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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0kjd | Right triangle $ABC$ has side lengths $BC = 6$, $AC = 8$, and $AB = 10$. A circle centered at $O$ is tangent to line $BC$ at $B$ and passes through $A$. A circle centered at $P$ is tangent to line $AC$ at $A$ and passes through $B$. What is $OP$?
(A) $\frac{23}{8}$ (B) $\frac{29}{10}$ (C) $\frac{35}{12}$ (D) $\frac{73}... | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
0cna | In triangle $ABC$, sides $AB$ and $BC$ have equal lengths. Point $D$ inside the triangle is chosen so that $\angle ADC = 2\angle ABC$. Prove that the distance from point $B$ to the external bisector line of angle $\angle ADC$ is twice smaller than $AD + DC$. (S. Berlov) | [
"Let $\\ell$ be the external angle bisector of the angles adjacent to $\\angle ADC$, and let $K$ be the projection of $B$ onto $\\ell$. Let points $B'$ and $C'$ be the reflections of $B$ and $C$ with respect to $\\ell$, respectively. Then $BB' = 2BK$—that is, twice the distance from $B$ to $\\ell$. Moreover, point ... | Russia | Euler olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English; Russian | proof only | null | |
0ear | Samo wrote down a 3-digit odd positive integer on a piece of paper. He then told Peter what the last digit of this number was. Peter immediately concluded that the number Samo wrote down is not prime. What was this last digit?
(A) 1
(B) 3
(C) 5
(D) 7
(E) 9 | [
"Peter can only arrive at this conclusion if the last digit is even (and the number is divisible by $2$) or equal to $5$ (and the number is divisible by $5$). All other digits can be the last digit of a three-digit prime, e.g. $101$, $103$, $107$ and $109$ are prime. Samo's number was odd, so the last digit must ha... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | MCQ | C | |
0hok | Problem:
If $x$ and $y$ are two positive numbers less than $1$, prove that
$$
\frac{1}{1-x^{2}} + \frac{1}{1-y^{2}} \geq \frac{2}{1-x y}
$$ | [
"Solution:\nFirst we use the inequality $a + b \\geq 2 \\sqrt{a b}$ and get\n$$\n\\frac{1}{1-x^{2}} + \\frac{1}{1-y^{2}} \\geq \\frac{2}{\\sqrt{(1-x^{2})(1-y^{2})}}\n$$\nNow we notice that\n$$\n(1-x^{2})(1-y^{2}) = 1 + x^{2} y^{2} - x^{2} - y^{2} \\leq 1 + x^{2} y^{2} - 2 x y = (1-x y)^{2}\n$$\nwhich implies that\n... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0bw8 | Find the numbers of the solutions $(x, y)$ of the equation
$$
4x^3 + 20x^2 + 33x = 2y^2 - 18,
$$
with $x < 2017$, $x, y \in \mathbb{Z}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 63 | |
085k | Problem:
Se $a$ è un intero positivo minore di $100$, per quanti valori di $a$ il sistema
$$
\begin{cases}
x^{2} = y + a \\
y^{2} = x + a
\end{cases}
$$
ha soluzioni intere? | [
"Solution:\n\nLa risposta è $19$. Sottraendo membro a membro le due equazioni si ha $x^{2} - y^{2} = y - x$, ovvero $(x - y)(x + y + 1) = 0$.\n\nSupponiamo che ad annullarsi sia il primo fattore, cioè che $y = x$: sostituendo, $x$ deve soddisfare l'equazione $x^{2} - x - a = 0$. Quindi $a$ deve essere tale che\n$$\... | Italy | Progetto Olimpiadi di Matematica 2007 GARA di SECONDO LIVELLO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 19 | |
042r | In a plane rectangular coordinate system $xOy$, circle $\Omega$ passes through points $(0, 0)$, $(2, 4)$, $(3, 3)$. Then the maximum of the distance from a point on circle $\Omega$ to the origin is ______. | [
"Denote $A(2, 4)$, $B(3, 3)$. Then circle $\\Omega$ passes through points $O$, $A$ and $B$. Note that $\\angle OBA = 90^\\circ$ (the slopes of lines $OB$ and $AB$ are $1$ and $-1$, respectively), so $OA$ is a diameter of circle $\\Omega$. Consequently, the maximum of the distance from a point on circle $\\Omega$ to... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 2*sqrt(5) | |
0060 | Sea $ABC$ un triángulo con $\hat{A} = 45°$ tal que la bisectriz de $\hat{A}$, la mediana desde $B$ y la altura desde $C$ concurren en un punto. Calcular la medida del ángulo $\hat{B}$. | [] | Argentina | Argentina 2008 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | Spanish | proof and answer | 22.5° | |
0inu | Problem:
Circle $\omega$ has radius $5$ and is centered at $O$. Point $A$ lies outside $\omega$ such that $OA = 13$. The two tangents to $\omega$ passing through $A$ are drawn, and points $B$ and $C$ are chosen on them (one on each tangent), such that line $BC$ is tangent to $\omega$ and $\omega$ lies outside triangle... | [] | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 31 | |
0f8k | Problem:
If rationals $x$, $y$ satisfy $x^{5} + y^{5} = 2x^{2}y^{2}$ show that $1 - x y$ is the square of a rational. | [
"Solution:\nPut $y = k x$, then $x^{5}(1 + k^{5}) = 2k^{2}x^{4}$, so $x = \\dfrac{2k^{2}}{1 + k^{5}}$, $y = \\dfrac{2k^{3}}{1 + k^{5}}$ and $1 - x y = \\dfrac{(1 - k^{5})^{2}}{(1 + k^{5})^{2}}$. $x$ and $y$ are rational, so $\\dfrac{1 - k^{5}}{1 + k^{5}}$ is rational."
] | Soviet Union | 22nd ASU | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
0dpm | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that both equations
$$
f(x+1) = 1 + f(x) \text{ and } f(x^4 - x^2) = f(x)^4 - f(x)^2
$$
simultaneously hold for all real $x$. ($\mathbb{R}$ is the set of real numbers.) | [
"Answer: $f(x) = x$ for all real $x$.\nFrom the first equation we have by induction that\n$$\nf(x+n) = f(x) + n \\quad (1)\n$$\nfor every real $x$ and integer $n$. Since $x^4 - x^2 = (x^2 - \\frac{1}{2})^2 - \\frac{1}{4}$, for every $y \\ge -\\frac{1}{4}$ there is $x$ such that $y = x^4 - x^2$, implying\n$$\nf(y) =... | Silk Road Mathematics Competition | SILK ROAD MATHEMATICS COMPETITION XVII | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | f(x) = x for all real x | |
046o | Does there exist a sequence of pairwise distinct integers $a_1, a_2, \dots$ that satisfies both of the following conditions?
a. For all positive integers $k$, we have $a_{k^2} > 0$ and $a_{k^2+k} < 0$.
b. For all positive integers $n$, we have $|a_{n+1} - a_n| \le 2023\sqrt{n}$. | [
"*Proof.* Such a sequence does not exist. We prove this by contradiction. Suppose there exists such a sequence. Take a positive integer $N$ satisfying\n$$\n\\frac{1}{N+1} + \\frac{1}{N+2} + \\dots + \\frac{1}{N^2} > 2024.\n$$\nSuch an $N$ exists because\n$$\n\\sum_{k=N+1}^{N^2} \\frac{1}{k} \\ge \\int_{N+1}^{N^2+1}... | China | China-TST-2023B | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
0cnu | A positive integer $m$ is chosen so that the sum of all the digits of $8^m$ (in its decimal representation) equals to $8$. Determine if the last digit of $8^m$ can appear to be $6$. | [
"**Первое решение.** Предположим, что сумма цифр числа $8^m$ при некотором $m > 1$ равна $8$, и оно оканчивается на $6$. Число $2^m$ не может оканчиваться на $06$ или на $26$, так как в этом случае оно не делится на $4$. Следовательно, оно оканчивается на $16$ (иначе сумма цифр будет больше $8$), и поэтому имеет де... | Russia | Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English; Russian | proof and answer | No, it cannot be 6. | |
0jy2 | Problem:
In the game Kayles, there is a line of bowling pins, and two players take turns knocking over one pin or two adjacent pins. The player who makes the last move (by knocking over the last pin) wins.
Show that the first player can always win no matter what the second player does.
(Two pins are adjacent if they a... | [
"Solution:\n\nOn their first move, the first player knocks over the middle pin (if the number of pins is odd) or two pins (if the number is even). Then, they simply mirror what the second player does, and they will have a move as long as the second player does."
] | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0chk | A natural number $n \ge 2$ is called *special* if there exist $n$ natural numbers whose sum is equal to their product.
a) Prove that $5$ is a special number.
b) Determine how many special numbers are in the set $\{2, 3, \dots, 2024\}$. | [
"a) Since $1 + 1 + 1 + 3 + 3 = 1 \\cdot 1 \\cdot 1 \\cdot 3 \\cdot 3$, there exist $5$ odd numbers whose sum is equal to their product, so $5$ is special.\n\nb) If $n$ is a special number, then there exist the odd numbers $a_1, a_2, \\dots, a_n$ such that $a_1 + a_2 + \\dots + a_n = a_1 a_2 \\dots a_n$.\nLet's assu... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 505 | |
0l0j | Three different pairs of shoes are placed in a row so that no left shoe is next to a right shoe from a different pair. In how many ways can these six shoes be lined up?
(A) 60 (B) 72 (C) 90 (D) 108 (E) 120 | [
"**Answer (A):** There are $\\binom{6}{3} = 20$ arrangements of the letters LLLRRR representing the positions of 3 left shoes and 3 right shoes in the row of 6 shoes. Of the 20, any sequence containing LRL or RLR will violate the condition given in the problem. There are 8 arrangements that avoid these two sequence... | United States | AMC 10 B | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | A | |
03o8 | Problem:
A polynomial $c_{d}x^{d} + c_{d - 1}x^{d - 1} + \dots + c_{1}x + c_{0}$ with degree $d$ is reflexive if there is an integer $n \geq d$ such that $c_{i} = c_{n - i}$ for every $0 \leq i \leq n$, where $c_{i} = 0$ for $i > d$. Let $\ell \geq 2$ be an integer and $p(x)$ be a polynomial with integer coefficients. ... | [
"Solution:\nLet $d$ be the degree of $p$ and let $k$ be any non-negative integer. We will choose \n$$q(x) = \\frac{x^{d + k + \\ell}p\\left(\\frac{1}{x}\\right) - p(x)}{x - 1},$$ \n$$r(x) = \\frac{p(x) - x^{d + k}p\\left(\\frac{1}{x}\\right)}{x - 1}.$$ \nFirst, we must show that both $q$ and $r$ are integer polyn... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0bvf | Let $ABC$ and $ADC$ be isosceles right triangles with $AB = BC = CD = DA$ and $B \neq D$. Consider $E \in (CD)$ and $F \in AD$ such that $EC = AF$ and $A \in (DF)$. Denote $\{G\} = EF \cap AC$. Find the measures of the angles in triangle $EGB$.
Sorin Peligrad | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | The triangle EGB is right isosceles: angle at G is 90 degrees, and the angles at E and at B are 45 degrees each. | |
0cb6 | Determine the twice differentiable functions $f: \mathbb{R} \to \mathbb{R}$ that verify the relation $(f'(x))^2 + f''(x) \le 0$, for all $x \in \mathbb{R}$. | [
"Let $f: \\mathbb{R} \\to \\mathbb{R}$ be a function that verifies the conditions of the statement. Let us define the twice differentiable function $g: \\mathbb{R} \\to \\mathbb{R}$ by $g(x) = e^{f(x)}$, $x \\in \\mathbb{R}$. We have $g''(x) = e^{f(x)} \\left( (f'(x))^2 + f''(x) \\right) \\le 0$, for all $x \\in \\... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | proof and answer | All constant functions. | |
01hm | Let $\Gamma$ be a circle in the plane and $S$ be a point on $\Gamma$. Two brothers, Mario and Luigi, drive around the circle $\Gamma$ with their go-karts. They both start at $S$ at the same time, they both drive for exactly 6 minutes at constant speed counterclockwise around the track. During these 6 minutes, Luigi mak... | [
"Without loss of generality, we assume that $\\Gamma$ is the unit circle and $S = (1, 0)$. Three points are marked with bananas:\n(i) After 45 seconds, Luigi has passed through an arc with a subtended angle of $45^\\circ$ and is at the point $(\\sqrt{2}/2, \\sqrt{2}/2)$, whereas Mario has passed through an arc with... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 3 | |
0e5v | There are 63 rows of seats on an airplane. Each row has 6 seats, and they are marked with the letters $A$, $B$, $C$, $D$, $E$ and $F$. On a flight, there were 193 passengers on the airplane. Prove that on this flight there were two rows in which the seats that were occupied were marked with the same letters. | [
"Proof by contradiction: Suppose that there were no two rows in which the occupied seats were marked with the same letters. The number of subsets of the set $\\{A, B, C, D, E, F\\}$ is equal to $2^6 = 64$, which means that there was exactly one subset of $\\{A, B, C, D, E, F\\}$ that did not correspond to the occup... | Slovenia | Selection Examinations for the IMO 2012 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0cj4 | Let $n \ge 3$ be an integer and $M = \{z_1, z_2, \dots, z_n\}$ be a given set of complex numbers with the sum of the elements being different from zero. Starting from the set $M$, we repeatedly apply the following transformation: at one step, we replace each of the $n$ elements of the set from the previous step with th... | [] | Romania | 75th NMO | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | No | |
0c12 | Let $ABC$ be an acute-angled triangle with circumcenter $O$ and orthocenter $H$. Denote $P$ as the midpoint of the segment $AO$ and $S$ as the intersection of the perpendicular bisector of the segment $AO$ with $BC$. The circumcircle of the triangle $APS$ intersects the segment $OH$ at $Z$. Prove that
$$
\vec{OZ} = \fr... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
03jj | Problem:
For what values of $b$ do the equations: $1988 x^{2} + b x + 8891 = 0$ and $8891 x^{2} + b x + 1988 = 0$ have a common root? | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | -10879 and 10879 | |
05yb | Problem:
2024 élèves, tous de taille différente, doivent se placer en file indienne. Cependant, chaque élève ne souhaite pas avoir à la fois devant lui et derrière lui un élève plus petit que lui. Combien y a-t-il de façons de former une telle file indienne? | [
"Solution:\n\nRegardons l'élève de plus grande taille : s'il n'est pas placé au tout début ou à la toute fin de la file, il est entre deux élèves plus petits que lui. Ainsi l'élève le plus grand doit être placé à la fin ou au début.\n\nVia ce raisonnement, et en regardant les petits cas on peut conjecturer que si o... | France | Préparation Olympique Française de Mathématiques - ENVOI 5 : Pot-POURRI | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{2023} | |
0anm | Problem:
Exactly one of the following people is lying. Determine the liar.
Bee said, "Cee is certainly not a liar."
Cee said, "I know Gee is lying."
Dee said, "Bee is telling the truth."
Gee said, "Dee is not telling the truth."
