfs | Given two distinct square-free positive integers $a$ and $b$, show that
$$
|\{n\sqrt{a}\} - \{n\sqrt{b}\}| > c n^{-3},
$$
for some positive constant $c$ and all positive integers $n$. | [
"Assume, without any loss, that $a > b$, consider a positive integer $n$, and let $k_n = \\lfloor n\\sqrt{a} \\rfloor - \\lfloor n\\sqrt{b} \\rfloor$. Since $n(\\sqrt{a} - \\sqrt{b})$ is irrational, it follows that $\\{n\\sqrt{a}\\} \\neq \\{n\\sqrt{b}\\}$, so\n$$\n\\begin{aligned}\n0 < |\\{n\\sqrt{a}\\} - \\{n\\sq... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
06gt | $E$ and $F$ are points on side $AB$ of $\triangle ABC$ such that $AE = EF = FB$. $D$ is a point on the line $BC$ such that $BC$ is perpendicular to $ED$, $AD$ is perpendicular to $CF$. Suppose $\angle CFA = 3\angle BDF$. Determine the value of $\frac{DB}{DC}$. | [
"We have $\\frac{DB}{DC} = \\frac{7}{2}$.\nFirstly, since $\\angle EDB = 90^\\circ$ and $BF = EF$, we have $DF = BF$. This implies $\\angle FDB = \\angle FBD = \\theta$, and hence $\\angle DFE = 2\\theta$. It follows that $C$ lies on the segment $BD$, and\n$$\n\\angle CDF = 3\\theta - 2\\theta = \\theta.\n$$\nLet $... | Hong Kong | Year 2012 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 7/2 | |
09vp | For a positive integer $n$, let $d(n)$ be the number of positive divisors of $n$. Determine the positive integers $k$ for which there exist positive integers $a$ and $b$ satisfying
$$
k = d(a) = d(b) = d(2a + 3b).
$$ | [
"For $i \\ge 0$, let $a = 2 \\cdot 5^i$ and $b = 3 \\cdot 5^i$. Then both $a$ and $b$ have $2(i+1)$ divisors. Moreover, we have $2a+3b = 4 \\cdot 5^i + 9 \\cdot 5^i = 13 \\cdot 5^i$, which also has $2(i+1)$ divisors. Therefore all even values of $k$ satisfy the condition in the problem.\n\nNow suppose that $k$ is o... | Netherlands | IMO Team Selection Test 3, June 2020 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All even positive integers | |
05uu | Problem:
Soit $a, b, c \geqslant 1$ des entiers tels que $a^{b} \mid b^{c}$ et $a^{c} \mid c^{b}$. Montrer que $a^{2} \mid b c$. | [
"Solution:\n\nNous allons regarder les valuations $p$-adiques de chacun des nombres premiers $a$, $b$ et $c$. En effet, il suffit de montrer que pour tout nombre premier $p$, on a $2 v_{p}(a) = v_{p}\\left(a^{2}\\right) \\leqslant v_{p}(b c) = v_{p}(b) + v_{p}(c)$.\n\nSoit $p$ un nombre premier. L'hypothèse que $a^... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0hwu | Problem:
Let $ABC$ be a triangle. The incircle, centered at $I$, touches side $BC$ at $D$. Let $M$ be the midpoint of the altitude from $A$ to $BC$. Lines $MI$ and $BC$ meet at $E$. Prove that $BD = CE$. | [
"Solution:\n\nLet $E'$ be the point on side $BC$ such that $BD = E'C$. Then it is well known that $E'$ is the tangency point of the $A$-excircle $\\Omega$ of $\\triangle ABC$ with $BC$. Let $H$ be the foot of the altitude from $A$ to $BC$, and let $K$ be the point on the incircle $\\omega$ opposite $D$.\n\nCircles ... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0bsg | Let $n \in \mathbb{N}$, $n \geq 2$, and $a_1, a_2, ..., a_n \in (0, \infty)$ such that $a_1 \cdot a_2 \cdot ... \cdot a_n = 1$. Prove that the function
$$
f : (0, \infty) \to \mathbb{R}, f(x) = (1 + a_1^x) \cdot (1 + a_2^x) \cdot \dots \cdot (1 + a_n^x)
