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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0div | $$
u_1 = a, \ u_{n+1} = u_n + f(u_n)$$
for $n = 1, 2, \dots$ with $f(x)$ is the product of all of digits of $x$. Prove that there exist $N$ such that $u_n = u_N$ for any $n \geq N$. | [
"Let $u_1 = a$ be a positive integer. Define $u_{n+1} = u_n + f(u_n)$, where $f(x)$ is the product of all digits of $x$.\n\nObserve that if $u_n$ contains a digit $0$, then $f(u_n) = 0$, so $u_{n+1} = u_n$. Thus, the sequence becomes constant from that point onward.\n\nSuppose $u_n$ does not contain any digit $0$. ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof only | null | |
0jfq | Problem:
Find the number of ordered triples of integers $(a, b, c)$ with $1 \leq a, b, c \leq 100$ and $a^{2} b + b^{2} c + c^{2} a = a b^{2} + b c^{2} + c a^{2}$. | [
"Solution:\n\nThis factors as $(a-b)(b-c)(c-a)=0$. By the inclusion-exclusion principle, we get $3 \\cdot 100^{2} - 3 \\cdot 100 + 100 = 29800$."
] | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 29800 | |
00rz | For any set of points $A_1, A_2, ..., A_n$ on the plane, one defines $r(A_1, A_2, ..., A_n)$ as the radius of the smallest circle that contains all of these points. Prove that if $n \ge 3$, there are indices $i, j, k$ such that
$$
r(A_1, A_2, ..., A_n) = r(A_i, A_j, A_k).
$$ | [
"We start with a lemma.\n**Lemma.** If the triangle $ABC$ is acute, $r(A, B, C)$ is its circumradius and if it is obtuse, $r(A, B, C)$ is half the length of its longest side.\n**Proof.**\nLet us do the acute case first. The circumcircle contains the vertices, so $r(A, B, C)$ is not greater than the circumradius. No... | Balkan Mathematical Olympiad | BMO 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0hdp | Equation $\frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_m} - \frac{1}{y_1} - \frac{1}{y_2} - \dots - \frac{1}{y_n} = \frac{577}{408}$ has an infinite number of natural solutions $x_1, x_2, \dots, x_m, y_1, y_2, \dots, y_n$ for some non-negative integers $m, n$. Prove that an equation $\frac{1}{x_1} + \frac{1}{x_2}... | [
"We prove the claim by induction on $m+n$. When $m+n=1$, if $a>0$ we have $m=1$ and $n=0$, therefore given equation becomes $\\frac{1}{x_1} = a$ and has no more than one solution. Similarly $a<0$.\n\nSuppose our claim is true for $m+n<t$, and that equation $\\frac{1}{x_1} + \\frac{1}{x_2} + \\dots + \\frac{1}{x_m} ... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
020h | Problem:
Find the greatest positive integer $N$ with the following property: there exist integers $x_{1}, \ldots, x_{N}$ such that $x_{i}^{2}-x_{i} x_{j}$ is not divisible by $1111$ for any $i \neq j$. | [
"Solution:\nWe prove that the greatest $N$ with the required property is $N=1000$. First note that $x_{i}^{2}-x_{i} x_{j}=x_{i}\\left(x_{i}-x_{j}\\right)$, and that the prime factorisation of $1111$ is $11 \\cdot 101$.\n\nWe first show that we can find $1000$ integers $x_{1}, x_{2}, \\ldots, x_{1000}$ such that $x_... | Benelux Mathematical Olympiad | 8th Benelux Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 1000 | |
0004 | Sean $ABCD$ un rectángulo con $AB > BC$, y $O$ el punto de intersección de sus diagonales $AC$ y $BD$. La bisectriz del ángulo $B\hat{A}C$ corta a $BD$ en $E$. Llamamos $M$ al punto medio de $AB$. Se traza por $E$ la perpendicular a $AB$, que corta a $AB$ en $F$; se traza por $E$ la perpendicular a $AE$, que corta a $A... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | español | proof and answer | 2704/75 · a^2 | |
0auq | Problem:
Find the value of $\cot \left(\cot^{-1} 2 + \cot^{-1} 3 + \cot^{-1} 4 + \cot^{-1} 5\right)$. | [
"Solution:\n\nUsing the identity\n$$\n\\cot^{-1} x + \\cot^{-1} y = \\cot^{-1}\\left(\\frac{xy - 1}{x + y}\\right)\n$$\nthe expression $\\cot^{-1} 2 + \\cot^{-1} 3 + \\cot^{-1} 4 + \\cot^{-1} 5$ can be simplified to $\\cot^{-1} \\frac{9}{7} + \\cot^{-1} \\frac{11}{7}$. Thus, we have\n$$\n\\cot \\left(\\cot^{-1} 2 +... | Philippines | 18th PMO National Stage Oral Phase | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 5/14 | |
09pa | Let $A_n$ denote the number of subsets of $\{1, 2, \dots, n\}$ that contain no two elements differing by $1$, and let $B_n$ denote the number of subsets that contain no two elements differing by $2$. Prove that $A_{2025} < B_{1013}^2$.