(a) Bee
(b) Cee
(c) Dee
(d) Gee | [] | Philippines | Qualifying Round | [
"Discrete Mathematics > Logic"
] | null | MCQ | d | |
06q3 | Let $S=\{x_{1}, x_{2}, \ldots, x_{k+\ell}\}$ be a $(k+\ell)$-element set of real numbers contained in the interval $[0,1]$; $k$ and $\ell$ are positive integers. A $k$-element subset $A \subset S$ is called nice if
$$
\left|\frac{1}{k} \sum_{x_{i} \in A} x_{i}-\frac{1}{\ell} \sum_{x_{j} \in S \backslash A} x_{j}\right|... | [] | IMO | 49th International Mathematical Olympiad Spain | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
030g | Problem:
No paralelogramo $A B C D$, o ângulo $B A D$ é agudo e o lado $A D$ é menor que o lado $A B$. A bissetriz do ângulo $\angle B A D$ corta o lado $C D$ em $E$. Por $D$ se traça uma perpendicular a $A E$ que corta $A E$ em $P$ e $A B$ em $F$. Traçamos por $E$ uma perpendicular a $A E$ que corta o lado $B C$ em $... | [
"Solution:\n\na) De $\\angle D A P=\\angle P A F$ e $A P \\perp D F$, segue que o triângulo $A D F$ é isósceles e que $P$ é o ponto médio de $D F$. Como $E Q$ e $D F$ são perpendiculares a $A E$, temos que $D F \\| E Q$ e $\\angle C E Q=\\angle C D F=\\angle D F A$. No paralelogramo $A B C D$, $\\angle D A B=\\angl... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | CQ/AD = 1/2; AD = 40/3 cm | |
0e3x | Find all integers $x$, such that $9x^2 - 40x + 39$ is a power of a prime. (A positive integer $m$ is a power of a prime, if $m = p^a$ for some prime number $p$ and some non-negative integer $a$.) | [
"Let $9x^2 - 40x + 39 = p^n$ for some prime $p$ and some non-negative integer $n$. From\n$$\np^n = 9x^2 - 40x + 39 = (9x - 13)(x - 3)\n$$\nit follows that $9x - 13 = p^k$ and $x - 3 = p^l$ or $9x - 13 = -p^k$ and $x - 3 = -p^l$ for some integers $k$ and $l$, where $0 \\le l < k$ and $n = k + l$.\n\nFirst, let us so... | Slovenia | National Math Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | x = -4, 1, 4, 5 | |
02z5 | Problem:
As progressões geométricas $a_{1}, a_{2}, a_{3}, \ldots$ e $b_{1}, b_{2}, b_{3}, \ldots$ possuem a mesma razão, com $a_{1}=27$, $b_{1}=99$ e $a_{15}=b_{11}$. Encontre $o$ valor de $a_{9}$. | [
"Solution:\n\nSeja $r$ o valor da razão comum das duas progressões geométricas. Temos\n$$\n\\begin{aligned}\na_{15} & =a_{1} r^{14} \\\\\n& =27 r^{14} \\\\\nb_{11} & =b_{1} r^{10} \\\\\n& =99 r^{10}\n\\end{aligned}\n$$\nPortanto, $27 r^{14}=99 r^{10}$ e daí $3 r^{4}=11$. Finalmente\n$$\n\\begin{aligned}\na_{9} & =a... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | proof and answer | 363 | |
0h23 | Suppose that for two natural numbers $m, n$ the following equality holds
$$
m+n = [m,n] + (m,n),
$$
Where $[m,n]$ and $(m,n)$ are the least common multiple and the greatest common divisor of $m,n$ respectively. Prove that one number is divisible by another. | [
"Let $d = (m,n)$, then $m = ad$, $n = bd$, and using the formula $mn = [m,n] \\cdot (m,n)$ we get, that $[m,n] = abd$. This implies that $abd + d = ad + bd \\Leftrightarrow d(a-1)(b-1) = 0$, which is possible if either $a=1$ or $b=1$ and the result follows."
] | Ukraine | 51st Ukrainian National Mathematical Olympiad, 3rd Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof only | null | |
0b5g | Let $AA'$, $BB'$, $CC'$ be the altitudes from the vertices of an acute-angled triangle $ABC$. Points $E$ and $F$ lie on the segments $CB'$ and $BC'$ respectively, such that
$$
B'E \cdot C'F = BF \cdot CE.
$$
Prove that the quadrilateral $AEA'F$ is cyclic. | [
"The given relation writes as $\\frac{BF}{FC'} = \\frac{B'E}{EC}$. Let $M$ be a point on the side $BC$ such that $FM \\parallel CC'$. Then $\\frac{BM}{MC} = \\frac{BF}{FC'} = \\frac{B'E}{EC}$, implying $ME \\parallel BB'$. Hence the angles $\\angle AEM$, $\\angle AFM$ and $\\angle AA'M$ are right angles, therefore ... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0iuh | Problem:
Bob is writing a sequence of letters of the alphabet, each of which can be either uppercase or lowercase, according to the following two rules:
- If he had just written an uppercase letter, he can either write the same letter in lowercase after it, or the next letter of the alphabet in uppercase.
- If he had ... | [
"Solution:\n\nThe smallest possible sequence from $a$ to $z$ is $a A B C D \\ldots Z z$, which has 28 letters. To insert 4 more letters, we can either switch two (not necessarily distinct) letters to lowercase and back again (as in $a A B C c C D E F f F G H \\ldots Z z$), or we can insert a lowercase letter after ... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 376 | |
0202 | Problem:
A sequence $a_{1}, a_{2}, \ldots, a_{n}, \ldots$ of natural numbers is defined by the rule
$$
a_{n+1}=a_{n}+b_{n} \quad(n=1,2, \ldots)
$$
where $b_{n}$ is the last digit of $a_{n}$. Prove that such a sequence contains infinitely many powers of 2 if and only if $a_{1}$ is not divisible by 5 . | [
"Solution:\nFirst we can observe that:\n- If $a_{1}$ is divisible by $5$, then $a_{n}=a_{2}=0 \\pmod{10}$ for all $n \\geq 2$.\n- If $a_{1}$ is not divisible by $5$, then for $n \\geq 2$: $a_{n}$ is even, the sequence $b_{n}$ is periodic, its period is a cyclic permutation of $(2,4,8,6)$, and $a_{n+4}=a_{n}+20$.\n\... | Benelux Mathematical Olympiad | 4th Benelux Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0dhm | Let $ABC$ be an acute, non-isosceles triangle which circumcircle $(O)$. Take $M$ on segment $BC$ and $J$ on $(O)$ such that $\angle BAJ = \angle CAM$. Take a point $D$ on the segment $AM$ and denote $O_1, O_2$ as the circumcenters of triangles $ABD$ and $ACD$. The line $O_1O_2$ meets $BC$ at $T$ and $G$ is the second i... | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations >... | English | proof only | null | |
05f2 | Problem:
Trouver tous les polynômes $P$ à coefficients réels tels que pour tous réels $x, y$,
$$
x P(x)+y P(y) \geqslant 2 P(x y)
$$ | [
"Solution:\nSi $P(X) = c X$ avec $c \\geqslant 0$, $x P(x) + y P(y) = c(x^{2} + y^{2}) \\geqslant 2 c x y = 2 P(x y)$ par inégalité de la moyenne, les polynômes de la forme $c X$ avec $c \\geqslant 0$ conviennent, montrons que ce sont les seuls.\n\nSupposons $P$ non constant. En évaluant en $(x, 0)$ l'inégalité, on... | France | ENVOI 2 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | P(x) = c x with c ≥ 0 | |
0hlr | Problem:
Bob has five airplane tickets with prices $\$ 100, \$ 120, \$ 140, \$ 160$, and $\$ 180$. Bob gives an offer to Aerith: she can distribute his tickets among two bags, after which, without looking inside, Bob will randomly choose a bag and a ticket from it for Aerith to keep. What strategy should Aerith use to... | [
"Solution:\n\nLet the first bag be the one with less tickets. Let $a < b$ and $S, T$ be the numbers and prices of tickets in the respective bags. We have $a + b = 5$ and $S + T = \\$ 700$. The expected values of random tickets for each bag is then $S / a$ and $T / b$, the average of which is\n$$\n\\frac{S / a + T /... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | Put the most expensive ticket alone in one bag and all the remaining tickets in the other bag. | |
0ckd | Let $a_1, a_2, \dots, a_n, \dots$ be a sequence of positive real numbers. For every positive integer $n$, write
$$
s_n = a_1 + a_2 + \dots + a_n \quad \text{and} \quad \sigma_n = \frac{a_1}{1+a_1} + \frac{a_2}{1+a_2} + \dots + \frac{a_n}{1+a_n}.
$$
Prove that, if the $s_n$ form an unbounded sequence, then so do the $\... | [
"Clearly, the $s_n$ and the $\\sigma_n$ both form strictly increasing sequences. We will construct a sequence of positive integers $n_1 < n_2 < \\dots < n_k < \\dots$ such that $\\sigma_{n_k} > k a_1/(1+a_1)$ for all $k \\ge 2$. The conclusion then follows at once.\nLet $n_1$ be any positive integer, then use unbou... | Romania | 75th NMO Selection Tests | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
050n | A finite grid is covered with $1 \times 2$ cards in such a way that the edges of the cards match with the lines of the grid, no card lies over the edge of the grid, and every square is covered by exactly two cards. Prove that one can remove some of the cards in such a way that every square will be covered by exactly on... | [
"Choose any square covered by two cards, and choose one of these cards. Move that card to a neighbouring square, and choose the other card that is covering that square. From there we move to the next square, etc., until we return to the first square. We cannot return to any other square visited previously, since in... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07jx | Initially, numbers $2, 3, \ldots, 99$ are written on the board. In each step, one of the following operations would be performed;
i. We choose an integer $i$, $2 \le i \le 89$ and if numbers $i$ and $i+10$ are both on the board, we remove both of them from the board;
ii. We choose an integer $i$, $2 \le i \le 98$, $1... | [
"Consider a $10 \\times 10$ table and number its cells sequentially and row by row from one to $100$. The problem is equivalent to this: *if we remove two opposite corners of this table, determine the number of dominoes can be placed in the table without any overlap.* Using coloring, it follows that at least two ce... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 96 | |
0k5j | For which positive integers $b > 2$ do there exist infinitely many positive integers $n$ such that $n^2$ divides $b^n + 1$?