$$
is increasing. | [
"Multiplying out yields\n$$\n\\begin{aligned}\nf(x) &= (1 + a_1^x) \\cdot (1 + a_2^x) \\cdot \\dots \\cdot (1 + a_n^x) \\\\\n&= 1 + \\sum_{1 \\leq i \\leq n} a_i^x + \\sum_{1 \\leq i < j \\leq n} a_i^x a_j^x + \\dots + a_1^x a_2^x \\dots a_n^x.\n\\end{aligned}\n$$\n\nSince $a_1 \\cdot a_2 \\cdot \\dots \\cdot a_n =... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof only | null | |
0hs7 | Problem:
Prove that the product of 6 consecutive positive integers is never equal to $n^{5}$ for some positive integer $n$. | [
"Solution:\n\nIt is easy to show that at least one of 6 consecutive numbers is relatively prime to each of the others (3 of these numbers are odd, exactly one of these odd numbers is divisible by 3, and for each prime $p > 3$ at most one is divisible by $p$). Assume that $x-2, x-1, x, x+1, x+2, x+3$, $x \\geq 3$ ar... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
01bc | A rectangle of size $m \times n$ consisting of $m \cdot n$ squares of size $1 \times 1$ is given. Compute the sum of areas of all subrectangles consisting of some of the $m \cdot n$ squares. | [
"Answer:\n$$\n\\frac{m(m+1)(m+2)n(n+1)(n+2)}{36}\n$$\n\nFor each square $1 \\times 1$ we will count by how many rectangles it is consisted. Take a square from $i$-th column and $j$-th row. It is consisted by $i(m - i + 1)j(n - j + 1)$ rectangles, as we may choose one of $i$ lines on the left side of the square to b... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | m(m+1)(m+2)n(n+1)(n+2)/36 | |
0797 | $f: \mathbb{N} \to \mathbb{N}$ is an increasing function and $n$ is a natural number. Suppose that there exist prime numbers $p_1, p_2, \ldots, p_n$ and natural numbers $s_1, s_2, \ldots, s_n$ such that for every $1 \le i \le n$ the set $\{f(p_i r + s_i) \mid r = 1, 2, \ldots \}$ is an infinite arithmetic progression. ... | [
"Suppose that $\\{f(p_i r + s_i) \\mid r \\in \\mathbb{N}\\}$ is an arithmetic progression with common difference $d_i$. Since $f$ is increasing, $d_i$ must be positive. (Even if for one $i$, we have $d_i = 0$, the problem is easy.) So we know that $f(p_i r_1 + s_i) - f(p_i r_2 + s_i) = d_i(r_1 - r_2)$.\n\nNow by u... | Iran | 27th Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0ekr | Problem:
Poišči vse pare naravnih števil $a$ in $b$, za katere velja
$$
v = a b - 2 a - 4 b
$$
kjer je $v$ najmanjši skupni večkratnik števil $a$ in $b$. | [
"Solution:\n\nNaj bo $d$ največji skupni delitelj števil $a$ in $b$ in naj bosta $m$ in $n$ taki naravni števili, da je $a = d m$ in $b = d n$. Tedaj sta števili $m$ in $n$ tuji in velja $v = d m n$. Z upoštevanjem teh zvez iz dane enačbe sledi\n$$\nd m n = d^2 m n - 2 d m - 4 d n,\n$$\nod koder po krajšanju z $d$ ... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof and answer | (7, 7), (6, 12), (10, 5), (16, 4) | |
08rh | Find all real numbers $x$, $y$ and $z$ such that all the following equations hold.
$$
x^2 - 3y - z = -8
$$
$$
y^2 - 5z - x = -12
$$
$$
z^2 - x - y = 6
$$ | [
"By adding the three equations and simplifying, we get\n$$\n(x - 1)^2 + (y - 2)^2 + (z - 3)^2 = 0.\n$$\nSo it is necessary that $(x, y, z) = (1, 2, 3)$. And this satisfies the given equations."
] | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (1, 2, 3) | |
0fpr | De entre todas las permutaciones $(a_1, a_2, \dots, a_n)$ del conjunto $\{1, 2, \dots, n\}$, $(n \ge 1$ entero), se consideran las que cumplen que $2(a_1 + a_2 + \dots + a_m)$ es divisible por $m$, para cada $m = 1, 2, \dots, n$. Calcular el número total de estas permutaciones. | [
"Sea $\\mathcal{P}_n$ el conjunto de permutaciones de $\\{1, 2, \\dots, n\\}$ que cumplen las condiciones del enunciado. El problema consiste en calcular $|\\mathcal{P}_n|$. Observemos que, para cualquier $n$, las condiciones se cumplen siempre para $m=1$, para $m=2$ y para $m=n$, de manera que $\\mathcal{P}_1, \\m... | Spain | LII Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Number Theory > Divisibility / Factorization"
] | Spanish | proof and answer | 3 * 2^{n-2} | |
03ky | Problem:
Let $m$ be a positive integer. Define the sequence $a_{0}, a_{1}, a_{2}, \ldots$ by $a_{0}=0$, $a_{1}=m$, and $a_{n+1}=m^{2} a_{n}-a_{n-1}$ for $n=1,2,3, \ldots$. Prove that an ordered pair $(a, b)$ of non-negative integers, with $a \leq b$, gives a solution to the equation
$$
\frac{a^{2}+b^{2}}{a b+1}=m^{2}
... | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0en1 | Let $A$, $B$, $C$ be three distinct points in the plane for which $AB = AC$. Describe the locus of the points $P$ for which $\angle APB = \angle APC$. | [
"Clearly every point on the perpendicular bisector of $BC$ satisfies the requirements of the problem. We now show that no other point $P$ does.\nLet $D$ be the midpoint of $BC$, and suppose that there exists a point $P$ not on the line $AD$ such that $\\angle APB = \\angle APC$. Let $Q$ be the reflection of $P$ abo... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof and answer | The perpendicular bisector of BC. | |
0by3 | a) Show that there exist functions $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ such that $f \circ g = g \circ f$, $f \circ f = g \circ g$ and $f(x) \ne g(x)$, for all $x \in \mathbb{R}$.
b) Prove that if $f : \mathbb{R} \to \mathbb{R}$ and $g : \mathbb{R} \to \mathbb{R}$ are continuous and have... | [
"a) The functions defined by $f(x) = \\begin{cases} -1, & x < 0 \\\\ 1, & x \\ge 0 \\end{cases}$, $g(x) = \\begin{cases} 1, & x < 0 \\\\ -1, & x \\ge 0 \\end{cases}$ fulfill the given conditions.\n\nb) The function $h = f - g$ is continuous and doesn't vanish, so $h(x) > 0$, for all $x \\in \\mathbb{R}$, or $h(x) <... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof only | null | |
03ee | In the house of the wealthy Lady Gilmore, one of her most precious possessions has been stolen: her pearl necklace. The task of unraveling the mystery fell on the shoulders of Inspector Goodenough. He had the following information: on the day of the theft, $7$ of Lady Gilmore's servants, whom we will refer to as $A$, $... | [
"We will first prove the following lemma.\n\n**Lemma.** Let $X$, $Y$, $Z$ and $T$ be four of the servants. If it is known that the pairs $X, Y$; $Y, Z$; $Z, T$ and $T, X$ were in the room together at some point, then one of the pairs $X, Z$ and $Y, T$ also detected each other.\n\n**Proof of Lemma.** Let us assume, ... | Bulgaria | Autumn tournament | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Other"
] | English | proof and answer | A | |
0hna | Problem:
Show that for each $n \geq 17$ one can cut a square into $n$ smaller squares. | [
"Solution:\nOne can cut a square into $n$ smaller squares for all $n \\geq 6$. Here is how.\nFor $n=4$\n\n\nFor $n=6$\n\n\nFor $n=8$\n\n\nSuppose we know how to cut a square into $n$ smaller squares. If we take one of the square \"pieces\" a... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fea | Problem:
Un triángulo tiene sus vértices en cada uno de los ejes de un sistema de coordenadas cartesianas en el espacio; ninguno está en el origen, ni dos de ellos coinciden el mismo eje. Demostrad que el triángulo es acutángulo. | [
"Solution:\n\nSean $A$, $B$, $C$ los vértices del triángulo. Denotamos, respectivamente, $x$, $y$, $z$ las distancias de los vértices al origen de coordenadas; y, también respectivamente, $a$, $b$, $c$ las longitudes de los lados opuestos a los vértices.\n\nBasta probar que uno de los ángulos es agudo. Probaremos q... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0j5g | Problem:
Let $R$ be a semicircle with diameter $X Y$. A trapezoid $A B C D$ in which $A B$ is parallel to $C D$ is circumscribed about $R$ such that $A B$ contains $X Y$. If $A D=4$, $C D=5$, and $B C=6$, determine $A B$. | [
"Solution:\nWe claim that $A B = A D + B C$.\nLet $O$ denote the center of $R$. Since $D A$ and $D C$ are both tangent to $R$, we have $\\angle A D O = \\angle O D C$. Since $C D$ is parallel to $A B$, we also have $\\angle O D C = \\angle D O A$. Thus $\\angle A D O = \\angle D O A$, and it follows that $A D = A O... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10 | |
0jhr | Problem:
In equilateral triangle $ABC$, a circle $\omega$ is drawn such that it is tangent to all three sides of the triangle. A line is drawn from $A$ to point $D$ on segment $BC$ such that $AD$ intersects $\omega$ at points $E$ and $F$. If $EF = 4$ and $AB = 8$, determine $|AE - FD|$. | [
"Solution:\n\nAnswer: $\\frac{4}{\\sqrt{5}}$ OR $\\frac{4 \\sqrt{5}}{5}$ Without loss of generality, $A, E, F, D$ lie in that order. Let $x = AE$, $y = DF$.\n\nBy power of a point, $x(x+4) = 4^{2} \\Longrightarrow x = 2 \\sqrt{5} - 2$, and $y(y+4) = (x+4+y)^{2} - (4 \\sqrt{3})^{2} \\Longrightarrow y = \\frac{48 - (... | United States | HMMT November | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasin... | null | proof and answer | 4√5/5 | |
046q | Suppose $P_1, \dots, P_n$ are $n$ points inside triangle $ABC$, such that any three points among $P_1, \dots, P_n, A, B, C$ are not collinear. Prove: It is possible to divide $\triangle ABC$ into a union of $2n+1$ smaller triangles such that each smaller triangle has its vertices among $P_1, \dots, P_n, A, B, C$, and t... | [
"*Proof.* For the edge $BC$, a partial order $\\prec_A$ can be defined on the set $P = P_1, \\cdots, P_n$ as follows:\n$$\nP_i \\prec_A P_j \\iff \\begin{aligned}[t] & P_i \\text{ is an interior point of the } \\triangle P_j BC \\\\ & \\text{the ray } P_j P_i \\text{ intersects with the segment } BC. \\end{aligned}... | China | 2023 Chinese IMO National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
06cz | Find all prime numbers $p$ and $q$ such that $\frac{(7^p - 2^p)(7^q - 2^q)}{pq}$ is an integer. | [
"The solutions are $(p, q) = (5, 5), (5, 11), (5, 61)$ and their permutations.\nIf $p \\mid 7^p - 2^p$, then $p \\mid 7 - 2 = 5$ by the Fermat little theorem. This implies $p = 5$. If we further have $q \\mid 7^q - 2^q$, then $q = 5$ for the same reason. So $p = q = 5$ is a solution. If $q \\ne 5$, then $q \\mid 7^... | Hong Kong | CHKMO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | The pairs are (5,5), (5,11), (11,5), (5,61), (61,5). | |
0hfi | Let $ABCD$ be a quadrilateral with equal angles $B$ and $D$. Circles $\omega_1, \omega_2$ are symmetric with respect to $AC$; $\omega_1$ passes through $B$ and intersects $AB$ and $BC$ for the second time at $K$ and $L$, respectively, and $\omega_2$ passes through $D$ and intersects $CD$ and $AD$ for the second time at... | [
"Let $D'$ be symmetric to $D$ with respect to $AC$. If $D'$ coincides with $B$, then $N$ and $M$ are symmetric to $K$ and $L$, respectively, with respect to $AC$, and the problem clearly holds. Let $D' \\neq B$. Since $\\omega_2$ is symmetric to $\\omega_1$ with respect to $AC$, If we denote by $M'$ and $N'$ the po... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07es | There are five keys that assigned to lamps of a room. Each key is connected to one or more lamps and by switching a key, all lamps which is connected to that key will be switched. The set of connected lamps to each key is distinct with another. If all lamps are turned off at the beginning, prove that there are 3 keys t... | [
"Let $a_1, a_2, \\dots, a_n$ denote the lamps, and $b_1, b_2, \\dots, b_5$ denote the keys. First, it is easy to see that $n \\ge 3$, because otherwise there are at most 4 subsets of lamps and it's in contradiction with the assumption of the problem. Construct a $10 \\times n$ table that each row corresponds to a t... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0kyo | Let $\alpha$ be the radian measure of the smallest angle in a $3$-$4$-$5$ right triangle. Let $\beta$ be the radian measure of the smallest angle in a $7$-$24$-$25$ right triangle. In terms of $\alpha$, what is $\beta$?