(Batbayasgalan Balkhuu and Nursoltan Khavalbolot) | [
"We begin by observing that for all integers $n, m \\ge 1$, the inequality\n$$\nA_{n+m} \\le A_n \\cdot A_m\n$$\nholds. This is because any subset of $\\{1, 2, \\dots, n+m\\}$ avoiding adjacent elements can be formed by taking a subset of $\\{1, 2, \\dots, n\\}$ and a subset of $\\{n+1, \\dots, n+m\\}$, each with n... | Mongolia | MMO2025 Round 4 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
04lx | Let $a$, $b$ and $c$ be positive real numbers. Prove that
$$
\frac{3}{2} < \frac{4a+b}{a+4b} + \frac{4b+c}{b+4c} + \frac{4c+a}{c+4a} < 9. \quad (\text{Japan 2009})
$$ | [
"Notice that\n$$\n\\frac{4a+b}{4(4a+b)} < \\frac{4a+b}{a+4b} < \\frac{4(a+4b)}{a+4b} = 4,\n$$\nand analogously\n$$\n\\frac{1}{4} < \\frac{4b+c}{b+4c} < 4,\n$$\n$$\n\\frac{1}{4} < \\frac{4c+a}{c+4a} < 4.\n$$\nWithout loss of generality we may assume that $a \\le b \\le c$. Hence we have\n$$\n\\frac{4c+a}{c+4a} = \\f... | Croatia | Croatian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
08wx | Let $n, k$ be positive integers satisfying $n \ge k$.
There is a group consisting of $n$ people. Each person of this group belongs to one and only one of $k$ clubs, called club $C_1, C_2, \dots, C_k$. Each club has at least one member belonging to it. Prove that it is possible to distribute $n^2$ pieces of cake to thes... | [
"Let for each $i$, $1 \\le i \\le k$, $x_i$ be the number of people belonging to the club $C_i$.\nIf we set $a_i = x_i + 2(x_{i+1} + x_{i+2} + \\cdots + x_k)$ for each $i$, then we claim that $a_1, a_2, \\cdots, a_k$ satisfy all the conditions of the problem. The condition $a_i > 0$ is obvious. For $1 \\le i \\le k... | Japan | Japan 2013 Final Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03mq | The quadrilateral $ABCD$ is inscribed in a circle. The point $P$ lies in the interior of $ABCD$, and $\angle PAB = \angle PBC = \angle PCD = \angle PDA$. The lines $AD$ and $BC$ meet at $Q$, and the lines $AB$ and $CD$ meet at $R$. Prove that the lines $PQ$ and $PR$ form the same angle as the diagonals of $ABCD$. | [
"Let $\\Gamma$ be the circumcircle of quadrilateral $ABCD$. Let $\\alpha = \\angle PAB = \\angle PBC = \\angle PCD = \\angle PDA$ and let $T_1, T_2, T_3$ and $T_4$ denote the circumcircles of triangles $APD$, $BPC$, $APB$ and $CPD$, respectively. Let $M$ be the intersection of $T_1$ with line $RP$ and let $N$ be th... | Canada | Kanada 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ang | Problem:
The length of a leg of a right triangle is $5$ while the length of the altitude to its hypotenuse is $4$. Find the length of the other leg. | [] | Philippines | Area Stage | [
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 20/3 | |
05ky | Problem:
Trouver tous les entiers $n \geqslant 1$ tels que $2^{n}+12^{n}+2014^{n}$ soit un carré parfait. | [