This problem is sort of the union of IMO 1990/3 and IMO 2000/5. | [
"The answer is any $b$ such that $b+1$ is not a power of 2. In the forwards direction, we first prove more carefully the following claim.\n\n**Claim.** If $b+1$ is a power of 2, then the only $n$ which is valid is $n=1$.\n\n*Proof.* Assume $n > 1$ and let $p$ be the smallest prime dividing $n$. We cannot have $p = ... | United States | USA TSTST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All integers b > 2 such that b + 1 is not a power of 2. | |
0dxw | Problem:
V ravnini leži 16 črnih točk, kot prikazuje slika. Najmanj koliko izmed teh točk moramo pobarvati rdeče, da ne bo obstajal kvadrat z oglišči v preostalih črnih točkah in s stranicami, vzporednimi koordinatnima osema? Odgovor utemelji.
 | [
"Solution:\n\n1. način\n\nPobarvati moramo najmanj štiri točke. V mrežo namreč lahko vrišemo štiri disjunktne kvadrate kot na prvi sliki. Ker mora biti v vsakem vsaj eno oglišče pobarvano z rdečo, potrebujemo vsaj 4 rdeče točke.\n\nPokažimo, da je to tudi dovolj. Pobarvajmo točke kot prikazuje druga slika. V mreži ... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 4 | |
0gzk | Solve the equation:
$$
(x + 1)^5 + (x + 1)^4(x - 1) + (x + 1)^3(x - 1)^2 + \\
+ (x + 1)^2(x - 1)^3 + (x + 1)(x - 1)^4 + (x - 1)^5 = 0.
$$ | [
"**Answer:** $x = 0$.\n\nMultiplying both sides by $2 = ((x + 1) - (x - 1))$ yields $(x+1)^6 - (x-1)^6 = 0$ or equivalently $(x + 1)^2 = (x - 1)^2$. Solving this equation, we obtain the unique solution $x = 0$."
] | Ukraine | 50th Mathematical Olympiad in Ukraine, Third Round (January 23, 2010) | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | x = 0 | |
02ay | Problem:
A soma dos algarismos de um número - Denotemos por $s(n)$ a soma dos algarismos do número $n$. Por exemplo $s(2345)=2+3+4+5=14$. Observemos que:
$40-s(40)=36=9 \times 4 ; 500-s(500)=495=9 \times 55 ; 2345-s(2345)=2331=9 \times 259$.
a. O que podemos afirmar sobre o número $n-s(n)$ ?
b. Usando o item anterio... | [
"Solution:\n\n(a) Observe esses dois exemplos:\n$$\n\\underbrace{2000}_{2 \\cdot 10^3}-\\underbrace{s(2000)}_{2}=1998, \\underbrace{60000}_{6 \\cdot 10^4}-\\underbrace{s(60000)}_{6}=59994\n$$\nA partir deles é fácil entender que se $a$ é um algarismos entre 1 e 9 , então $s\\left(a \\cdot 10^{k}\\right)=a$.\nDaí te... | Brazil | Nível 3 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | n − s(n) is divisible by 9; s(s(s(2^2009))) = 5 | |
0fgl | Problem:
En un triángulo acutángulo $ABC$ la bisectriz interior del ángulo $A$ corta a $BC$ en $L$ y corta la circunferencia circunscrita de $ABC$ de nuevo en $N$. Trazamos perpendiculares desde $L$ a $AB$ y $AC$, con pies $K$ y $M$, respectivamente. Demostrar que el cuadrilátero $AKNM$ y el triángulo $ABC$ tienen la ... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0hhf | For arbitrary positive numbers $a, b, c$, solve the system of equations:
$$
\begin{cases} ax^3 + by = cz^5, \\ az^3 + bx = cy^5, \\ ay^3 + bz = cx^5. \end{cases}
$$ | [
"**Answer:** $(0, 0, 0)$ and $(t, t, t)$, where $t = \\pm\\sqrt{\\frac{a+\\sqrt{a^2+4bc}}{2c}}$.\n\nSuppose that $x < y < z$ (one of the inequalities may not be strict). Then subtract the third equation from the first: $a(x^3 - y^3) + b(y - z) < 0 \\le c(z^5 - x^5)$ – a contradiction. If we assume that $x < z < y$ ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | (0, 0, 0) and (t, t, t), where t = ±sqrt((a + sqrt(a^2 + 4bc)) / (2c)) | |
0fhn | Problem:
En una reunión hay 201 personas de 5 nacionalidades diferentes. Se sabe que, en cada grupo de 6, al menos 2 tienen la misma edad. Demostrar que hay al menos 5 personas del mismo país, de la misma edad y del mismo sexo. | [
"Solution:\n\nSi en cada grupo de 6 personas, 2 son de la misma edad, sólo puede haber 5 edades diferentes, ya que, si hubiese 6 edades diferentes, eligiendo una persona de cada edad tendríamos 6 personas de edades distintas contra la hipótesis.\n\nComo $201 = 2 \\cdot 100 + 1$, debe haber al menos 101 personas del... | Spain | OME 29 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
049p | Let $\overline{AD}$ and $\overline{BE}$ be altitudes of the triangle $ABC$. Given $|AE| = 5$, $|CE| = 3$ and $|CD| = 2$, determine $|BD|$. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 10 | |
0dl0 | Let $x_1$, $x_2$, $x_3$ and $y_1$, $y_2$, $y_3$ are 6 positive real numbers such that
$$
x_1 + x_2 + x_3 = y_1y_2y_3 \text{ and } y_1 + y_2 + y_3 = x_1x_2x_3.
$$
Find the minimum value of $T = x_1y_1 + x_2y_2 + x_3y_3$. | [
"Using AM-GM, we have\n$$\nx_1 + x_2 + x_3 \\ge 3\\sqrt[3]{x_1x_2x_3} \\implies y_1y_2y_3 \\ge 3\\sqrt[3]{x_1x_2x_3}.\n$$\nSimilarly, $x_1x_2x_3 \\ge 3\\sqrt[3]{y_1y_2y_3}$. Multiplying these inequalities, side-by-side, we get\n$$\n\\sqrt[3]{(x_1x_2x_3 \\cdot y_1y_2y_3)^2} \\ge 9 \\implies x_1x_2x_3 \\cdot y_1y_2y_... | Saudi Arabia | Saudi Booklet | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 9 | |
0a8g | Problem:
Let $f$ be a bounded real function defined for all real numbers and satisfying for all real numbers $x$ the condition
$$
f\left(x+\frac{1}{3}\right)+f\left(x+\frac{1}{2}\right)=f(x)+f\left(x+\frac{5}{6}\right)
$$
Show that $f$ is periodic. (A function $f$ is bounded, if there exists a number $L$ such that $|f... | [
"Solution:\n\nLet $g(6x) = f(x)$. Then $g$ is bounded, and\n$$\n\\begin{gathered}\ng(t+2) = f\\left(\\frac{t}{6} + \\frac{1}{3}\\right), \\quad g(t+3) = f\\left(\\frac{t}{6} + \\frac{1}{2}\\right) \\\\\ng(t+5) = f\\left(\\frac{t}{6} + \\frac{5}{6}\\right), \\quad g(t+2) + g(t+3) = g(t) + g(t+5) \\\\\ng(t+5) - g(t+3... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 15 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
05av | Find the smallest positive integer $n$ for which there exist two distinct pairs of positive integers $(x, y)$ such that $(x^2 - 1)(y^2 - 1) = n$ and $x \le y$. | [
"Notice that for $n = 360$, the pairs $x = 2, y = 11$ and $x = 4, y = 5$ satisfy the condition.\n\nWe will show that for smaller numbers $n$, there do not exist two distinct suitable pairs $(x, y)$. If $x = 1$, then $(x^2 - 1)(y^2 - 1)$ is not positive. Thus, we can assume that $1 < x \\le y$. Now, as $x$ or $y$ in... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 360 | |
03vr | Given $x$, $y$, $z \in (0, 1)$ satisfying
$$
\sqrt{\frac{1-x}{yz}} + \sqrt{\frac{1-y}{zx}} + \sqrt{\frac{1-z}{xy}} = 2,
$$
find the maximum value of $xyz$. (Posed by Tang Lihua) | [
"Denote $u = \\sqrt[6]{xyz}$. Then by the given condition and mean inequality,\n$$\n\\begin{aligned}\n2u^3 &= 2\\sqrt{xyz} = \\frac{1}{\\sqrt{3}}\\sum \\sqrt{x(3-3x)} \\\\\n&\\le \\frac{1}{\\sqrt{3}}\\sum \\frac{x+(3-3x)}{2} = \\frac{3\\sqrt{3}}{2} - \\frac{1}{\\sqrt{3}}(x+y+z) \\\\\n&\\le \\frac{3\\sqrt{3}}{2} - \... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 27/64 | |
08t2 | In some junior high school, a group of students are asked to plant tulip bulbs. Each participating student will be required to plant at least one bulb. And students in the same grade will be planting the same number of bulbs. There are 6 possibilities for the number less than 100 of bulbs to be planted, and the smalles... | [
"Since the smallest and the next to smallest possible numbers of the bulbs to be planted are $52$ and $64$, and since each participating student is asked to plant at least $1$ bulb, the total number of participating students should be $52$, and the number of students in the grade with the smallest number of partici... | Japan | Japan Junior Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (12, 12, 28) | |
0bhm | Let $n \in \mathbb{N}^*$ and $(G, \cdot)$ be a group with the property that there exists an endomorphism $f : G \to G$, so that
$$
f(x^n y^{n+1}) = x^{n+1} y^n \text{ for every } x, y \in G
$$ | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0k20 | Problem:
Pascal has a triangle. In the $n$th row, there are $n+1$ numbers $a_{n, 0}, a_{n, 1}, a_{n, 2}, \ldots, a_{n, n}$ where $a_{n, 0}=a_{n, n}=1$. For all $1 \leq k \leq n-1$, $a_{n, k}=a_{n-1, k}-a_{n-1, k-1}$. What is the sum of all numbers in the 2018th row? | [
"Solution:\n\nAnswer: 2\nIn general, the sum of the numbers on the $n$th row will be\n$$\n\\sum_{k=0}^{n} a_{n, k}=a_{n, 0}+\\sum_{k=1}^{n-1}\\left(a_{n-1, k}-a_{n-1, k-1}\\right)+a_{n, n}=a_{n, 0}+\\left(a_{n-1, n-1}-a_{n-1,0}\\right)+a_{n, n}=2\n$$"
] | United States | HMMT November 2018 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | final answer only | 2 | |
05xt | Problem:
Soit $a$, $b$ et $c$ trois réels strictement positifs. Démontrer que
$$
4\left(a^{3}+b^{3}+c^{3}+3\right) \geqslant 3(a+1)(b+1)(c+1) .
