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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0a20 | Let $\triangle ABC$ be a triangle with orthocentre $H$ and circumcircle $\Gamma$. Let $D$ be the reflection of $A$ across the point $B$, and let $E$ be the reflection of $A$ across the point $C$. Let $M$ be the midpoint of segment $DE$.
Prove that the tangent to $\Gamma$ at $A$ is perpendicular to $HM$. | [
"Let $A'$ be the reflection of $H$ across the midpoint of $BC$ and $A''$ the reflection of $H$ across $BC$. Then by angle chasing, we find that $\\angle BHC = \\angle ABC + \\angle BCA = 180^\\circ - \\angle CAB$. This angle is also equal to $\\angle BA'C$ and $\\angle BA''C$. Therefore, both $A'$ and $A''$ lie on ... | Netherlands | IMO Team Selection Test 2 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0h7z | In a triangle $ABC$ with sides $BC > AC > AB$, consider the angles between the height and median, being built from one top. Find out, on which top this angle is the largest of the three. | [
"Let us denote in a standard way the sides of triangle $a, b, c$. Then under condition, $a > b > c$. Let us denote the height and median from the top $B$ as $h_b = BH$ and $m_b = BM$ accordingly (fig. 48). Then the interesting for us angle is $\\angle MBH$, while $\\cos\\angle MBH = \\frac{h_b}{m_b}$. Similarly, we... | Ukraine | UkraineMO | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | Vertex B | |
01gt | On a plane, there are given points $A_0, B_0, C_0$ (not necessarily distinct) such that $A_0B_0 + B_0C_0 + C_0A_0 = 1$. Points $A_1, B_1, C_1$ (not necessarily distinct) are chosen in such a way that $A_1B_1 = A_0B_0$ and $B_1C_1 = B_0C_0$. Points $A_2, B_2, C_2$ are chosen as a permutation of points $A_1, B_1, C_1$. F... | [
"Answer: $\\frac{1}{3}$ and $3$.\n\nDenote the lengths $A_0B_0, B_0C_0, C_0A_0$ by $x, y, z$ in non-increasing order. Similarly, denote the lengths $A_1B_1, B_1C_1, C_1A_1$ by $x', y', z'$ in non-increasing order, and the lengths $A_3B_3, B_3C_3, C_3A_3$ by $x'', y'', z''$ in non-increasing order. (As permuting the... | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof and answer | 1/3 and 3 | |
0ff8 | Problem:
Dados el triángulo equilátero $ABC$, de lado $a$, y su circunferencia circunscrita, se considera el segmento de círculo limitado por la cuerda $AB$ y el arco (de $120^{\circ}$) con los mismos extremos. Al cortar este segmento circular con rectas paralelas al lado $BC$, queda determinado sobre cada una de ella... | [
"Solution:\n\nDenotemos con $B'$ $C'$ a uno cualquiera de esos segmentos, y completemos, con un punto $A'$ sobre la cuerda $AB$, el triángulo $A' C' B'$ (equilátero) de lados paralelos a los del triángulo $ABC$. La longitud $B' C'$ será máxima cuando la altura del $A' C' B'$ sea máxima, lo que corresponde evidentem... | Spain | Olimpiadas Matemáticas Españolas | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a/3 | |
07va | Suppose $x_1, x_2, \dots, x_n$ are complex numbers. Prove that
$$
\sum_{i,j=1}^{n} |x_i - x_j|^2 \le \sum_{i,j=1}^{n} |x_i + x_j|^2,
$$
with equality iff $x_1 + x_2 + \dots + x_n = 0$. | [
"Note that $|a+b|^2 - |a-b|^2 = 4 \\operatorname{Re}(\\bar{a}b)$ for any complex numbers $a, b$. Hence,\n$$\n\\begin{align*}\n\\sum_{i,j=1}^{n} |x_i + x_j|^2 - \\sum_{i,j=1}^{n} |x_i - x_j|^2 &= 4 \\sum_{i,j=1}^{n} \\operatorname{Re} (\\bar{x}_i x_j) = 4 \\operatorname{Re} \\sum_{i,j=1}^{n} \\bar{x}_i x_j \\\\\n&= ... | Ireland | IRL_ABooklet | [
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
05y6 | Problem:
Soit $ABC$ un triangle rectangle en $B$ avec $BC < BA$. Soit $D$ le point du segment $[AB]$ tel que $BD = BC$. La perpendiculaire à $(AC)$ passant par $D$ intersecte $(AC)$ en $E$. Soit $B'$ le symétrique de $B$ par rapport à $(CD)$. Montrer que $(EC)$ est la bissectrice de l'angle $\widehat{BEB'}$. | [
"Solution:\n\n\n\nOn remarque que le cercle de diamètre $[CD]$ apparaît assez naturellement. En effet, on a des angles droits $\\widehat{DBC} = \\widehat{CBD} = \\widehat{DEC} = 90^\\circ$, les points $B$, $B'$, et $E$ sont sur le cercle de diamètre $[DC]$, autrement dit $C, B, D, E, B'$ so... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0icj | Problem:
Let $S=\{p_{1} p_{2} \cdots p_{n} \mid p_{1}, p_{2}, \ldots, p_{n}$ are distinct primes and $p_{1}, \ldots, p_{n}<30\}$. Assume $1$ is in $S$. Let $a_{1}$ be an element of $S$. We define, for all positive integers $n$:
$$
\begin{gathered}
a_{n+1}=a_{n} /(n+1) \quad \text{ if } a_{n} \text{ is divisible by } n+... | [
"Solution:\nIf $a_{1}$ is odd, then we can see by induction that $a_{j}=(j+1) a_{1}$ when $j$ is even and $a_{j}=a_{1}$ when $j$ is odd (using the fact that no even $j$ can divide $a_{1}$). So we have infinitely many $j$'s for which $a_{j}=a_{1}$.\n\nIf $a_{1}>2$ is even, then $a_{2}$ is odd, since $a_{2}=a_{1} / 2... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 512 | |
0j21 | Problem:
Let
$$
\begin{gathered}
e^{x}+e^{y}=A \\
x e^{x}+y e^{y}=B \\
x^{2} e^{x}+y^{2} e^{y}=C \\
x^{3} e^{x}+y^{3} e^{y}=D \\
x^{4} e^{x}+y^{4} e^{y}=E .
\end{gathered}
$$
Prove that if $A, B, C$, and $D$ are all rational, then so is $E$. | [
"Solution:\nWe can express $x+y$ in two ways:\n$$\n\\begin{aligned}\n& x+y=\\frac{A D-B C}{A C-B^{2}} \\\\\n& x+y=\\frac{A E-C^{2}}{A D-B C}\n\\end{aligned}\n$$\n(We have to be careful if $A C-B^{2}$ or $A D-B C$ is zero. We'll deal with that case later.) It is easy to check that these equations hold by substitutin... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0kjv | A quadratic polynomial $p(x)$ with real coefficients and leading coefficient $1$ is called *disrespectful* if the equation $p(p(x)) = 0$ is satisfied by exactly three real numbers. Among all the disrespectful quadratic polynomials, there is a unique such polynomial $\tilde{p}(x)$ for which the sum of the roots is maxim... | [] | United States | AMC 10 A | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | MCQ | A | |
09pf | Find all quintuples of positive integers $(a, b, c, m, n)$ such that $m \le 2n - 2$ and
$$
a^n + b^n = (a, b)^m (a + b), \quad b^n + c^n = (b, c)^m (b + c), \quad c^n + a^n = (c, a)^m (c + a).
$$
Here, $(x, y)$ denotes the greatest common divisor of $x$ and $y$.
(Bilegdemberel Bat-Amgalan) | [] | Mongolia | MMO2025 Round 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | All solutions have a = b = c = t and n ≥ 2. If t > 1, then m = n − 1. If t = 1, then any m with 1 ≤ m ≤ 2n − 2 works. | |
0cbj | Determine all prime numbers $p, q < 2023$ such that $q \mid p^2 + 8$ and $p \mid q^2 + 8$. | [
"If one of the numbers is $2$, then $p = q = 2$, and we can assume $p, q \\ge 3$. Since $(p, p^2 + 8) = (q, q^2 + 8) = 1$, we have: $q \\mid p^2 + 8$ and $p \\mid q^2 + 8 \\Rightarrow pq \\mid (p^2 + 8)(q^2 + 8) \\Rightarrow pq \\mid 8(p^2 + q^2 + 8) \\Rightarrow pq \\mid (p^2 + q^2 + 8)$.\n\nFor a fixed $k \\in \\... | Romania | THE Eighteenth IMAR Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysi... | null | proof and answer | (p, q) = (2, 2), (17, 3), (3, 17) | |
0e2n | For what positive integers $n \ge 3$ does there exist a convex $n$-gon which can be divided into finitely many parallelograms? | [
"We will show that an $n$-gon with this property exists for $n$ even, but not for $n$ odd.\nAssume that a convex $n$-gon can be divided into finitely many parallelograms. Denote one of the sides by $a$. There exists a parallelogram with one side lying on $a$. Denote the opposite side of this parallelogram by $b_1$.... | Slovenia | Selection Examinations for the IMO | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | n is even | |
0dq3 | An odd integer $a > 1$ is given. Initially, Basil chooses an even positive integer $b$ such that $b < a$ and tells it to Pete. Basil then writes down three integers on a blackboard. After that, Pete makes a sequence of moves. By a move, Pete can either add $a$ to one of the numbers on the blackboard, add $b$ to the sec... | [
"**Answer.** $a = \\frac{4^n - 1}{3}$ for an integer $n > 1$ and $a = \\frac{2p-1}{3}$, where $p$ is an odd prime number (necessarily having remainder 2 when divided by 3).\n\nWe start with describing the set of pairs $(a, b)$ for which Pete can win.\n\n**Lemma.**\n(a) If $\\gcd(2a + b + 1, a - b) > 1$, then Pete c... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory >... | English | proof and answer | All odd a > 1 for which Basil cannot prevent Pete's win are a = (4^n − 1)/3 for an integer n > 1, and a = (2p − 1)/3 where p is an odd prime with p ≡ 2 (mod 3). | |
0263 | Problem:
O valor de $\left(\sqrt{1+\sqrt{1+\sqrt{1}}}\right)^4$ é:
(a) $\sqrt{2}+\sqrt{3}$
(b) $\frac{1}{2}(7+3 \sqrt{5})$
(c) $1+2 \sqrt{3}$
(d) 3
(e) $3+2 \sqrt{2}$ | [
"Solution:\n$$\n\\left(\\sqrt{1+\\sqrt{1+\\sqrt{1}}}\\right)^4 = (1+\\sqrt{2})^2 = 1+2 \\sqrt{2}+2 = 3+2 \\sqrt{2}\n$$\nA opção correta é (e)."
] | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | e | |
0dl1 | Let $ABC$ be a triangle inscribed in circle $(O)$ with angle $\angle A = 45^\circ$ and $AB < AC$. Let $AD, AH$ be the angle bisector and altitude of triangle $ABC$ with $D, H$ are on $BC$. Suppose that $OD$ intersects $AH$ at $E$ and $K$ is the circumcenter of triangle $EBC$. Prove that $HK \parallel AD$. | [
"Let $P, N$ be the projections of $O, K$ on $AE$ and $M, T$ be the midpoints of $BC$ and the minor arc $BC$ of $(O)$. Let $AH = h$ and $R, R'$ be the radii of $(O), (K)$ respectively.\n\nBy Thales' theorem, we have\n$$\n\\frac{HE}{HA} = \\frac{MO}{MT} = \\frac{MO}{OT - OM} = \\sqrt{2} + 1 \... | Saudi Arabia | Saudi Booklet | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
08tf | Let $N$ be a positive integer. Suppose some collection of integers are written on a blackboard satisfying the following properties.
* Every number $k$ written satisfies $1 \le k \le N$.
* Every $k$ with $1 \le k \le N$ is written at least once.
* The sum of all the numbers written is even.
Prove that by marking some of... | [
"Suppose we line up the numbers written on the blackboard in a non-increasing order and represent them as $a_1, a_2, \\dots, a_m$, with $a_k \\ge a_{k+1}$ for each $k$. We mark each of the numbers $a_1, a_2, \\dots, a_m$ by $\\bigcirc$ or $\\times$ in order in the following way. Start with $a_1$, and at each step c... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic... | null | proof only | null | |
0cph | Let $\triangle ABC$ be an acute-angled triangle. Points $P$ and $Q$ are chosen on the extensions of its altitudes $BB_1$ and $CC_1$ along the points $B_1$ and $C_1$ respectively so that $\angle PAQ = 90^\circ$. Let $AF$ be the altitude of triangle $APQ$. Prove that $\angle BFC = 90^\circ$. (A. Polyanskiy)
Дан остроуго... | [
"Точки $B_1$ и $C_1$ лежат на окружности, построенной на $BC$ как на диаметре. Для решения достаточно доказать, что $F$ также лежит на этой окружности, то есть достаточно доказать, что четырёхугольник $CB_1FC_1$ — вписанный.\n\nТак как $\\angle AB_1P = \\angle AFP = 90^\\circ$, то точки $B_1$ и $F$ лежат на окружно... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English, Russian | proof only | null | |
0hec | We enumerate all prime numbers in ascending order: $p_1 = 2, p_2 = 3, p_3 = 5, \dots$. Find all positive integer $n$, for which $p_1!+p_2!+\dots+p_n!=a^b$, for some positive integer $a, b>1$, where $k!$ denotes the product of all integers from 1 to $k$. | [
"Directly checking yields:\n$$\np_1! = 2, \\quad p_1! + p_2! = 8 = 2^3, \\quad p_1! + p_2! + p_3! = 128 = 2^7\n$$\nFor $n \\ge 4$, $p_1! + p_2! + \\dots + p_n! = 128 + 7! + 11! + \\dots + p_n!$ is a number in which all summands are divisible by $2^5$, except $p_4! = 7! = 2^4 \\cdot 315$, which is divisible by $2^4$... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 2, 3 | |
0dxa | Problem:
Poišči vse pare naravnih števil $a$ in $b$, ki zadoščajo enačbi $a^{2}-5 a b+24=0$. | [
"Solution:\n\n1. Enačbo preoblikujemo v $a(a-5 b)=-24$ in jo pomnožimo z $-1$, da dobimo $a(5 b-a)=24$. Ker sta $a$ in $5 b-a$ celi števili, je $a$ pozitiven delitelj števila $24$. Torej imamo naslednje možnosti: $a=1$ in $5 b-a=24$, $a=2$ in $5 b-a=12$, $a=3$ in $5 b-a=8$, $a=4$ in $5 b-a=6$, $a=6$ in $5 b-a=4$, $... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | Pairs: (1, 5), (4, 2), (6, 2), (24, 5) | |
06yn | Let $n$ be a positive integer. We say that a polynomial $P$ with integer coefficients is $n$-good if there exists a polynomial $Q$ of degree 2 with integer coefficients such that $Q(k)(P(k)+Q(k))$ is never divisible by $n$ for any integer $k$.
Determine all integers $n$ such that every polynomial with integer coefficie... | [
"First, observe that no polynomial is $1$-good (because $Q(X)(P(X)+Q(X))$ always has roots modulo $1$) and the polynomial $P(X)=1$ is not $2$-good (because $Q(X)(Q(X)+1)$ is always divisible by $2$).\n\nNow, if $P$ is $d$-good with some $Q$, then $Q \\cdot (P+Q)$ has no roots $\\bmod d$. Therefore, it certainly has... | IMO | IMO2024 Shortlisted Problems | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Discrete Mathematics > ... | English | proof and answer | all positive integers n ≥ 3 | |
03qk | Suppose that $\alpha$ and $\beta$ are different real roots of the equation $4x^2 - 4tx - 1 = 0$ ($t \in \mathbb{R}$). $[\alpha, \beta]$ is the domain of the function $f(x) = \frac{2x-t}{x^2+1}$.
(1) Find $g(t) = \max f(x) - \min f(x)$.
(2) Prove that for $u_i \in (0, \frac{\pi}{2})$ ($i = 1, 2, 3$), if $\sin u_1 + \s... | [
"(1) Let $\\alpha \\le x_1 < x_2 \\le \\beta$, then\n$$4x_1^2 - 4tx_1 - 1 \\le 0, \\quad 4x_2^2 - 4tx_2 - 1 \\le 0.$$\nTherefore,\n$$\n4(x_1^2 + x_2^2) - 4(t(x_1 + x_2) - 2) \\le 0, \\\\\n2x_1x_2 - t(x_1 + x_2) - \\frac{1}{2} < 0.\n$$\nBut\n$$\nf(x_2) - f(x_1) = \\frac{2x_2 - t}{x_2^2 + 1} - \\frac{2x_1 - t}{x_1^2 ... | China | China Mathematical Competition (Hainan) | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | g(t) = 8 sqrt(t^2 + 1) (2 t^2 + 5) / (16 t^2 + 25), and for acute u1, u2, u3 with sin u1 + sin u2 + sin u3 = 1: 1/g(tan u1) + 1/g(tan u2) + 1/g(tan u3) < (3/4) sqrt(6). | |
0ej9 | Problem:
Kvadrat s stranico dolžine $2$ je razdeljen na $4$ trikotnike (glej sliko). Vsi $3$ osenčeni trikotniki imajo enako ploščino. Koliko je ploščina belega trikotnika?
(A) $\frac{1+\sqrt{5}}{2}$
(B) $\frac{8}{5}$
(C) $2$
(D) $3 \sqrt{5}-5$
(E) $6-2 \sqrt{5}$
 | [
"Solution:\n\nOznačimo z $A, B, C$ in $D$ oglišča kvadrata in dodatno z $E$ in $F$ oglišči belega trikotnika (glej sliko).\n\n\n\nOznačimo $x=|AE|$. Ker imata pravokotna trikotnika $AED$ in $FCD$ eno od stranic enako stranici kvadrata in imata enaki ploščini, je $|CF|=|AE|=x$. Če upoštevamo... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
03kr | Problem:
The closed interval $A = [0, 50]$ is the union of a finite number of closed intervals, each of length $1$. Prove that some of the intervals can be removed so that those remaining are mutually disjoint and have total length $\geq 25$.