"Solution:\n\nRegardons l'expression modulo $3$ : $2^{n}+12^{n}+2014^{n} \\equiv (-1)^{n}+1 \\pmod{3}$. Comme un carré n'est jamais congru à $2$ modulo $3$, on en déduit que $n$ est impair.\n\nRegardons ensuite l'expression modulo $7$ :\n$$\n2^{n}+12^{n}+2014^{n} \\equiv 2^{n}+(-2)^{n}+5^{n} \\equiv 5^{n} \\pmod{7}... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | no positive integers n | |
0k2i | Problem:
Points $E, F, G, H$ are chosen on segments $AB, BC, CD, DA$, respectively, of square $ABCD$. Given that segment $EG$ has length $7$, segment $FH$ has length $8$, and that $EG$ and $FH$ intersect inside $ABCD$ at an acute angle of $30^{\circ}$, then compute the area of square $ABCD$. | [
"Solution:\n\nRotate $EG$ by $90^{\\circ}$ about the center of the square to $E'G'$ with $E' \\in AD$ and $G' \\in BC$. Now $E'G'$ and $FH$ intersect at an angle of $60^{\\circ}$. Then consider the translation which takes $E'$ to $H$ and $G'$ to $I$. Triangle $FHI$ has $FH = 8$, $HI = 7$ and $\\angle FHI = 60^{\\ci... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 784/19 | |
03p5 | Let $\triangle ABC$ be a triangle. Points $D$ and $E$ are on sides $AB$ and $AC$, respectively, and point $F$ is on line segment $DE$. Let $\frac{AD}{AB} = x$, $\frac{AE}{AC} = y$, $\frac{DF}{DE} = z$. Prove that
(1) $S_{\triangle BDF} = (1-x)y zS_{\triangle ABC}$ and $S_{\triangle CEF} = x(1-y)(1-z)S_{\triangle ABC}$... | [
"Connect $BE$ and $CD$. Then we have\n\n(1) $S_{\\triangle BDF} = zS_{\\triangle BDE} = z(1-x)S_{\\triangle ABE}$\n$= z(1-x)yS_{\\triangle ABC}$ and\n$S_{\\triangle CEF} = (1-z)S_{\\triangle CDE}$\n$= (1-z)(1-y)S_{\\triangle ACD}$\n$= (1-z)(1-y)xS_{\\triangle ABC}$.\n\n(2) From (1) we get\n... | China | China Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gt1 | Find all pairs of integers $(a, b)$ satisfying
$$
a^7(a-1) = 19b(19b+2).
$$ | [
"$(a, b) = (0, 0)$ and $(a, b) = (1, 0)$.\nBy adding $1$ to both sides of the given equation, we can rewrite it as\n$$\na^8 - a^7 + 1 = (19b + 1)^2.\n$$\nLeft hand side of this equation has the following factorization:\n$$\na^8 - a^7 + 1 = (a^8 - a^7 + a^6) - (a^6 - 1) = (a^2 - a + 1) \\cdot (a^6 - (a+1)(a^3-1)).\n... | Turkey | Team Selection Test | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | [(0, 0), (1, 0)] | |
08ph | Problem:
Determine all four-digit numbers $\overline{a b c d}$ such that
$$
(a+b)(a+c)(a+d)(b+c)(b+d)(c+d)=\overline{a b c d}
$$ | [
"Solution:\nDepending on the parity of $a, b, c, d$, at least two of the factors $(a+b), (a+c), (a+d), (b+c), (b+d), (c+d)$ are even, so that $4 \\mid \\overline{a b c d}$.\nWe claim that $3 \\mid \\overline{a b c d}$.\nAssume $a+b+c+d \\equiv 2 \\pmod{3}$. Then $x+y \\equiv 1 \\pmod{3}$, for all distinct $x, y \\i... | JBMO | Junior Balkan Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2016 | |
02z0 | Problem:
No cubo $ABCD EFGH$, cuja aresta mede $6~\mathrm{cm}$, o ponto $M$ é ponto médio de $\overline{EF}$.
a) Determine a área do triângulo $AMH$.