$$ | [
"Solution:\nDans cette inégalité, les termes en $a$, $b$ et $c$ sont additionnés les uns aux autres (et de degré $3$) dans le membre de gauche, tandis qu'ils sont multipliés les uns aux autres (et chacun de degré $1$) dans le membre de droite. Une première idée est donc d'utiliser l'inégalité arithmético-géométriqu... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
03mn | Let $a_1$, $a_2$, $\dots$, $a_n$ be positive real numbers whose product is $1$. Show that the sum
$$
\frac{a_1}{1+a_1} + \frac{a_2}{(1+a_1)(1+a_2)} + \frac{a_3}{(1+a_1)(1+a_2)(1+a_3)} + \dots + \frac{a_n}{(1+a_1)(1+a_2)\dots(1+a_n)}
$$
is greater than or equal to $\frac{2^n - 1}{2^n}$. | [
"Note that for every positive integer $m$,\n$$\n\\begin{aligned}\n\\frac{a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} &= \\frac{1+a_m}{(1+a_1)(1+a_2)\\cdots(1+a_m)} - \\frac{1}{(1+a_1)(1+a_2)\\cdots(1+a_m)} \\\\\n&= \\frac{1}{(1+a_1)\\cdots(1+a_{m-1})} - \\frac{1}{(1+a_1)\\cdots(1+a_m)}.\n\\end{aligned}\n$$\nTherefore, if we... | Canada | Kanada 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0l7u | The set of points in 3-dimensional coordinate space that lie in the plane $x + y + z = 75$ whose coordinates satisfy the inequalities
$$
x - yz < y - zx < z - xy
$$
forms three disjoint convex regions. Exactly one of those regions has finite area. The area of this finite region can be expressed in the form $a\sqrt{b}$,... | [] | United States | 2025 AIME I | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | 510 | |
02l2 | Problem:
Para a escola de bicicleta - Cátia sai da escola todos os dias no mesmo horário e volta para casa de bicicleta. Quando ela pedala a $20~\mathrm{km}/\mathrm{h}$, ela chega em casa às $4:30$ horas da tarde. Se ela pedalar a $10~\mathrm{km}/\mathrm{h}$, ela chega em casa às $5:15$ horas da tarde. A qual velocida... | [
"Solution:\n\nSeja $t$ o tempo que ela gasta pedalando a $20~\\mathrm{km}/\\mathrm{h}$. Pedalando a $10~\\mathrm{km}/\\mathrm{h}$, ela faz o percurso no dobro do tempo que pedalando a $20~\\mathrm{km}/\\mathrm{h}$, isto é, $2t$. No entanto, como ela demora 45 minutos a mais, temos:\n$$\n2t - t = 45 \\Longrightarrow... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 12 km/h | |
0fgg | Problem:
A cada vértice de un pentágono le asignamos un número entero, de forma que la suma de los cinco enteros sea positiva. Si tres vértices consecutivos tienen números asignados $x, y, z$, respectivamente, y es $y<0$, entonces se permite hacer la siguiente operación: los números $x, y, z$ se sustituyen respectivam... | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | yes | |
06cu | Prove that for any integer $n$, $n^{30} - n^{14} - n^{18} + n^2$ is divisible by $46410$. | [
"Note that $46410 = 2 \\times 3 \\times 5 \\times 7 \\times 13 \\times 17$. It suffices to check\n$$\nm = n^{30} - n^{14} - n^{18} + n^2\n$$\nis divisible by each of $p = 2, 3, 5, 7, 13, 17$.\nIf $p \\mid n$, then we must have $p \\mid m$. If $p \\nmid n$, then we have\n$$\nn^{p-1} \\equiv 1 \\pmod{p}\n$$\nby the F... | Hong Kong | HKG TST | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
07g8 | Prove that we can color all $n$-element subsets of the set $\{1, 2, \dots, 3n\}$ with eight colors, so that there are no three subsets of the same color such that every two of them have at most one element in common. | [
"Let $F$ be the family of all $n$-element subsets of $\\{1, 2, \\dots, 3n\\}$. For every $S \\subset \\{1, 2, 3, 4\\}$, denote the collection of all elements of $F$ whose intersection with $\\{1, 2, 3, 4\\}$ is $S$ by $F_S$. More formally we have\n$$\nF_S = \\{A \\in F \\mid A \\cap \\{1, 2, 3, 4\\} = S\\}.\n$$\n\n... | Iran | 38th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0go5 | A circle that passes through the vertex $A$ of a rectangle $ABCD$ intersects the side $AB$ at a second point $E$ different from $B$. A line passing through $B$ is tangent to this circle at a point $T$, and the circle with center $B$ and passing through $T$ intersects the side $BC$ at the point $F$. Show that if $\angle... | [
"Let $G$ be the point of intersection of the lines $EF$ and $DC$. Since $\\angle CFG = \\angle BFE = \\angle CDF$, the lines $DF$ and $EG$ are perpendicular.\n\n\n\nOn the other hand, $BA \\cdot BE = BT^2 = BF^2$ implies that the triangles $BAF$ and $BFE$ are similar and $\\angle BAF = \\an... | Turkey | 15th Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
01tz | The excircle $\omega_A$ of a triangle $ABC$ touches the side $BC$ at $P$. Let $I_1$ be the center of the excircle of the triangle $ABP$ touching the side $BP$, and $I_2$ be the center of the excircle of the triangle $APC$ touching the side $PC$.