Note. For $a \leq b$, the closed interval $[a, b] := \{ x \in \mathbb{R} : ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | proof only | null | |
07c4 | A circle passing through vertices $B$ and $C$ of triangle $ABC$ intersects sides $AC$ and $AB$ at points $D$ and $E$, respectively. If $P$ is the intersection point of $BD$ and $CE$, $H$ is the foot of the perpendicular line from $P$ to $AC$ and $M$ and $N$ are the midpoints of $BC$ and $AP$, prove that triangles $MNH$... | [
"Let $K$ and $T$ be the reflections of $P$ with respect to $M$ and $H$, respectively. According to Thales' Theorem, triangles $AKT$ and $MNH$ are similar. On the other hand, for triangles $ABD$ and $AEC$, $\\angle EBD = \\angle ECD$ and $\\angle BAC$ appears in both triangles; therefore, these two triangles are sim... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
012w | Problem:
Let $ABCD$ be a rectangle and $BC = 2 \cdot AB$. Let $E$ be the midpoint of $BC$ and $P$ an arbitrary inner point of $AD$. Let $F$ and $G$ be the feet of perpendiculars drawn correspondingly from $A$ to $BP$ and from $D$ to $CP$. Prove that the points $E, F, P, G$ are concyclic. | [
"Solution:\n\nFrom rectangular triangle $BAP$ we have $BP \\cdot BF = AB^{2} = BE^{2}$. Therefore the circumference through $F$ and $P$ touching the line $BC$ between $B$ and $C$ touches it at $E$.\n\nAnalogously, the circumference through $P$ and $G$ touching the line $BC$ between $B$ and $C$ touches it at $E$. Bu... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ewr | Problem:
Find the smallest value $x$ such that, given any point inside an equilateral triangle of side $1$, we can always choose two points on the sides of the triangle, collinear with the given point and a distance $x$ apart. | [
"Solution:\nAnswer: $2/3$.\n\nLet $O$ be the center of $ABC$. Let $AO$ meet $BC$ at $D$, let $BO$ meet $CA$ at $E$, and let $CO$ meet $AB$ at $F$. Given any point $X$ inside $ABC$, it lies in one of the quadrilaterals $AEOF$, $CDOE$, $BFOD$. Without loss of generality, it lies in $AEOF$. Take the line through $X$ p... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Tr... | null | proof and answer | 2/3 | |
03zy | Let $a_1, a_2, \dots, a_n$ be $n$ non-negative real numbers.
$$
\frac{1}{1+a_1} + \frac{a_1}{(1+a_1)(1+a_2)} + \dots + \frac{a_1a_2\dots a_{n-1}}{(1+a_1)(1+a_2)\dots(1+a_n)} \le 1.
$$ | [
"Let $a_0 = 1$. We prove the following identity:\n$$\n\\sum_{k=1}^{n} \\prod_{j=1}^{k} \\frac{a_{j-1}}{1+a_j} = 1 - \\prod_{j=1}^{n} \\frac{a_j}{1+a_j} \\qquad \\textcircled{1}\n$$\nby induction on $n$.\n\nIt is evident that $\\textcircled{1}$ is true for $n = 1$. Suppose that $\\textcircled{1}$ is true for $n-1$, ... | China | China Girls' Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0aqr | Problem:
Find the solution set to the equation $\left(x^{2}-5 x+5\right)^{x^{2}-9 x+20}=1$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | [1, 2, 3, 4, 5] | |
06tc | In Lineland there are $n \geqslant 1$ towns, arranged along a road running from left to right. Each town has a left bulldozer (put to the left of the town and facing left) and a right bulldozer (put to the right of the town and facing right). The sizes of the $2n$ bulldozers are distinct. Every time when a right and a ... | [
"Let $T_{1}, T_{2}, \\ldots, T_{n}$ be the towns enumerated from left to right. Observe first that, if town $T_{i}$ can sweep away town $T_{j}$, then $T_{i}$ also can sweep away every town located between $T_{i}$ and $T_{j}$.\nWe prove the problem statement by strong induction on $n$. The base case $n=1$ is trivial... | IMO | 56th International Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Logic"
] | English | proof only | null | |
06os | Let $ABCD$ be a trapezoid with parallel sides $AB > CD$. Points $K$ and $L$ lie on the line segments $AB$ and $CD$, respectively, so that $AK / KB = DL / LC$. Suppose that there are points $P$ and $Q$ on the line segment $KL$ satisfying
$$
\angle APB = \angle BCD \quad \text{and} \quad \angle CQD = \angle ABC.
$$
Prove... | [
"Because $AB \\parallel CD$, the relation $AK / KB = DL / LC$ readily implies that the lines $AD$, $BC$ and $KL$ have a common point $S$.\n\nConsider the second intersection points $X$ and $Y$ of the line $SK$ with the circles $(ABP)$ and $(CDQ)$, respectively. Since $APBX$ is a cyclic quad... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadr... | English | proof only | null | |
086b | Problem:
Sia $n \geq 2$ un numero intero. Coloriamo tutte le caselle di una scacchiera $n \times n$ in rosso o blu in modo che ogni quadrato $2 \times 2$ contenuto nella scacchiera abbia esattamente due caselle rosse e due blu.
Quante sono le colorazioni possibili?
NOTA: due colorazioni che si ottengono l'una dall'al... | [
"Solution:\n\nFissiamo una colorazione della prima riga della scacchiera. Essa può essere fatto in $2^{n}$ modi (2 scelte per il colore di ciascuna delle $n$ caselle). Mostriamo ora se e in quanti modi una colorazione della prima riga può essere completata ad una colorazione dell'intera scacchiera soddisfacendo le ... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{n+1} - 2 | |
001j | En una competencia de gimnasia deportiva de 50 participantes, cada participante está identificado con un número del 1 al 50. La competencia tiene 13 jueces, y cada uno de ellos ordena a los participantes de mejor a peor, a su criterio. Luego le asigna 1 punto al mejor, 2 al segundo, ..., 50 al último. Resultó que para ... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | español | proof and answer | Yes. The total score of participant k is 357 minus k, for k from 1 to 50. | |
0hhw | There are $n$ lamps and $2024$ switches in a room. Each lamp is connected to exactly $1000$ switches. When a switch is pressed, the state of each connected lamp changes from "ON" to "OFF" or vice versa. It is known that by pressing some of the switches, all lamps can be turned on. Prove that this can be achieved by pre... | [
"Obviously, each switch should not be pressed more than once. Let us divide all switches into two groups: Group I - those switches that must be pressed to turn all lamps on, and Group II - all other switches. Since each lamp is connected to an even number of switches, if all switches in Group II are pressed, then a... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0k9z | Problem:
In $\triangle ABC$, the external angle bisector of $\angle BAC$ intersects line $BC$ at $D$. $E$ is a point on ray $\overrightarrow{AC}$ such that $\angle BDE = 2 \angle ADB$. If $AB = 10$, $AC = 12$, and $CE = 33$, compute $\frac{DB}{DE}$. | [
"Solution:\n\nLet $F$ be a point on ray $\\overrightarrow{CA}$ such that $\\angle ADF = \\angle ADB$. $\\triangle ADF$ and $\\triangle ADB$ are congruent, so $AF = 10$ and $DF = DB$. So, $CF = CA + AF = 22$. Since $\\angle FDC = 2 \\angle ADB = \\angle EDC$, by the angle bisector theorem we compute $\\frac{DF}{DE} ... | United States | HMMT November 2019 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2/3 | |
0bh0 | Let $f : [0, 1] \to [0, 1]$ be a continuous function and $x_0 \in [0, 1]$. Define the sequence $(x_n)_{n \in \mathbb{N}}$ by
$$
x_{n+1} = \int_{0}^{\frac{1}{n+1}} (x_0 + x_1 + \dots + x_n) f(x) \, dx.
$$
Prove that the sequence $(x_n)_{n \in \mathbb{N}}$ is convergent. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
05yh | Problem:
Une suite olympique est une suite $s_{1}, s_{2}, \ldots, s_{2023}$ dont chacun des 2023 termes est égal à 1 ou à -1. Une suite peu croissante est une suite d'entiers $t_{1}, t_{2}, \ldots, t_{n}$ telle que $1 \leqslant t_{1}<t_{2}<\cdots<t_{n} \leqslant 2023$, et dont chacune des $n-1$ différences $t_{i+1}-t_... | [
"Solution:\n\nUne étude des petits cas nous pousse à nous intéresser à la suite $\\left(s_{i}\\right)$ dont les termes valent $+1,+1,-1,-1,+1,+1,-1,-1, \\ldots$, et qui semble être un «pire» cas. On la subdivise en 1011 blocs de longueur 2, notés $B_{1}, B_{2}, \\ldots, B_{1011}$, et un bloc $B_{1012}$ de longueur ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 507 | |
0fal | Problem:
A cinema has its seats arranged in $n$ rows $ imes$ $m$ columns. It sold $mn$ tickets but sold some seats more than once. The usher managed to allocate seats so that every ticket holder was in the correct row or column. Show that he could have allocated seats so that every ticket holder was in the correct row... | [
"Solution:\n\nSuppose it is not possible. Take any person, label him $P_1$. Suppose he should be in seat $S_1$. If seat $S_1$ is vacant, then we can just move him to $S_1$, so $S_1$ must be occupied by someone. Call him $P_2$. Continue, so that we get a sequence $P_1$, $P_2$, $P_3$, ... where $P_i$ should be in the... | Soviet Union | 1st CIS | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1 | |
09ke | For a positive integer $n$, let $a_n = \frac{(2n)!}{(n!)^3}$. Here $k! = 1 \times 2 \times 3 \times \cdots \times k$.