b) Determine o volume da pirâmide $AMHE$. (O volume de uma pirâmide pode ser calculado pela terça parte do produto entre a área da base e a a... | [
"Solution:\n\na) $\\overline{AH}$ é a diagonal de uma face, ou seja, $AH = 6 \\sqrt{2}~\\mathrm{cm}$. $\\overline{MH}$ e $\\overline{AM}$ são hipotenusas de triângulos cujos catetos medem $6~\\mathrm{cm}$ e $3~\\mathrm{cm}$, ou seja, $MH = AM = 3\\sqrt{5}~\\mathrm{cm}$. Traçando a altura $MM'$ do triângulo $AMH$, r... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) 9√6 cm²; b) 18 cm³; c) √6 cm | |
0jvh | Problem:
Rachel has two indistinguishable tokens, and places them on the first and second square of a $1 \times 6$ grid of squares. She can move the pieces in two ways:
- If a token has a free square in front of it, then she can move this token one square to the right.
- If the square immediately to the right of a tok... | [
"Solution:\n\nWe put a marker on $(i, j)$ when a token is on the $i$th and $j$th square and $i > j$. When the token in front/behind moves one step forward to a blank square, move the marker rightward/upward one unit correspondingly. When a \"leapfrog\" happens, the marker moves from $(x-1, x)$ to $(x, x+1)$. We can... | United States | HMMT November 2016 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions"
] | null | proof and answer | 42 | |
0243 | Problem:
No seguinte tabuleiro, devemos colocar todos os números, desde $1$ até $25$, seguindo as seguintes regras:
- Em cada fila, os $5$ números colocados devem formar uma sequência estritamente crescente quando lidas da esquerda para a direita.
- Em cada coluna, os $5$ números colocados devem formar uma sequência e... | [
"Solution:\n\na) Vamos colocar as seguintes letras para representar os números colocados no tabuleiro, da maneira mostrada na seguinte figura:\n\n| $M$ | $N$ | $P$ | $Q$ | $E$ |\n| :---: | :---: | :---: | :---: | :---: |\n| $R$ | $S$ | $T$ | $D$ | |\n| $U$ | $V$ | $C$ | | |\n| $W$ | $B$ | | | |\n| $A$ | | |... | Brazil | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | An example main diagonal is 5, 9, 12, 14, 15; the smallest diagonal entry is always at least 5, the largest is always at least 15, and the minimal possible diagonal sum is 55. | |
0ani | Problem:
A polyhedron has $30$ faces and $62$ edges. How many vertices does the polyhedron have?
(a) $61$
(b) $34$
(c) $46$
(d) $77$ | [] | Philippines | QUALIFYING STAGE | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | null | MCQ | b | |
0764 | Problem:
In a triangle $A B C$, let $D$ be a point on the segment $B C$ such that $A B + B D = A C + C D$. Suppose that the points $B$, $C$ and the centroids of triangles $A B D$ and $A C D$ lie on a circle. Prove that $A B = A C$. | [
"Solution:\n\nLet $G_{1}, G_{2}$ denote the centroids of triangles $A B D$ and $A C D$. Then $G_{1}, G_{2}$ lie on the line parallel to $B C$ that passes through the centroid of triangle $A B C$. Therefore $B G_{1} G_{2} C$ is an isosceles trapezoid. Therefore it follows that $B G_{1} = C G_{2}$. This proves that $... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof only | null | |
01vy | For all positive real numbers $x$, $y$, $z$ prove the inequality
$$
x^2z(4x - 3y) + y^2x(4y - 3z) + z^2y(4z - 3x) > 0.
$$ | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0825 | Problem:
Siano $a < b < c$ interi positivi tali che $a^{2} + b^{2} + c^{2}$ ha lo stesso numero di cifre decimali di $a + b + c$. Qual è il massimo valore che può assumere $c$?