Prove that the circumcircle of the triangle $I_1I_2P$ touches the circle $... | [
"We show that the circumcircle of the triangle $I_1I_2P$ touches the line $BC$ at the point $P$, then, in particular, it follows that this circle touches the circle $\\omega_A$. We use the following well-known lemma.\n\n**Lemma.** Let excircle touch the side $BC$ of the triangle $ABC$ at $A_1$ and touch the prolong... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05bz | A point $E$ is chosen on the side $AB$ of a rectangle $ABCD$ ($E \neq A, E \neq B$). The line segments $BD$ and $CE$ intersect at point $F$. Among the triangles $ADE$, $DEF$, $DCF$, $BCF$ and $BEF$, there are exactly two pairs of triangles with equal area (the order of components in a pair is not taken into account). F... | [
"Denote the area of a figure $K$ by $S_K$. As\n$$\nS_{DEF} + S_{DCF} = S_{CDE} = \\frac{1}{2} S_{ABCD} = S_{BCD} = S_{BCF} + S_{DCF},\n$$\nwe have $S_{DEF} = S_{BCF}$ (Fig. 39). Hence ($DEF$, $BCF$) is one pair of triangles with equal area regardless of the choice of point $E$.\n\nIn order to have exactly two such ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | proof and answer | sqrt(2)/2 | |
0k2l | Problem:
At lunch, Abby, Bart, Carl, Dana, and Evan share a pizza divided radially into $16$ slices. Each one takes one slice of pizza uniformly at random, leaving $11$ slices. The remaining slices of pizza form "sectors" broken up by the taken slices, e.g. if they take five consecutive slices then there is one sector... | [
"Solution:\n\nConsider the more general case where there are $N$ slices and $M>0$ slices are taken. Let $S$ denote the number of adjacent pairs of slices of pizza which still remain. There are $N-M$ slices and a sector of $k$ slices contributes $k-1$ pairs to $S$. Hence the number of sectors is $N-M-S$. We compute ... | United States | HMMT November 2018 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 11/3 | |
0gpd | Let $A$, $B$, $C$, $A'$, $B'$, $C'$ be distinct points on the plane satisfying $ABC \cong A'B'C'$ and the point $G$ be the centroid of the triangle $ABC$. If the circle of center $A'$ passing through $G$ and the circle of diameter $[AA']$ intersect at point $A_1$, the circle of center $B'$ passing through $G$ and the c... | [
"**Lemma:** Let $X$, $Y$, $Z$, $T$ be points on a plane. Then\n$$\nXY^2 + YZ^2 + YT^2 + TX^2 \\ge XZ^2 + YT^2.\n$$\n*Proof:* Let $x = \\overrightarrow{XY}$, $y = \\overrightarrow{YZ}$, $z = \\overrightarrow{ZT}$. Note that $\\overrightarrow{XZ} = x + y$, $\\overrightarrow{YT} = y + z$ and $\\overrightarrow{XT} = x ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | English | proof only | null | |
00dq | Let $n \ge 4$ and $k$ be positive integers. We consider $n$ lines on the plane such that no two of them are parallel and no three of them intersect in a single point. On each of the $\frac{n(n-1)}{2}$ intersection points of these lines there are $k$ coins. Ana and Beto play the following game: each player, in their tur... | [
"We will prove that Ana has a winning strategy if and only if the total number of coins is odd. It is easy to see that this happens if and only if $k$ is odd and $n \\equiv 2$ or $3$ \\pmod{4}$.\n\nFor the rest of the solution we think of the lines as numbered from $1$ to $n$ and denote by $p_{ij}$ the intersection... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | Ana wins if and only if the total number of coins is odd, equivalently when k is odd and n is congruent to 2 or 3 modulo 4; otherwise Beto wins. | |
0esu | Determine the smallest positive integer $n$ whose prime factors are all greater than $18$, and that can be expressed as $n = a^3 + b^3$ with positive integers $a$ and $b$. | [
"We can factorise $n$ as\n$$\nn = a^3 + b^3 = (a + b)(a^2 - ab + b^2).\n$$\nThe first factor $a + b$ has to be at least $19$, since $n$ would otherwise contain a prime factor that is smaller than $18$. Setting $a + b = s$, we obtain\n$$\na^2 - ab + b^2 = a^2 - a(s - a) + (s - a)^2 = 3a^2 - 3as + s^2 = 3\\left(a - \... | South Africa | The South African Mathematical Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1843 | |
0g39 | Problem:
Das Dorf Roche hat 2020 Einwohner. Eines Tages macht der berühmte Mathematiker Georges de Rham die folgenden Beobachtungen:
- Jeder Dorfbewohner kennt einen weiteren mit dem gleichen Alter.
- In jeder Gruppe von 192 Personen aus dem Dorf gibt es mindestens drei mit demselben Alter.
Zeige, dass es eine Gruppe ... | [
"Solution:\n\nWir beweisen, dass höchstens 95 verschiedene Alter vorkommen. Nehme an es treten 96 oder mehr verschiedene Alter auf. Da jeder Dorfbewohner einen mit dem selben Alter kennt, können wir Paare mit 96 verschiedenen Altern bilden, total gibt es also eine Gruppe von 192 Dorfbewohnern, in welchen keine drei... | Switzerland | Vorrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
08rv | There is a village with a population of $2007$. This village has no name. You are God of this village and you want villagers to decide the name of this village. Every villager has one idea of the village's name.
Each villager can send a letter to each villager (including himself). And every villager can send any numbe... | [
"If $0 \\le T \\le 668$, we will prove that there exists an instruction which fulfills the conditions. Give the following instruction to every villager.\n\nDefine today as 0th day. All the villagers must prepare a notebook and a memo pad.\n\nToday, each villager $p$ should write the idea of the village's name $m$ i... | Japan | Japan 2007 | [
"Discrete Mathematics > Logic",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 668 | |
0dhh | Let $\sigma(n)$ denote the sum of the divisors of $n$. Prove that there exist infinitely many integers $n$ such that $\sigma(n) > 3n$. Prove also that $\sigma(n) < n(1+\log_2 n)$. | [
"a. To show that there exist infinitely many integers $n$ such that $\\sigma(n) > 3n$:\n\nLet $n = p^k$, where $p$ is a prime and $k \\geq 1$. Then\n$$\n\\sigma(n) = 1 + p + p^2 + \\cdots + p^k = \\frac{p^{k+1} - 1}{p - 1}.\n$$\n\nLet $p = 2$, $n = 2^k$:\n$$\n\\sigma(2^k) = 2^{k+1} - 1.\n$$\n\nFor $k \\geq 2$,\n$$\... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03ki | Problem:
Let $n$ be a fixed positive integer. Show that for only nonnegative integers $k$, the diophantine equation
$$
x_{1}^{3} + x_{2}^{3} + \cdots + x_{n}^{3} = y^{3k+2}
$$
has infinitely many solutions in positive integers $x_{i}$ and $y$. | [] | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0897 | Problem:
Quanti sono i numeri di 2 cifre tali che, se si sottrae la somma delle cifre dal numero di partenza, si ottiene 45?
(A) 0
(B) 1
(C) 9
(D) 10
(E) 20. | [
"Solution:\n\nLa risposta è (D). Un numero che abbia $a$ come cifra delle decine e $b$ come cifra delle unità si può esprimere come $10a + b$; l'esercizio chiede di contare i numeri di due cifre per cui $10a + b - (a + b) = 45$, ovvero tali che $9a = 45$. Tali numeri sono tutti quelli che hanno cifra delle decine 5... | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0ecl | In a football tournament participated $n$ teams. Each team played exactly one match with each other team. There was a total of $2015n$ matches played in the tournament. How many teams participated in the tournament?
(A) 2015 (B) 4029 (C) 4030 (D) 4031
(E) It is impossible to determine. | [
"There was a total of $\\binom{n}{2}$ matches played in the tournament, thus $\\binom{n}{2} = \\frac{n(n-1)}{2} = 2015n$. From this we deduce $n = 4031$."
] | Slovenia | National Math Olympiad 2015 – First Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
026a | Problem:
Cartões premiados - Uma loja distribui 9999 cartões entre os seus clientes. Cada um dos cartões possui um número de 4 algarismos, entre 0001 e 9999. Se a soma dos primeiros 2 algarismos for igual à soma dos 2 últimos, o cartão é premiado. Por exemplo, o cartão 0743 é premiado. Prove que a soma dos números de ... | [
"Solution:\n\nObserve que se o cartão $a b c d$ é premiado então o cartão $c d a b$ também é premiado, por exemplo: 2341 e 4123 são ambos premiados. Assim sempre que $a b \\neq c d$ temos dois cartões premiados cuja soma é\n$$\na b c d + c d a b = (a b \\times 100 + c d) + (c d \\times 100 + a b) = 101(a b + c d)\n... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
01gc | Let $a$, $b$ and $c$ be positive real numbers such that $a > b + c$. Prove that
$$
(a^2 - b^2 - c^2)(a^5 - b^5 - c^5) \le (a^3 - b^3 - c^3)(a^4 - b^4 - c^4).
$$ | [
"We shall apply the following inequality due to Aczél:\n**Lemma.** Let $n \\in \\mathbb{Z}_+$, let $t, u \\in \\mathbb{R}$ and let $x, y \\in \\mathbb{R}^n$ so that $t^2 > |x|^2$ and $u^2 > |y|^2$. Then\n$$\n(tu - x \\cdot y)^2 \\ge (t^2 - |x|^2)(u^2 - |y|^2).\n$$\nLet us momentarily assume that this lemma is true.... | Baltic Way | Baltic Way 2020 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
09si | Problem:
Voor een positief geheel getal $n$ definiëren we $D_{n}$ als het grootste getal dat een deler is van $a^{n}+(a+1)^{n}+(a+2)^{n}$ voor alle positieve gehele $a$.