(1)
Prove $a_n > a_{n+1}$ for all $n \ge 3$.
(2)
Find all $n$ such that $a_n$ is a whole number. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | n = 1, 2 | |
0hei | Determine all possible positive integers $m$ and $n$, that satisfy the following:
$$
(m+n)! = 2m! \cdot n!
$$
where $k!$ denotes the product $1 \cdot 2 \cdot \dots \cdot k$, where $k$ is a positive integer. | [
"Without loss of generality, assume that $m \\ge n$. Then, if $n > 1$ equation can be written the following way:\n$$\n1 \\cdot 2 \\cdot 3 \\cdots m \\cdot (m+1) \\cdot (m+2) \\cdots (m+n) = 2 \\cdot 1 \\cdot 2 \\cdot 3 \\cdots m \\cdot 1 \\cdot 2 \\cdot 3 \\cdots n \\Rightarrow \\\\ (m+1) \\cdot (m+2) \\cdots (m+n)... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | m = 1, n = 1 | |
09uk | There are $13$ distinct multiples of $7$ that consist of two digits. You want to create a longest possible chain consisting of these multiples, where two multiples can only be adjacent if the last digit of the left multiple equals the first digit of the right multiple. You can use each multiple at most once. For exampl... | [
"B) $7$"
] | Netherlands | First Round, January 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | English | MCQ | B | |
09kt | Suppose that there are $n$ students standing in a line, and each student randomly raises either their left or right hand, but not both. Let $P_n$ denote the probability that in every group of three students standing in a row, at least one student raised their right hand. Prove
$$
P_n < \left(\frac{12}{13}\right)^{n-2}.... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0699 | The sequence $\alpha_v$ satisfies the recurrence relation: $\alpha_1 = 1$ and $\alpha_v = 5\alpha_{v-1} + 3^{v-1}$, $v \ge 2$. Determine the general term $\alpha_v$ and the greatest power of $2$ which divides the term $a_k$, where $k=2^{2019}$. | [
"$$\n\\alpha_v = 5^{v-1} \\cdot \\left[ 1 + \\frac{3}{5} + \\left(\\frac{3}{5}\\right)^2 + \\dots + \\left(\\frac{3}{5}\\right)^{v-1} \\right] = \\frac{1}{2}(5^v - 3^v),\\ v = 1, 2, \\dots\n$$\nNow for $k=2^{2019}$, we have: $2a_k = 5^{2019} - 3^{2019} = 2 \\cdot (5+3)(5^2+3^2) \\dots (5^{2018} + 3^{2018})$, and he... | Greece | 36th Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | α_v = (5^v − 3^v)/2 for v ≥ 1; and for k = 2^2019, the greatest power of 2 dividing a_k is 2^2021. | |
0awb | Problem:
120 unit cubes are put together to form a rectangular prism whose six faces are then painted. This leaves 24 unit cubes without any paint. What is the surface area of the prism? | [
"Solution:\n\nLet the length, width and height of the rectangular prism made by the 24 cubes without paint be denoted by $\\ell, w, h$ (necessarily positive integers), respectively. Then, those of the prism made by the 120 cubes have measures $\\ell+2, w+2$ and $h+2$, respectively. Hence, $\\ell w h=24$ and $(\\ell... | Philippines | 18th PMO National Stage Oral Phase | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Surface Area",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 148 | |
0euy | Let $A_1, A_2, \dots, A_n$ be sets. For a subset $X$ of $\{1, 2, \dots, n\}$, let
$$
N(X) = \{i \in \{1, 2, \dots, n\} - X : A_i \cap A_j \neq \emptyset \text{ for all } j \in X\}.
$$
Prove that for every integer $3 \le m \le n - 2$, there exists a subset $X$ of $\{1, 2, \dots, n\}$ such that $|X| = m$ and $|N(X)| \neq... | [
"Let $G$ be a graph on vertices $v_1, v_2, \\dots, v_n$ such that two vertices $v_i$ and $v_j$ are adjacent if and only if $A_i \\cap A_j \\neq \\emptyset$ and $i \\neq j$. For a set $X$ of vertices of $G$, let $N(X)$ be the set of all vertices adjacent to every vertex in $X$.\nSuppose, on the contrary, that $n \\g... | South Korea | 25th Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | null | proof only | null | |
07ui | If $(ABC)$ denotes the area of $ABC$ prove that
$$
(ABC) = \frac{a^2}{2(\cot B + \cot C)}.
$$
Deduce or prove otherwise that if $ABC$ is acute-angled, then
$$
\cos A \cos B \cos C \le \frac{1}{8},
$$
with equality iff the triangle is equilateral. | [
"To prove $(ABC) = \\frac{a^2}{2(\\cot B + \\cot C)}$, we recall that $\\sin(B + C) = \\sin(A)$ because $\\angle A + \\angle B + \\angle C = 180^\\circ$ and obtain\n$$\n\\begin{aligned}\n\\cot B + \\cot C &= \\frac{\\cos B \\sin C + \\cos C \\sin B}{\\sin B \\sin C} = \\frac{\\sin(B + C)}{\\sin B \\sin C} \\\\\n&= ... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0f9z | Problem:
The numbers $1, 2, 3, \ldots, n$ are written on a blackboard (where $n \geq 3$). A move is to replace two numbers by their sum and non-negative difference. A series of moves makes all the numbers equal $k$. Find all possible $k$. | [
"Solution:\n\nIf a prime $p$ divides $a + b$ and $a - b$, then it divides $2a$ and $2b$, so if $p$ is odd, it divides $a$ and $b$. Thus if an odd prime $p$ divides $k$, then it must divide all the original numbers including $1$. So $k$ must be a power of $2$. Note that $k, k \\rightarrow 0, 2k \\rightarrow 2k, 2k$ ... | Soviet Union | 25th ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All k of the form 2^m with m an integer and 2^m ≥ n. | |
03uc | Find the array of prime numbers $(a, b, c)$ satisfying conditions as follows:
(1) $a < b < c < 100$, where $a$, $b$, $c$ are all prime numbers;
(2) $a + 1$, $b + 1$, $c + 1$ constitute a geometric progression. | [
"From condition (2), we get\n$$\n(a+1)(c+1) = (b+1)^2.\n$$\nSet $a+1 = n^2 x$, $c+1 = m^2 y$, with no square factor larger than 1 in $x$, $y$, then we could get that $x = y$. This is due to the fact that from above, we have\n$$\n(mn)^2 x y = (b+1)^2,\n$$\nwhich means $mn \\mid (b+1)$. Set $b+1 = mn \\cdot w$, then ... | China | China Southeastern Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | (2, 5, 11), (5, 11, 23), (7, 11, 17), (5, 17, 53), (11, 23, 47), (2, 11, 47), (17, 23, 31), (7, 23, 71), (31, 47, 71), (17, 41, 97), (71, 83, 97) | |
00e6 | The vertices of a regular hexagon are marked on a blackboard. Ana draws some segments that are either sides or diagonals of the hexagon, in any way she wants to (she can even decide not to draw any segment at all, or to draw the 15 possible segments).
Afterwards, Beto writes a positive integer on each vertex, in such a... | [
"a. Beto can proceed as follows. First he picks a unique prime number for each segment drawn by Ana, and he assigns that prime number to both endpoints of this segment. Then, the number he writes on each vertex is the product of all prime numbers assigned to that vertex (if there are none, we consider the product t... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes | |
0ifr | Problem:
The L shape made by adjoining three congruent squares can be subdivided into four smaller L shapes.
Each of these can in turn be subdivided, and so forth. If we perform 2005 successive subdivisions, how many of the $4^{2005}$ L's left at the end will be in the same orientation as the ... | [
"Solution:\n$4^{2004} + 2^{2004}$\n\nAfter $n$ successive subdivisions, let $a_{n}$ be the number of small L's in the same orientation as the original one; let $b_{n}$ be the number of small L's that have this orientation rotated counterclockwise $90^{\\circ}$; let $c_{n}$ be the number of small L's that are rotate... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 4^{2004} + 2^{2004} | |
0fe9 | Problem:
Cuatro bolas negras y cinco bolas blancas se colocan, en orden arbitrario, alrededor de una circunferencia.
Si dos bolas consecutivas son del mismo color, se inserta una nueva bola negra entre ellas. En caso contrario, se inserta una nueva bola blanca.