(A) $\frac{\alpha}{3}$ (B) $\alpha - \frac{\pi}{8}$ (C) $\frac{\pi}{2} - 2\alpha$ (D) $\frac{\alpha}... | [
"Because $\\alpha$ and $\\beta$ are the smallest angles in these triangles, $\\sin \\alpha = \\frac{3}{5}$, $\\cos \\alpha = \\frac{4}{5}$, $\\sin \\beta = \\frac{7}{25}$, and $\\cos \\beta = \\frac{24}{25}$. By a Double Angle Formula,\n$$\n\\sin(2\\alpha) = 2 \\sin \\alpha \\cdot \\cos \\alpha = 2 \\cdot \\frac{3}... | United States | AMC 12 A | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | MCQ | C | |
0789 | Let $ABC$ be an isosceles triangle with $AB = AC$. Suppose $P, Q, R$ are points on segments $AC, AB, BC$ respectively such that $AP = QB$, $\angle PBC = 90 - \angle BAC$ and $RP = RQ$. Let $O_1, O_2$ be the circumcenters of $\triangle APQ$ and $\triangle CRP$. Prove that $BR = O_1O_2$. | [] | India | EGMO TST Day 2 | [
"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 > Angle chasing",
"Geometry > Plane Geometry... | null | proof only | null | |
0cxo | The squares $OAB C$ and $OA_{1}B_{1}C_{1}$ are situated in the same plane and are directly oriented. Prove that the lines $AA_{1}$, $BB_{1}$, and $CC_{1}$ are concurrent. | [
"Let $AA_{1} \\cap CC_{1} = \\{M\\}$. We have $\\triangle AOA_{1} \\equiv \\triangle COC_{1}$ (S.A.S.), then $\\widehat{AA_{1}O} \\equiv \\widehat{MA_{1}O} \\equiv \\widehat{CC_{1}O}$. It follows that $A_{1}C_{1}MO$ is cyclic, hence $\\widehat{A_{1}MC} \\equiv \\widehat{A_{1}OC_{1}} = 90^{\\circ}$, that is $A_{1}M ... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry"
] | English | proof only | null | |
05yl | Problem:
Soit $n \geqslant 3$ un entier. Pour chaque couple de nombres premiers $p$ et $q$ tels que $p<q \leqslant n$, Morgane a écrit la somme $p+q$ au tableau. Elle note ensuite $\mathcal{P}(n)$ le produit de toutes ces sommes. Par exemple, $\mathcal{P}(5)=(2+3) \times (2+5) \times (3+5)=280$.
Trouver toutes les va... | [
"Solution:\n\nSoit $n$ une solution éventuelle, et soit $r$ le plus grand nombre premier tel que $r \\leqslant n$. Puisque $r$ divise $n!$, il divise $\\mathcal{P}(n)$, donc il existe deux nombres premiers $p$ et $q$ tels que $p<q \\leqslant n$ et $r$ divise $p+q$. Par maximalité de $r$, on sait que $p+q<2r$, de so... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 7 | |
06p5 | Prove that, for every positive integer $n$, there exists an integer $m$ such that $2^{m}+m$ is divisible by $n$. | [
"We will prove by induction on $d$ that, for every positive integer $N$, there exist positive integers $b_{0}, b_{1}, \\ldots, b_{d-1}$ such that, for each $i=0,1,2, \\ldots, d-1$, we have $b_{i}>N$ and\n$$\n2^{b_{i}}+b_{i} \\equiv i \\quad(\\bmod d) .\n$$\nThis yields the claim for $m=b_{0}$.\n\nThe base case $d=1... | IMO | IMO 2006 Shortlisted Problems | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0a3x | Determine all triples $(x, y, p)$ of positive integers such that $p$ is a prime number, $x^2 = p - 1$ and $y^2 = 2p^2 - 1$. | [
"Note that $(2, 7, 5)$ is such a triple. We show that it is the only one.\nLet $(x, y, p)$ be any such triple. We first compute\n$$\n(y + x)(y - x) = y^2 - x^2 = (2p^2 - 1) - (p - 1) = 2p^2 - p = p(2p - 1). \\quad (2)\n$$\nIn particular, either $p \\mid x + y$ or $p \\mid x - y$ must hold.\n\nSuppose $p \\mid x + y... | Netherlands | IMO Team Selection Test 3 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (2, 7, 5) | |
02qt | Problem:
Uma tira de papel retangular, branca de um lado e cinza do outro, foi dobrada como na figura. Qual é a medida do ângulo $\alpha$ ?