a. Bewijs dat voor elke positieve gehele $n$ het getal $D_{n}$ van de vorm $3^{k}$ is met $k \geq 0$.
b. Bewijs dat er voor elke $k \geq 0$ een pos... | [
"Solution:\n\na. Zij $p$ een priemgetal en stel dat $p$ een deler is van $D_{n}$. Dan is $p$ een deler van\n$$\n\\left((a+1)^{n}+(a+2)^{n}+(a+3)^{n}\\right)-\\left(a^{n}+(a+1)^{n}+(a+2)^{n}\\right)=(a+3)^{n}-a^{n}\n$$\nvoor alle positieve gehele $a$. Kies nu $a=p$, dan $p \\mid (p+3)^{n}-p^{n}$, oftewel $(p+3)^{n}-... | Netherlands | IMO-selectietoets | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
07pj | Find all positive integers $n$ for which both $837 + n$ and $837 - n$ are cubes of positive integers. | [
"We need to find all positive integers $n$ for which there exist positive integers $x, y$ so that $837 + n = x^3$ and $837 - n = y^3$. Adding these equations gives\n$$\n1674 = x^3 + y^3 = (x + y)(x^2 - x y + y^2).\n$$\nLet $u = x + y$ and $v = x^2 - x y + y^2 = (x + y)^2 - 3 x y = u^2 - 3 x y$, then $3 x y = u^2 - ... | Ireland | Ireland | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Me... | null | proof and answer | 494 | |
03vu | For a chessboard of the size $2008 \times 2008$, in each case (they all have different colors) write one of the letters $C, G, M, O$. If every $2 \times 2$ square contains all these four letters, we call it a "harmonic chessboard." How many different harmonic chessboard are there? (Posed by Zuming Feng) | [
"There are $12 \\times 2^{2008} - 24$ harmonic chessboards. We first prove the following claim:\n\nIn every harmonic chessboard, at least one of the following two situations occurs: (1) each line is composed of just two letters, in an alternative way; (2) each column is composed of just two letters, in an alternati... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 12 * 2^2008 - 24 | |
097t | Problem:
Rezolvați în $\mathbb{R}$ ecuația $\sqrt{8 x^{2}+10 x-3}-\sqrt{8 x+12}=3+\sqrt{4 x+8}-\sqrt{4 x^{2}+7 x-2}$. | [
"Solution:\n\nEcuația din enunț este echivalentă cu ecuația $\\sqrt{(4 x-1)(2 x+3)}+\\sqrt{(4 x-1)(x+2)}-2 \\sqrt{2 x+3}-2 \\sqrt{x+2}=3$.\nVom avea $D V A=\\left[\\frac{1}{4} ;+\\infty\\right)$.\n\nÎn continuare scriem ecuația sub forma\n$$\n\\sqrt{4 x-1}(\\sqrt{2 x+3}+\\sqrt{x+2})-2(\\sqrt{2 x+3}+\\sqrt{x+2})=3 \... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2 | |
0gr6 | Let $x_0, x_1, \dots, x_{2017}$ be a non-decreasing sequence of positive integers. Suppose that $x_0 = 1$ and the subsequence $x_1, x_2, \dots, x_{2017}$ contains exactly 25 distinct positive integers. Show that
$$
\sum_{i=2}^{2017} x_i (x_i - x_{i-2}) \ge 623.
$$
Find the total number of such sequences in the case of ... | [
"Let us solve more general problem: Let $x_0, x_1, \\dots, x_n$ be a non-decreasing sequence of positive integers. Suppose that $x_0 = 1$ and the subsequence $x_1, x_2, \\dots, x_{2017}$ contains exactly $m$ distinct positive integers. Show that\n$$\n\\sum_{i=2}^{n} x_i (x_i - x_{i-2}) \\ge m^2 - 2.\n$$\nSince the ... | Turkey | 25th Turkish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof and answer | binom(1992, 23) | |
0k03 | Problem:
Regular octagon CHILDREN has area $1$. Determine the area of quadrilateral $LINE$. | [
"Solution:\n\nSuppose that the side length $CH = \\sqrt{2} a$, then the area of the octagon is $((2+\\sqrt{2}) a)^2 - 4 \\cdot \\frac{1}{2} a^2 = (4+4 \\sqrt{2}) a^2$, and the area of $LINE$ is $(\\sqrt{2} a)((2+\\sqrt{2}) a) = (2+2 \\sqrt{2}) a^2$, which is exactly one-half of the area of the octagon. Therefore th... | United States | HMMT November | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 1/2 | |
0g71 | (1) 令 $ABC$ 為銳角三角形, 其中 $AB < AC$。令 $\Omega$ 為 $\triangle ABC$ 的外接圓。令 $B_0$ 為 $AC$ 中點, $C_0$ 為 $AB$ 中點, $\triangle AB_0C_0$ 的外接圓為 $\Omega_1$。令 $\omega$ 為一過 $B_0$ 和 $C_0$, 且與 $\Omega$ 切於異於 $A$ 的點 $X$ 的圓。令 $a$ 為 $\Omega$ 和 $\Omega_1$ 的公切線, $x$ 則為 $\Omega$ 和 $\omega$ 的公切線。試證: $a, x$ 和 $B_0C_0$ 三線共點。
(2) 續(1), 令 $D$ 為 $A$ ... | [
"(1) 注意到 $a$ 是 $\\Omega$ 和 $\\Omega_1$ 的根軸 (radical axis), $x$ 是 $\\Omega$ 和 $\\omega$ 的根軸, $B_0C_0$ 則為 $\\Omega_1$ 和 $\\omega$ 的根軸。基於三圓所決定的三根軸共點, 得: $a, x$ 和 $B_0C_0$ 三線共點。\n\n(2) 令 $O$ 為 $\\triangle ABC$ 外心, $A_0$ 為 $BC$ 中點, $Q$ 則為 $A_0$ 對 $B_0C_0$ 的垂足。注意到 $\\angle WAO = \\angle WQO = \\angle WXO = 90^\\circ$, 故 ... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > ... | null | proof only | null | |
0fua | Problem:
Für die reellen Zahlen $a$, $b$, $c$, $d$ gelten die Gleichungen
$$
\begin{array}{ll}
a=\sqrt{45-\sqrt{21-a}}, & b=\sqrt{45+\sqrt{21-b}} \\
c=\sqrt{45-\sqrt{21+c}}, & d=\sqrt{45+\sqrt{21+d}}
\end{array}
$$
Zeige, dass gilt $a b c d=2004$. | [
"Solution:\ndurch zweimaliges Quadrieren folgt für $a$ die Gleichung $(a^{2}-45)^{2}+a-21=0$, daher ist $a$ eine Nullstelle des Polynoms\n$$\nP(x)=x^{4}-90 x^{2}+x+2004\n$$\nDasselbe gilt für $b$. Analog findet man, dass $c$ und $d$ Nullstellen des Polynoms $x^{4}-90 x^{2}-x+2004$ sind, folglich sind $-c$ und $-d$ ... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 2004 | |
0aw8 | Problem:
For each $x \in \mathbb{R}$, let $\{x\}$ be the fractional part of $x$ in its decimal representation. For instance, $\{3.4\} = 3.4 - 3 = 0.4$, $\{2\} = 0$, and $\{-2.7\} = -2.7 - (-3) = 0.3$. Find the sum of all real numbers $x$ for which $\{x\} = \frac{1}{5} x$. | [] | Philippines | 19th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 15/2 | |
0cbo | a) Solve the equation in the set of real numbers $[x]^2 - x = -0.99$.
b) Show that, for every $a \le -1$, the equation $[x]^2 - x = a$ has no real solutions. | [
"a) The equation is written equivalently $[x]^2 - [x] = \\{x\\} - 0.99$, thus $\\{x\\} - 0.99 \\in \\mathbb{Z}$.\nSince $0 \\le \\{x\\} < 1$, we deduce that $-0.99 \\le \\{x\\} - 0.99 < 0.01$ therefore $\\{x\\} - 0.99 = 0$ so $\\{x\\} = 0.99$. Also, from $[x]^2 - [x] = 0$ we deduce that $[x] = 0$ or $[x] = 1$, so $... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) x ∈ {0.99, 1.99}. b) For every a ≤ −1, there are no real solutions. | |
0c6s | Find all functions $f : \mathbb{R} \to \mathbb{R}$ satisfying the conditions:
(i) $f(f(x^2) + y + f(y)) = x^2 + 2f(y)$;
(ii) $x \le y$ implies $f(x) \le f(y)$;
for all real numbers $x$ and $y$. | [
"We prove that the only function satisfying the two conditions is $f(x) = x$.\nWe prove that $f$ is an injection. If we put $y = 0$ in (i) we get $f(f(x^2) + f(0)) = x^2 + 2f(0)$, for any $x$, or equivalently,\n$$\nf(f(a) + f(0)) = a + 2f(0) \\qquad (1)\n$$\nfor any number $a \\ge 0$. From (1), $f$ is an injection ... | Romania | The DANUBE Mathematical Competition | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | f(x) = x | |
0858 | Problem:
Silvia ha $2006$ tessere identiche a forma di triangolo equilatero e vuole disporle tutte sul tavolo senza sovrapporle e in modo che ciascuna abbia esattamente due lati in comune con altre due tessere. Può riuscire nel suo intento? Poteva riuscirci l'anno scorso, quando aveva $2005$ tessere?