Se retiran las bolas negras y blancas previas a la inserc... | [
"Solution:\n\nSi asignamos a cada bola negra el valor $1$ y a cada bola blanca el valor $-1$, se observa que dos bolas consecutivas se sustituyen por su producto.\n\nConsiderando el producto $P$ de los nueve valores antes y después de cada operación, vemos que el nuevo $P$ es igual al cuadrado del anterior $P$. Así... | Spain | null | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
08fw | Problem:
Su un'isola ci sono 2023 persone in fila indiana, ciascuna delle quali è un furfante o un cavaliere: i cavalieri dicono sempre la verità, mentre i furfanti mentono sempre. Se $i$ è dispari, la persona in posizione $i$-esima esclama: "Ci sono almeno $i$ furfanti"; se $i$ è pari, la persona in posizione $i$-esi... | [
"Solution:\n\nLa risposta è 1348. Se i furfanti sono in numero dispari, diciamo $2m+1$, allora le persone che dicono la verità sono tutte e sole quelle in posizioni dispari minori o uguali di $2m+1$, che sono $m+1$. Ci sarebbero quindi $2m+1$ furfanti e $m+1$ cavalieri, per un totale di $3m+2$ persone, ma ciò è imp... | Italy | Italian Mathematical Olympiad | [
"Discrete Mathematics > Logic"
] | null | proof and answer | 1348 | |
0f8u | Problem:
Find the smallest positive integer $n$ for which we can find an integer $m$ such that $\left[ \dfrac{10^n}{m} \right] = 1989$. | [] | Soviet Union | 23rd ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 7 | |
0hvb | Problem:
Convex quadrilateral $ABCD$ with $BC = CD$ is inscribed in circle $\Omega$; the diagonals of $ABCD$ meet at $X$. Suppose $AD < AB$, the circumcircle of triangle $BCX$ intersects segment $AB$ at a point $Y \neq B$, and ray $\overrightarrow{CY}$ meets $\Omega$ again at a point $Z \neq C$. Prove that ray $\overr... | [
"Solution:\n\nThis is mostly just angle chasing. In this case $Y$ and $Z$ lie between $A$ and $B$, on the respective segment/arc. We'll prove $Y$ is the incenter of $\\triangle ZDB$; it will follow that ray $\\overrightarrow{DY}$ indeed internally bisects $\\angle ZDB$. It suffices to prove the following two facts:... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler ... | null | proof only | null | |
0gji | 記所有正實數所成的集合為 $\mathbb{R}_+$。試找出所有函數 $f : \mathbb{R}_+ \to \mathbb{R}_+$, 使得
$$
f(xy + x + y) + f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right) = 1
$$
$$
f(xy + x + y) + f\left(\frac{1}{x}\right) f\left(\frac{1}{y}\right) = 1 \quad \text{for every } x, y \in \mathbb{R}_+.
$$
對所有 $x, y \in \mathbb{R}_+$ 均成立。
Let $... | [
"$f(x) = \\frac{-1+\\sqrt{5}}{2}$ for all $x \\in \\mathbb{R}_+$, or $f(x) = \\frac{x}{x+1}$ for all $x \\in \\mathbb{R}_+$.\n\nDenote the functional equality by $P(x, y)$. Then $P((w+1)^{-1}, w)$ implies\n$$\nf(w+1) (f(w^{-1}) + 1) = 1. \\qquad (1)\n$$\nBy $P(xy + x + y, z)$ and (1), we have\n$$\n1 - f((x+1)(y+1)(... | Taiwan | IMO 2J, Independent Study 2 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Chinese; English | proof and answer | f(x) = (-1 + sqrt(5)) / 2 for all x > 0, or f(x) = x / (x + 1) for all x > 0 | |
0fsw | Problem:
Auf einer kreisförmigen Rennbahn ist an $n$ verschiedenen Positionen je ein Auto startbereit. Jedes von ihnen fährt mit konstantem Tempo und braucht eine Stunde pro Runde. Sobald das Startsignal ertönt, fährt jedes Auto sofort los, egal in welche der beiden möglichen Richtungen. Falls sich zwei Autos begegnen... | [
"Solution:\n\nNehme an, jedes Auto führt eine Fahne mit sich. Wenn sich zwei Autos begegnen, tauschen sie die Fahnen aus. Die Fahnen ändern also nie ihre Bewegungsrichtung und sind daher nach einer Stunde alle wieder an ihrem ursprünglichen Platz. Das bedeutet aber, dass auch die $n$ Autos wieder alle auf den Start... | Switzerland | IMO - Selektion | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
01jl | For a prime number $p$ and a polynomial $f$ with integer coefficients, define $\text{Im}(p, f)$ be the set of integers $a \in \{0, 1, \dots, p-1\}$ such that there exists an integer $x$, for which $f(x) - a$ is divisible by $p$.
Prove that there exist nonconstant polynomials $f$ and $g$ such that, for infinitely many ... | [
"We take $f(x) = (x^2 + 1)^2$ and $g(y) = -(y^2 + 1)^2$ and prove that if $p \\equiv 3 \\pmod 4$ then the equation $f(x) \\equiv g(y) \\pmod p$ has no solution. Famously, there are infinitely many primes congruent to $3$ modulo $4$.\n\nRecall the fact that if $p \\equiv 3 \\pmod 4$ then the only solution to the equ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof only | null | |
0747 | Let $ABC$ be a triangle in which $AB > AC$; $AM$ be the median and $AK$ be the angle bisector with $M, K$ on $BC$. Let $L$ be a point on $AM$ such that $KL$ is parallel to $AC$. Prove that $CL$ is perpendicular to $AK$. | [
"\n\nExtend $AK$ to meet the circum-circle of $ABC$ in $D$, and join $MD$. Let $P$ be the point of intersection of $AK$ and $CL$. Observe that $\\angle DMK = 90^\\circ$ and $D, M, O$ are collinear. We show that $DMK$ is similar to $CPK$, which proves that $CL$ is perpendicular to $AK$. It i... | India | Indija TS 2009 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07fo | Let $k \in \mathbb{Z}$ prove that there are infinitely many pairs of distinct positive integer numbers $n, m$ such that
$$
\begin{aligned}
n + S(2n) &= m + S(2m), \\
kn + S(n^2) &= km + S(m^2),
\end{aligned}
$$
where $S(n)$ is the sum of the digits of $n$ to base 10. | [
"Let $P_k$ be the set of solutions of\n$$\n\\begin{cases} n + S(n) = m + S(m), \\\\ kn + S(kn) = km + S(km). \\end{cases}\n$$\nWe want to map a single solution $(m_0, n_0) \\in P_k$, where $10 \\nmid m_0, n_0$, to infinite solutions like $(m_1, n_1)$, where $(m_1, n_1) \\in P_{k+1}$ or $(m_1, n_1) \\in P_{k-1}$ and... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
08hk | Problem:
The plane $\alpha$ is tangent in the points $A_{1}$, $A_{2}$ and $A_{3}$ to three spheres with different radii $R_{1}$, $R_{2}$ and $R_{3}$ respectively, situated in the same halfspace two by two exteriorly. The plane $\beta$ is parallel to the plane $\alpha$ and intersects all three spheres so that the circl... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | min(R1, R2, R3) | |
0eem | Let $ABC$ be an acute triangle and let $D$ be a point in the interior of the triangle, such that $\angle BAD = \angle DCB$ and $\angle CBD = \angle DAC$. Prove that the lines $AD$ and $BC$ are perpendicular. | [
"Let $E$ denote the intersection of the lines $AD$ and $BC$, let $F$ denote the intersection of $BD$ and $CA$ and let $G$ denote the intersection of $CD$ and $AB$. The equality $\\angle BAD = \\angle DCB$ implies that the triangles $GAD$ and $ECD$ are similar since they have two common angles. So,\n$$\n\\frac{|GD|}... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0j67 | Problem:
Let $\{a_{n}\}$ and $\{b_{n}\}$ be sequences defined recursively by $a_{0}=2 ; b_{0}=2$, and $a_{n+1}=a_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}-b_{n}$; $b_{n+1}=b_{n} \sqrt{1+a_{n}^{2}+b_{n}^{2}}+a_{n}$. Find the ternary (base 3) representation of $a_{4}$ and $b_{4}$. | [
"Solution:\n\nNote first that $\\sqrt{1+a_{n}^{2}+b_{n}^{2}}=3^{2^{n}}$. The proof is by induction; the base case follows trivially from what is given. For the inductive step, note that\n$$\n1+a_{n+1}^{2}+b_{n+1}^{2}=1+a_{n}^{2}(1+a_{n}^{2}+b_{n}^{2})+b_{n}^{2}-2 a_{n} b_{n} \\sqrt{1+a_{n}^{2}+b_{n}^{2}}+b_{n}^{2}(... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | a4 = 1000001100111222 (base 3); b4 = 2211100110000012 (base 3) | |
06wy | A $\pm 1$-sequence is a sequence of $2022$ numbers $a_{1}, \ldots, a_{2022}$, each equal to either $+1$ or $-1$. Determine the largest $C$ so that, for any $\pm 1$-sequence, there exists an integer $k$ and indices $1 \leqslant t_{1}<\ldots<t_{k} \leqslant 2022$ so that $t_{i+1}-t_{i} \leqslant 2$ for all $i$, and
$$
\l... | [
"First, we prove that this can always be achieved. Without loss of generality, suppose at least $\\frac{2022}{2}=1011$ terms of the $\\pm 1$-sequence are $+1$. Define a subsequence as follows: starting at $t=0$, if $a_{t}=+1$ we always include $a_{t}$ in the subsequence. Otherwise, we skip $a_{t}$ if we can (i.e. i... | IMO | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 506 | |
0bic | Determine the prime numbers $p$ and $q$ that satisfy the equality
$$
p^3 + 107 = 2q(17q + 24).