A) $110^\circ$
B) $115^\circ$
C) $120^\circ$
D) $125^\circ$
E) $130^\circ$ | [
"Solution:\n\nConsideremos o triângulo $ABC$ na figura ao lado. Ele é retângulo com $AB = 1\\ \\mathrm{cm}$ e $BC = 2\\ \\mathrm{cm}$, ou seja, um cateto é metade da hipotenusa. Segue que $D\\hat{C}B = A\\hat{C}B = 30^\\circ$ e, analogamente, $C\\hat{B}D = 30^\\circ$. Como a soma dos ângulos internos de um triângul... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | C | |
005j | Problem:
En cada casilla de un tablero de $101$ filas y $11$ columnas se escribe uno de los números $1, 2, \ldots, 11$. Si el número $a$ está inmediatamente a la izquierda de $b$ e inmediatamente arriba de $c$, entonces $a \leq b \leq c$. ¿Puede haber en el tablero exactamente una fila de números iguales? | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | No | |
0bws | Let $ABCD$ be a trapezium, $AD \parallel BC$, and let $E$ and $F$ be points on the sides $AB$ and $CD$, respectively. The circumcircle of the triangle $AEF$ meets the line $AD$ again at $A_1$, and the circumcircle of the triangle $CEF$ meets the line $BC$ again at $C_1$. Prove that the lines $A_1C_1$, $BD$, and $EF$ ar... | [
"*First solution.* Let the line $EF$ meet the circle $C_1BE$ again at $T$. Then\n$$\n\\angle (TC_1, C_1B) = \\angle (TE, EB) = \\angle (FE, EA) = \\angle (FA_1, A_1A),\n$$\nso $TC_1$ and $FA_1$ are parallel. Similarly,\n$$\n\\angle (TB, BC_1) = \\angle (TE, EC_1) = \\angle (FE, EC_1) = \\angle (FC, CC_1) = \\angle ... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06rm | Let $n \geq 1$ be an odd integer. Determine all functions $f$ from the set of integers to itself such that for all integers $x$ and $y$ the difference $f(x)-f(y)$ divides $x^{n}-y^{n}$. | [
"Obviously, all functions in the answer satisfy the condition of the problem. We will show that there are no other functions satisfying that condition.\nLet $f$ be a function satisfying the given condition. For each integer $n$, the function $g$ defined by $g(x)=f(x)+n$ also satisfies the same condition. Therefore,... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Diophantine Equations > Techniq... | null | proof and answer | All functions of the form f(x) = s · x^d + t, where s is either 1 or −1, t is any integer, and d is a positive integer dividing the given odd integer n. | |
0c7h | Let $m$ and $n$ be positive integers, and let $A_1, \dots, A_m$ be pairwise disjoint $n$-element sets of positive integers such that no member of $A_i$ is divisible by one of $A_{i+1}$, whatever $i$ (indices are reduced modulo $m$). Determine the largest number of ordered pairs $(a, b)$, where $a$ and $b$ are members o... | [
"The required maximum is $\\binom{m-1}{2}n^2$, and is achieved if, for instance\n$$\nA_k = \\{a^{(k-1)n+1}, a^{(k-1)n+2}, \\dots, a^{kn}\\}, \\quad k = 1, \\dots, m-1,\n$$\n\nFor brevity, an ordered pair $(a, b)$ satisfying the conditions in the statement will be called *suitable*. We show that the number of suitab... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | ((m-1)(m-2)/2) * n^2 | |
05pg | Problem:
Trouver tous les triplets d'entiers naturels $(x, y, z)$ tels que :
$$
x^{2}+y^{2}=3 \cdot 2016^{z}+77.
$$ | [
"Solution:\nSoit un triplet solution $(x, y, z)$. Que peut-on en dire de façon nécessaire ?\n\nSi $z=0$, on a $x^{2}+y^{2}=80$. On fait une table de congruence modulo $4$ : on montre ainsi que, si une somme de deux carrés est congrue modulo $4$, c'est que les deux carrés en question sont tous les deux pairs. On peu... | France | OCympiades Françaises de Mathématiques | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | {(4,8,0),(8,4,0),(35,70,1),(70,35,1),(14,77,1),(77,14,1)} |
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