(A) È impossibil... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Il testo chiede per quali $n$ è possibile affiancare $n$ tessere in una sequenza chiusa in cui ogni tessera ne ha altre due adiacenti. Facendo delle semplici prove si vede subito che i primi casi in cui è possibile sono $n=6$ (si ottiene un esagono in cui le tessere hann... | Italy | Progetto Olimpiadi di Matematica 2006 GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | MCQ | (C) | |
07ob | For what real numbers $\lambda$ are there positive numbers $a$, $b$, and $c$ such that
$$
a + \frac{1}{a} = 2(1 + \lambda),
$$
$$
c + \frac{1}{c} = 2(1 + \lambda),
$$
$$
b + \frac{1}{a} = 2(1 - \lambda),
$$
$$
c + \frac{1}{b} = 2(1 - \lambda)?
$$
For those values of $\lambda$, determine the corresponding values of $a$,... | [
"Suppose, for a fixed $\\lambda$, a solution to the four equations exists. Then, from the first equation, $2\\lambda = a + \\frac{1}{a} - 2 = \\frac{(a-1)^2}{a} \\ge 0$, and so $\\lambda \\ge 0$. From the first equation in the second row, $2(1 - \\lambda) = b + \\frac{1}{b} > 0$ and so $\\lambda < 1$. Hence, if a s... | Ireland | Ireland | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Two solutions: (1) lambda = 0 with a = b = c = 1; (2) lambda = 1/4 with a = 2, b = 1, c = 1/2. | |
0gmp | For a positive integer $n$, let $p(n)$ denote the number of sequences of positive integers the sum of whose terms is equal to $n$. Show that
$$
\frac{1 + p(1) + p(2) + \dots + p(n-1)}{p(n)} \le \sqrt{2n} .
$$ | [] | Turkey | IMO TEAM SELECTION EXAM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0688 | Find the number of ordered 6-tuples $(\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6)$ can be created, if the numbers $\alpha_1, \alpha_2, \alpha_3, \alpha_4, \alpha_5, \alpha_6$ can take the values $0$, $1$ and $2$ and the sum $\alpha_1 + \alpha_2 + \alpha_3 + \alpha_4 + \alpha_5 + \alpha_6$ is even. | [
"The sum $\\alpha_1 + \\alpha_2 + \\alpha_3 + \\alpha_4 + \\alpha_5 + \\alpha_6$ is even, if and only if, the number of $1$'s is even, that is $0$, $2$, $4$, $6$.\n\nIn the case of zero $1$'s, the possible selections are $2^6$, because for each $\\alpha_i$ we have $2$ selections, ($0$ or $2$).\n\nWhen we have two $... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics"
] | English | proof and answer | 365 | |
093p | Problem:
We call a positive integer $N$ contagious if there exist 1000 consecutive non-negative integers such that the sum of all their digits is $N$. Find all contagious positive integers. | [
"Solution:\nPart 1. We make the following observation:\n(T) Consider a block of 1000 consecutive non-negative integers. Then the last three digits of those numbers (prepended by zeros if needed) form a set $\\{000,001, \\ldots, 999\\}$.\nThus, given any such block, the sum of the last three digits alone equals $3 \... | Middle European Mathematical Olympiad (MEMO) | 14th Middle European Mathematical Olympiad 2020 | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | all integers N ≥ 13500 | |
075f | Problem:
Let $p_{1} < p_{2} < p_{3} < p_{4}$ and $q_{1} < q_{2} < q_{3} < q_{4}$ be two sets of prime numbers such that $p_{4} - p_{1} = 8$ and $q_{4} - q_{1} = 8$. Suppose $p_{1} > 5$ and $q_{1} > 5$. Prove that $30$ divides $p_{1} - q_{1}$. | [
"Solution:\n\nSince $p_{4} - p_{1} = 8$, and no prime is even, we observe that $\\{p_{1}, p_{2}, p_{3}, p_{4}\\}$ is a subset of $\\{p_{1}, p_{1} + 2, p_{1} + 4, p_{1} + 6, p_{1} + 8\\}$. Moreover $p_{1}$ is larger than $3$. If $p_{1} \\equiv 1 \\pmod{3}$, then $p_{1} + 2$ and $p_{1} + 8$ are divisible by $3$. Henc... | India | INMO | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0ic6 | Problem:
A swimming pool is in the shape of a circle with diameter $60$ ft. The depth varies linearly along the east-west direction from $3$ ft at the shallow end in the east to $15$ ft at the diving end in the west (this is so that divers look impressive against the sunset) but does not vary at all along the north-sou... | [
"Solution:\nTake another copy of the pool, turn it upside-down, and put the two together to form a cylinder. It has height $18$ ft and radius $30$ ft, so the volume is $\\pi (30\\ \\mathrm{ft})^{2} \\cdot 18\\ \\mathrm{ft} = 16200 \\pi\\ \\mathrm{ft}^{3}$; since our pool is only half of that, the answer is $8100 \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Volume"
] | null | final answer only | 8100 π ft^3 | |
002g | Hallar todos los números $n$ que se pueden expresar en la forma $n = k + 2\lfloor\sqrt{k}\rfloor + 2$, donde $k$ es un entero no negativo. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Español | proof and answer | All integers n with n ≥ 2 that are neither a perfect square nor one less than a perfect square; equivalently, n ∈ ⋃_{a≥1} [a^2 + 1, (a+1)^2 − 2]. | |
0c8n | Two congruent squares $ABCD$ and $EFGH$ are placed such that they have disjoint interiors, but $C$ is the midpoint of the line segment $EF$ and the points $B$, $F$, $G$ are collinear. The line $BC$ intersects $EH$ at $K$ and the line $AC$ intersects $GH$ at $M$. Let $L$ be the midpoint of $GH$, and let the parallel thr... | [
"a) Clearly, triangle $CFB$ is a right triangle, with $\\angle F = 90^\\circ$, hence it is similar to $CEK$. Moreover, since $CF = CE$ the two triangles are congruent, therefore $CK = CB = CL$. Using the symmetry of $EFGH$ across $CL$ it also follows that $CK = CN$.\n\nb) In the right triangle $CBF$ we have $CB = 2... | Romania | RMC 2020 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04q3 | By drawing lines parallel to each of the sides, an equilateral triangle of side length $n$ is divided into $n^2$ equilateral triangles of side length $1$. At most how many segments of length $1$ on the obtained grid can be coloured in red, so that no three red segments form an equilateral triangle? | [] | Croatia | Croatian Mathematical Society Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n(n+1) | |
0h91 | Find all positive integer numbers $x, y, p, n, k$ such that:
$$
\begin{cases}
5x + y = p^k, \\
5y + x = p^{k+n}.
\end{cases}
$$ | [
"Rewrite first equation in the form $25x + 5y = 5p^k$ and subtract the second equation: $24x = p^k(5 - p^n)$. Thus, it is obvious, that $5 - p^n > 0$, so there are only the following cases.\n\nCase 1. $p = 1$, $n \\in \\mathbb{N}$. It is obvious, that $5x + y \\ge 6$, so there are no solutions.\n\nCase 2. $p=2$, $n... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All solutions are p = 2, n = 1, k ≥ 3 with x = 2^{k-3} and y = 3·2^{k-3}. No other positive integer solutions exist. | |
0au1 | Problem:
What is the fourth smallest positive integer having exactly $4$ positive integer divisors, including $1$ and itself? | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | 14 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.