$$ | [
"For $q = 2$ we obtain $p = 5$, which is a prime. For $q \\ge 3$, reducing modulo $4$, we get $p^3 \\equiv 3 \\pmod 4$, which leads to $p \\equiv 3 \\pmod 4$. The equation reduces to $p^3 + 125 = 34q^2 + 48q + 18$, or, equivalently,\n$$\n(p+5)(p^2-5p+25) = 2[q^2 + (4q+3)^2].\n$$\nAs $p^2 - 5p + 25 \\equiv 3 \\pmod ... | Romania | 65th NMO Selection Tests for JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial o... | null | proof and answer | (p, q) = (5, 2) and (7, 3) | |
09t2 | Problem:
Laat $a$, $b$ en $c$ positieve gehele getallen zijn, allemaal verschillend, en veronderstel dat $p = a b + b c + c a$ een priemgetal is.
a) Bewijs dat $a^{2}$, $b^{2}$ en $c^{2}$ verschillende resten geven bij deling door $p$.
b) Bewijs dat $a^{3}$, $b^{3}$ en $c^{3}$ verschillende resten geven bij deling door... | [
"Solution:\na) We bewijzen dit uit het ongerijmde. Stel dat twee van $a^{2}$, $b^{2}$ en $c^{2}$ dezelfde rest geven bij deling door $p$, zeg dat $a^{2} \\equiv b^{2} \\bmod p$. Dan geldt $p \\mid a^{2}-b^{2} = (a-b)(a+b)$, dus $p \\mid a-b$ of $p \\mid a+b$. In het laatste geval geldt $p \\leq a+b \\leq c(a+b) < a... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0c4g | Problem:
În $n$ cutii transparente se introduc bile roşii şi bile albastre. Trebuie alese 50 de cutii astfel încât ele să conţină împreună cel puţin jumătate din bilele roşii şi cel puţin jumătate din bilele albastre. Este întotdeauna posibilă o asemenea alegere indiferent de numărul bilelor şi de modul în care au fos... | [
"Solution:\n\na. La 100 cutii: în 25 de cutii pun câte o bilă roşie, iar în celelalte 75 câte o bilă albastră. Pentru a alege cel puţin jumătate din bilele roşii trebuie alese 13 cutii care conţin bile roşii, iar pentru a alege cel puţin jumătate din bilele albastre, trebuie să alegem cel puţin 38 de cutii care con... | Romania | Al cincilea test de selecţie pentru OBMJ | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | a) No; b) Yes | |
0amm | Problem:
Find the value/s of $k$ so that the inequality $k\left(x^{2}+6x-k\right)\left(x^{2}+x-12\right)>0$ has solution set $(-4,3)$. | [] | Philippines | AREA STAGE | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | k < -9 | |
0g04 | Problem:
Den unten abgebildeten Spielstein nennen wir eine Treppe. Für welche Paare $(m, n)$ natürlicher Zahlen mit $m, n \geq 6$ ist es möglich, ein $m \times n$ Feld lückenlos und überlappungsfrei mit Treppen zu bedecken?
 | [
"Solution:\n\nComme l'escalier a une taille de 6, on conclue que $6 \\mid m n$ et donc, sans perte de généralité, $2 \\mid n$. Faisons une coloration par ligne en deux couleurs (noir et blanc) perpendiculaire au côté de longueur paire. Il y a donc autant de cases noires que de cases blanches. Or chaque escalier cou... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | (m, n) are exactly those pairs with m, n ≥ 6 such that either one side is a multiple of 12, or one side is a multiple of 3 and the other is a multiple of 4. Equivalently: (12a, b) with a ≥ 1, b ≥ 6; (3c, 4d) with c ≥ 2, d ≥ 3; and their permutations. | |
0464 | Does there exist an irrational number $x$ such that there are at most finitely many positive integers $n$ satisfying
$$
\{kx\} \geq \frac{1}{n+1}
$$
for every $k \in \{1, \dots, n\}$?
Note: Here, for a positive real number $y$, $\{y\}$ denotes the fractional part of $y$. | [
"*Proof* 1. Nonexistence. Assume that there exists such a positive integer $n$. Let $x$ be an irrational number. Since the terms of the sequence $\\{x\\}$, $\\{2x\\}$, $\\dots$ are distinct and densely distributed in the interval $(0, 1)$, there exist infinitely many positive integers $d$ satisfying\n$$\n\\{dx\\} =... | China | China-TST-2023B | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | No; such an irrational number does not exist. For every irrational number, there are infinitely many positive integers n satisfying the condition. | |
0gdz | 設實數 $u_1, ..., u_n$ ($n \ge 3$) 滿足：
$$
\sum_{i=1}^{n} u_i^{2018} = 1, \quad \sum_{i=1}^{n} u_i^{2019} = 0.
$$
證明：必存在 $1 \le k_1 < k_2 \le n$ 滿足：
$$
\sum_{i=1}^{n} u_i^{2020} \le |u_{k_1} u_{k_2}|.
$$
Let $n \ge 3$ and $u_1, ..., u_n$ be real numbers satisfying
$$ \sum_{i=1}^{n} u_i^{2018} = 1, \quad \sum_{i=1}^{n} u_i... | [
"將原不等式記為(*)。不失一般性，我們假設 $u_i \\le u_{i+1}$, $i = 1, ..., n-1$. 令\n$$\nP = \\{i : u_i > 0\\}, \\quad Q = \\{i : u_i \\le 0\\}.\n$$\n根據條件，顯然 $u_n \\in P$, $u_1 \\in N$, $u_1 < 0$, 且\n$$\n\\max_{i \\in P} u_i = u_n, \\quad \\max_{i \\in N} |u_i| = |u_1|. \\quad (1)\n$$\n\n首先，由 $\\sum_{i=1}^{n} u_{i}^{2019} = 0$ 我們可得到\n... | Taiwan | 2020 Taiwan IMO 1J | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
098k | Problem:
Pentru orice număr natural $m$ notăm cu $S(m)$ suma cifrelor numărului $m$. Calculați $S\left(S\left(S\left(2023^{2023}\right)\right)\right)$. | [
"Solution:\n\nPentru orice număr natural $m$ notăm prin $S(m)$ suma cifrelor și prin $N(m)$ numărul cifrelor ale numărului $m$. Avem $2023^{2023}<\\left(10^{4}\\right)^{2023}=10^{8092}$, ceea ce implică $N\\left(2023^{2023}\\right) \\leq 8092$. Numărul $2023^{2023}$ are nu mai mult de 8092 cifre, iar suma cifrelor ... | Moldova | Olimpiada Republicană la Matematică | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 7 | |
02yw | Problem:
Ache todos os valores de $x$ satisfazendo
$$
\frac{x+\sqrt{x+1}}{x-\sqrt{x+1}}=\frac{11}{5}
$$ | [
"Solution:\n\nA equação pode ser reescrita como\n$$\n\\begin{aligned}\n\\frac{x+\\sqrt{x+1}}{x-\\sqrt{x+1}} &= \\frac{11}{5} \\\\\n5x + 5\\sqrt{x+1} &= 11x - 11\\sqrt{x+1} \\\\\n16\\sqrt{x+1} &= 6x \\\\\n8\\sqrt{x+1} &= 3x\n\\end{aligned}\n$$\nElevando os membros da última equação ao quadrado, temos\n$$\n\\begin{al... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | 8 | |
03oo | $3n$ ($n$ is a positive integer) girl students took part in a summer camp. There were three girl students to be on duty every day. When the summer camp ended, it was found that any two of the $3n$ girl students had just one time to be on duty on the same day.
(1) When $n = 3$, is there any arrangement satisfying the r... | [
"(1) When $n = 3$, there is an arrangement satisfying the requirement. Now the concrete arrangement is given as follows (Denote the nine girl students by $1, 2, \\ldots, 9$):\n\n$$\n(1, 2, 3), (1, 4, 5), (1, 6, 7), (1, 8, 9), \\\\\n(2, 4, 6), (2, 7, 8), (2, 5, 9), (3, 4, 8), \\\\\n(3, 5, 7), (3, 6, 9), (4, 7, 9), (... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
048c | If one half of a working crew finishes one quarter of a job in one third of a day, how many crews does it take to finish 15 jobs in five days? | [] | Croatia | Hrvatska 2011 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 2 | |
0h79 | There are 22 cards, where the numbers $1, 2, \ldots, 22$ are written. Using these cards one formed 11 fractions. What is the greatest possible number of integer numbers among the fractions? | [
"The numbers $13$, $17$, $19$ may form an integer only if they stand at the numerator position and $1$ stands in the denominator position. Hence, at least one fraction cannot be integer. However, $10$ numbers may occur to be integer:\n$$\n\\frac{22}{11}, \\frac{14}{7}, \\frac{15}{5}, \\frac{21}{3}, \\frac{20}{10}, ... | Ukraine | 56th Ukrainian National Mathematical Olympiad, Third Round | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 10 | |
0bxw | Let $ABCDA'B'C'D'$, be a cube with side length $a$. Let $M$ and $P$ be the midpoints of the edges $[AB]$ and $[DD']$, respectively.
a) Prove that $MP \perp A'C$.
b) Find the distance between the lines $MP$ and $A'C$. | [
"Let $O$ be the midpoint of $[A'C]$.\n\na) Triangle $MA'C$ is isosceles, hence $MO \\perp A'C$.\nSimilarly $PA'C$ is isosceles, therefore $PO \\perp A'C$, yielding $A'C \\perp (PMO)$, hence $A'C \\perp MP$.\n\n\n\nb) Let $S$ be the midpoint of $[MP]$. The triangles $MA'C$ and $PA'C$ are con... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems"
] | English | proof and answer | a√2/4 | |
0b5l | Let $A$, $B$, $C$ be nodes of the lattice $\mathbb{Z} \times \mathbb{Z}$ such that inside the triangle $ABC$ lies a unique node $P$ of the lattice. Denote $E := AP \cap BC$. Determine $\max \frac{AP}{PE}$, over all such configurations. | [
"Let us build $E'$, the symmetrical of $E$ with respect to $P$. If $E$ is latticeal, then $E'$ is latticeal, and this implies $E' \\notin \\text{int}(ABC)$. We deduce that $E' = A$, otherwise $A \\in (E'P)$ and through a translation we get that there exists a latticeal point lying on $PE$, contradiction. So in this... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 5 | |
0ihm | Problem:
How many positive integers $x$ are there such that $3x$ has 3 digits and $4x$ has four digits? | [
"Solution:\n\nNote that $x$ must be between $250$ and $333$, inclusive. There are $84$ integers in that interval."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 84 | |
0em1 | Prove that
$$(x^3 + x^2 + 3)^2 > 4x^3(x - 1)^2$$
for all real $x$. | [
"First observe that if $x$ is negative the inequality is trivial since the left hand side is a square and hence larger than or equal to zero. The right hand side would be negative if $x$ is negative.\n\nNow assume $x \\in \\mathbb{R}^+$. Apply the AM-GM inequality in $x^3$ and $(x-1)^2$:\n$$\n\\begin{aligned}\n\\sq... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
09ci | $f: \mathbb{N} \rightarrow \mathbb{R}$ ба
$$
\forall n \in \mathbb{N} : \sqrt{f(n+2)+2} \leq f(n) \leq 2 \text{ тэнзэтгэл биш биелж байх бүх } f \text{ функцийг ол.}
$$ | [
"$0 \\leq f(n) \\leq 2$, $\\forall n \\in \\mathbb{N}$\n$$\n\\Rightarrow f(n) = 2 \\cos g(n),\\ g(n) \\in [0, \\frac{\\pi}{2}] \\text{ гэж үзэж болно.}\n$$\n$$\n\\sqrt{f(n+2)+2} \\leq f(n) \\text{ ба } \\cos 2\\alpha + 1 = 2\\cos^2 \\alpha\n$$\n$$\n\\Rightarrow \\cos \\frac{g(n+2)}{2} \\leq \\cos g(n) \\text{ ба } ... | Mongolia | ОУМО-53 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Mongolian | proof and answer | f(n) = 2 for all natural numbers n | |
081h | Problem:
Sia $ABC$ un triangolo e sia $\gamma$ la circonferenza inscritta in $ABC$. La circonferenza $\gamma$ è tangente al lato $AB$ nel punto $T$. Sia $D$ il punto di $\gamma$ diametralmente opposto a $T$, e sia $S$ il punto di intersezione della retta passante per $C$ e $D$ con il lato $AB$.
Dimostrare che $AT = SB... | [
"Solution:\n\nCon riferimento alla figura a fianco, tracciamo la retta $r$ passante per $D$ e parallela al lato $AB$. Siano $L$ ed $M$, rispettivamente, le intersezioni di $r$ con i lati $AC$ e $BC$. Denotiamo inoltre con $H$ e $K$, rispettivamente, i punti di tangenza di $\\gamma$ con i lati $AC$ e $BC$. Poiché su... | Italy | Gara Nazionale di Matematica | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
01it | The Perseverance is NASA's Mars rover exploring this planet. Every day it starts in home base and then goes north, south, east or west. Every one kilometer the robot makes one $90^\circ$ turn. Moreover, the robot doesn't want to check the same place twice during the day (except for the home base, which is always the st... | [
"Let's define the coordinate system with the origin point at the home base and vertical-horizontal axes. W.l.o.g. assume that the first move was east and the path had length of $n$. Then each odd move changed the $x$ coordinate of the robot by $1$ and each even move changed the $y$ coordinate by $1$.\n\nAt the end ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | All positive multiples of four | |
0le3 | Let $f: \mathbb{R} \to (0; +\infty)$ be a continuous function such that $\lim_{x \to -\infty} f(x) = \lim_{x \to +\infty} f(x) = 0$.
a) Prove that $f(x)$ obtains the maximum value on $\mathbb{R}$.
b) Prove that there exist two sequences $(x_n), (y_n)$ with $x_n < y_n$ for all positive integers $n$ such that they have... | [
"a) We have a well-known theorem: If the function $f(x)$ is continuous on the segment $[a; b]$ then $f(x)$ reaches the maximum value and minimum value at some point on that segment.\n\nConsider the number $f(0) > 0$, because $\\lim_{x \\to -\\infty} f(x) = \\lim_{x \\to +\\infty} f(x) = 0$ then there exist values $... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof only | null | |
0ccz | a) Let $A, B \in \mathcal{M}_n(\mathbb{C})$ be two matrices such that $A^2B = A$. Prove that
$$
(AB - BA)^2 = O_n.
$$
b) Show that, for any natural number $k \le n/2$, there are two matrices $A$, $B \in \mathcal{M}_n(\mathbb{C})$ with the property $A^2B = A$, such that $\mathrm{rank}(AB - BA) = k$. | [
"a) If $A$ is invertible or $A = O_n$ then clearly $AB - BA = O_n$. Assume $A \\ne O_n$, with $\\det(A) = 0$. Let $P \\in \\mathbb{C}[X]$ be the minimal polynomial of the matrix $A$. Since $P(0) = 0$ and $P \\ne X$, the polynomial $P$ has the form $P = X^k + a_{k-1}X^{k-1} + \\dots + a_1X$, where $2 \\le k \\le n$.... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
047x | Let $k$, $l$, $n$ be positive integers, and let $a_1, a_2, \dots, a_k \in \{1, 2, \dots, n\}$ satisfy the following three conditions:
(1) $n \ge 3$, $l \le n-2$, and $l-k \le \frac{n-3}{2}$;
(2) For each $t \in \{1, 2, \dots, l\}$, there exists a non-empty subset $I \subseteq \{1, 2, \dots, k\}$ such that
$$
\sum_{i \i... | [
"**Proof:** For a finite multiset $S$ of integers, let $\\sigma(S)$ denote the sum of all elements in $S$ (counting multiplicities), and let $\\Sigma(S) = \\{\\sigma(T) \\mid \\sigma(T) \\neq T \\subseteq S\\}$, where $\\Sigma(S)$ is treated as a set (ignoring multiplicities). Let $\\Sigma_n(S)$ denote the set of c... | China | China-TST-2025A | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0f91 | Problem:
$N$ is the set of positive integers. Does there exist a function $f: N \to N$ such that $f(n + 1) = f(f(n)) + f(f(n + 2))$ for all $n$? | [] | Soviet Union | 23rd ASU | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No | |
0le6 | Given an acute, scalene triangle $ABC$ with circumcircle $(O)$ and orthocenter $H$. Let $M$, $N$ and $P$ be the midpoints of $BC$, $CA$ and $AB$ and $D$, $E$ and $F$ be the feet of the altitudes from $A$, $B$ and $C$ of triangle $ABC$. Let $K$ be the reflection of $H$ through $BC$. Two lines $DE$, $MP$ intersect at $X$... | [
"a) First, applying Pascal's theorem for 6 points ($DPN$, $MEF$), we get the intersections of pairs of lines ($DE, MP$); ($DF, MN$); ($PF, NE$) collinear or $A$, $X$ and $Y$ are collinear.\n\n\n\nClearly, $180^\\circ - \\angle BAC = \\angle BHC = \\angle BKC$ so $K$ lies on $(O)$. Next, we ... | Vietnam | VMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geom... | English | proof only | null | |
07y1 | Prove that $1$ is the only positive value of $r$ for which there is a two-way infinite sequence $a(n)$ such that
$$
a(-n) = 1 - \sum_{k=0}^{n} r^k a(n-k), \ n = 0, \pm 1, \pm 2, \dots
$$
Determine $a(n)$, $n = 0, \pm1, \pm2, \dots$, when $r = 1$. | [
"Suppose, for some $r > 0$, a two-way infinite sequence $a(n)$ satisfies\n$$\na(-n) = 1 - \\sum_{k=0}^{n} r^k a(n-k), \\quad \\text{for all } n \\in \\mathbb{Z}. \\qquad (1)\n$$\n\nFor $n = 0$ this means $a(0) = 1 - a(0)$, hence $a(0) = \\frac{1}{2}$. For $n = \\pm 1$ we get\n$$\na(1) + a(-1) = 1 - r^{-1}a(0) \\qua... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Only r = 1 works. For r = 1, a(n) = 1/2^{|n|+1} for all integers n. | |
07wq | Let $a$, $b$, $c$ be positive real numbers with $a \le c$ and $b \le c$. Prove that
$$
(a + 10b)(b + 22c)(c + 7a) \ge 2024 \quad abc.
$$ | [
"By the AM-GM inequality for eleven numbers (or weighted AM-GM) we get $a + 10b \\ge 11\\sqrt[11]{ab^{10}}$. Similarly, we get\n$$\nb + 22c \\ge 23\\sqrt[23]{bc^{22}} \\quad \\text{and} \\quad c + 7a \\ge 8\\sqrt[8]{ca^7}.\n$$\nMultiplying these three inequalities gives\n$$\n(a + 10b)(b + 22c)(c + 7a) \\ge 2024 a^{... | Ireland | IRL_ABooklet_2024 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09w1 | We consider numbers with two digits (the first digit cannot be $0$). Such a number is called *vain* if the sum of the two digits is greater than or equal to the product of the two digits. For example, the number $36$ is *not* vain, as $3 + 6$ is smaller than $3 \cdot 6$.
How many numbers with two digits are vain?
A) $1... | [
"D) $27$"
] | Netherlands | First Round | [
"Discrete Mathematics > Combinatorics",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | D | |
0211 | Problem:
Let $n \geqslant 0$ be an integer, and let $a_{0}, a_{1}, \ldots, a_{n}$ be real numbers. Show that there exists $k \in \{0,1, \ldots, n\}$ such that
$$
a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{n} x^{n} \leqslant a_{0}+a_{1}+\cdots+a_{k}
$$
for all real numbers $x \in[0,1]$. | [
"Solution:\nThe case $n=0$ is trivial; for $n>0$, the proof goes by induction on $n$. We need to make one preliminary observation:\nClaim. For all reals $a, b$, $a+b x \\leqslant \\max \\{a, a+b\\}$ for all $x \\in[0,1]$.\nProof. If $b \\leqslant 0$, then $a+b x \\leqslant a$ for all $x \\in[0,1]$; otherwise, if $b... | Benelux Mathematical Olympiad | Benelux Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fpl | Un conjunto de números enteros positivos se llama fragante si contiene al menos dos elementos, y cada uno de sus elementos tiene algún factor primo en común con al menos uno de los elementos restantes. Sea $P(n) = n^2 + n + 1$. Determinar el menor número entero positivo $b$ para el cual existe algún número entero no ne... | [] | Spain | LVII Olimpiada Internacional de Matemáticas | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof and answer | 6 | |
0hg5 | Teacher wrote on the board 5 distinct numbers. After that Petrik counted the sums of each two of these numbers and wrote them on the left half of the board, Vasyl did the same for the sums of each three of these numbers and wrote them on the right half of the board. Could the teacher write such numbers so that the sets... | [
"It's enough to choose the following numbers: $-2, -1, 0, 1, 2$. Then we can write down the sets of integers, but we can also apply the following reasoning: for any two numbers, say, $a, b$, selected by Petrik, from one side exists pair of numbers $(-a, -b)$, whose sum is the opposite to the initial, and, from anot... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes; for example, −2, −1, 0, 1, 2. | |
0gja | Let $ABC$ be an acute, scalene triangle with orthocenter $H$. Let $l_a$ be the line through the reflection of $B$ with respect to $CH$ and the reflection of $C$ with respect to $BH$. Lines $l_b$ and $l_c$ are defined similarly. Suppose lines $l_a, l_b$, and $l_c$ determine a triangle. Prove that the orthocenter and cir... | [
"\nDenote by $A_b, A_c$ the reflections of $A$ in $BH$ and $CH$ respectively. $B_c, B_a$ and $C_a, C_b$ are defined similarly. By definition, $l_a = B_cC_b, l_b = C_aA_c, l_c = A_bB_a$. Let $A_1 = l_b \\cap l_c, B_1 = l_c \\cap l_a, C_1 = l_a \\cap l_b$ and let $H_1, O_1$ be the orthocentre... | Taiwan | IMO 3J, Independent Study 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle ... | Chinese; English | proof only | null | |
00dh | Determine the number of permutations $a_1, a_2, \ldots, a_{2021}$ of the numbers $2, 3, \ldots, 2022$ such that $a_k$ is divisible by $k$, for all $k = 1, 2, \ldots, 2021$. | [
"The answer is $13$.\n\nThere is an $m_0 \\in \\{1, 2, \\dots, 2021\\}$ such that $a_{m_0} = 2022$, $m_0$ being a divisor of $2022$.\n\nIf $m_0 = 1$, we have $a_1 = 2022$. In this case, we must have $a_k = k$, for all $k = 2, 3, \\dots, 2021$. Indeed, for a given $k \\in \\{2, 3, \\dots, 2021\\}$, suppose that $a_{... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 13 | |
01jj | For a positive integer $m$, let $S(m)$ denote the positive integer whose decimal representation is equal to the octal representation of $m$. For example, $S(64) = 100$, $S(100) = 144$, $S(2023) = 3747$. Two positive integers $x, y$ satisfy the relation
$$
\frac{S(x) + S(1)}{S(x + 1)} = \frac{S(y) + S(2023)}{S(y + 2023)... | [
"The answer is 2137.\n\nFirst of all, let's observe that $x = 63$, $y = 2137$ satisfies the given relation. In fact,\n$$\n\\frac{S(63) + S(1)}{S(64)} = \\frac{77 + 1}{100} = \\frac{78}{100} < \\frac{8}{10} \\quad \\text{and} \\\\\n\\frac{S(2137) + S(2023)}{S(4160)} = \\frac{4131 + 3747}{10100} = \\frac{7878}{10100}... | Baltic Way | Baltic Way 2023 Shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Other"
] | English | proof and answer | 2137 | |
00af | In basketball the free-throw rate (FRT) of a player is the ratio of the number of his successful free throws to the number of all of his free throws. After the first half of a game Mateo's FRT was less than $75\%$, and at the end of the game it was greater than $75\%$. Can one claim with certainty that there was a mome... | [
"The answer is yes for $75\\%$ and no for $60\\%$.\nLet the FRT was less than $75\\%$ after the first half but eventually greater than $75\\%$. Then there is a successful free throw in the second half such that after it the FRT became at least $75\\%$. Consider the first such free throw $S$. We claim that after $S$... | Argentina | Argentine National Olympiad 2016 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | Yes for 75%; No for 60% | |
00te | Angel has a warehouse, which initially contains $100$ piles of $100$ pieces of rubbish each. Each morning, Angel either clears every piece of rubbish from a single pile, or one piece of rubbish from each pile. However, every evening, a demon sneaks into the warehouse and adds one piece of rubbish to each non-empty pile... | [
"We will show that he can do so by the morning of day $199$ but not earlier.\n\nIf we have $n$ piles with at least two pieces of rubbish and $m$ piles with exactly one piece of rubbish, then we define the value of the pile to be\n$$\nV = \\begin{cases} n & m = 0, \\\\ n + \\frac{1}{2} & m = 1, \\\\ n + 1 & m \\ge 2... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 199 | |
0i80 | Problem:
There are 10 cities in a state, and some pairs of cities are connected by roads. There are 40 roads altogether. A city is called a "hub" if it is directly connected to every other city. What is the largest possible number of hubs? | [
"Solution:\n\nIf there are $h$ hubs, then $\\binom{h}{2}$ roads connect the hubs to each other, and each hub is connected to the other $10-h$ cities; we thus get $\\binom{h}{2} + h(10-h)$ distinct roads. So, $40 \\geq \\binom{h}{2} + h(10-h) = -h^{2}/2 + 19h/2$, or $80 \\geq h(19-h)$. The largest $h \\leq 10$ satis... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 6 | |
01cc | Consider four positive real numbers $a$, $b$, $c$ and $d$, satisfying
$$
a^2 + ab + b^2 = 3c^2 \quad \text{and} \quad a^3 + a^2b + ab^2 + b^3 = 4d^3.
$$
Prove that
$$
a + b + d \le 3c.
$$ | [
"Setting $x = \\frac{a+b}{2}$ and $y = \\frac{a-b}{2}$, we have $a = x + y$ and $b = x - y$. The given equations transform into\n$$\nc^2 = x^2 + \\frac{y^2}{3} \\quad (1)\n$$\n$$\nd^3 = x(x^2 + y^2), \\quad (2)\n$$\nand the inequality to be proved into $2x + d \\le 3c$.\n\nBy (1), we have $c \\ge x > 0$. Moreover, ... | Baltic Way | Baltic Way 2015 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
0ke6 | Problem:
Let $ABCD$ be a rectangle and $E$ be a point on segment $AD$. We are given that quadrilateral $BCDE$ has an inscribed circle $\omega_{1}$ that is tangent to $BE$ at $T$. If the incircle $\omega_{2}$ of $ABE$ is also tangent to $BE$ at $T$, then find the ratio of the radius of $\omega_{1}$ to the radius of $\om... | [
"Solution:\nLet $\\omega_{1}$ be tangent to $AD$, $BC$ at $R$, $S$ and $\\omega_{2}$ be tangent to $AD$, $AB$ at $X$, $Y$. Let $AX = AY = r$, $EX = ET = ER = a$, $BY = BT = BS = b$. Then noting that $RS \\parallel CD$, we see that $ABSR$ is a rectangle, so $r + 2a = b$. Therefore $AE = a + r$, $AB = b + r = 2(a + r... | United States | HMMT February | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | (3+sqrt(5))/2 |
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