(A) 9
(B) 10
(C) 18
(D) 30
(E) 31. | [
"Solution:\n\nLa risposta è (A). La condizione che $a + b + c$ abbia lo stesso numero di cifre di $a^{2} + b^{2} + c^{2}$ implica che\n$$\n\\frac{a^{2} + b^{2} + c^{2}}{a + b + c} < 10\n$$\nPertanto si deve avere che\n$$\na^{2} - 10a + b^{2} - 10b + c^{2} - 10c < 0\n$$\nil che è equivalente ad affermare che\n$$\n(a... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | A | |
00c5 | A company has $n$ employees. It is known that every employee works at least one of the 7 days of the week, except for an employee that does not work any of the 7 days. In addition, for every pair of these $n$ employees, there are at least 3 days of the week such that one of the two employees works that day but the othe... | [
"We will show that the maximum possible number of employees is $n = 16$.\n\nFirst, we prove that $n \\le 16$. We represent each possible weekly schedule for an employee with a 7-tuple of 0's and 1's whose coordinates correspond to the days of the week and where 1 stands for 'working day' and 0 stands for 'non worki... | Argentina | XXVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 16 | |
01ex | Prove that there exist only finitely many triples of positive integers $(n, a, b)$ such that:
$$
n! = 2^a - 2^b.
$$ | [
"Since $n!$ is divisible by $3^{\\lfloor n/3 \\rfloor}$, $(2^{a-b}-1)2^b = 2^a - 2^b$ is divisible by $3^{\\lfloor n/3 \\rfloor}$. From the Lifting the Exponent Lemma we obtain that $a-b$ is divisible by $3^{\\lfloor n/3 \\rfloor-1}$. So $a-b \\ge 3^{\\lfloor n/3 \\rfloor-1} \\ge 3^{n/3-2}$. Hence, the right hand s... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
07l5 | Prove that a line through the centroid of a triangle that bisects the area of the triangle is a median. | [
"Let $FH$, with $F$ on $AB$ and $H$ on $AC$, bisect the area of $\\triangle ABC$. Let the midpoints of $AB$ and $AC$ be $E$ and $D$ respectively and let the centroid of $\\triangle ABC$ be $G$. Assume $FH$ passes through $G$ and is not a median, i.e. $F \\ne E$ and $H \\ne D$.\n\nSince $CE$ bisects the area of $\\a... | Ireland | Irska | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
029x | Problem:
Sexta-feira 13 - Qual o número máximo de sexta-feiras 13 que podem ocorrer num ano não bissexto? Neste caso, qual é o $10^{\circ}$ dia do ano? | [
"Solution:\n\nDado que os dias da semana se repetem a cada 7 dias, então a diferença entre os dias da semana é dada pelo resto ao dividir por 7 o número de dias transcorridos.\n\nNa tabela seguinte temos:\n- na primeira linha o número de dias entre o dia 13 de um mês e o dia 13 do mês seguinte;\n- na segunda linha ... | Brazil | Nível 2 | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 3; Saturday | |
0j1p | Problem:
Paul Erdős was one of the most prolific mathematicians of all time and was renowned for his many collaborations. The Erdős number of a mathematician is defined as follows. Erdős has an Erdős number of $0$, a mathematician who has coauthored a paper with Erdős has an Erdős number of $1$, a mathematician who ha... | [
"Solution:\n\nAnswer: $4.65$ We'll suppose that each mathematician collaborates with approximately $20$ people (except for Erdős himself, of course). Furthermore, if a mathematician has Erdős number $k$, then we'd expect him to be the cause of approximately $\\frac{1}{2^{k}}$ of his collaborators' Erdős numbers. Th... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Other"
] | null | final answer only | 4.65 | |
07qj | Do there exist four polynomials $P_1(x), P_2(x), P_3(x), P_4(x)$ with real coefficients, such that the sum of any three of them always has a real root, but the sum of any two of them has no real root? | [
"There do not exist four such polynomials. We show this as follows. Suppose that there do exist four such polynomials. If a polynomial has no real roots, it is either positive for all real $x$, or else it is negative for all real $x$. Consider the complete graph with the four polynomials as vertices.\n\n![](attache... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | No; such four polynomials do not exist. | |
0jan | Problem:
Fix a positive integer $n$. Two players, Phil and Ellie, play the following game. First, Phil fills the squares of an $n \times n$ chessboard with nonnegative integers less than $n$. Then, Ellie chooses three squares making an L, as in any of the following pictures:
Ellie adds 1 to e... | [
"Solution:\n\na.\nPhil wins in this case. Consider the sum of all the numbers on the board. If Ellie adds 1 to three squares making an $\\mathrm{L}$, the sum increases by 3, and when a 12 is replaced by a 0, the sum decreases by 12. So the sum always increases and decreases by multiples of 3. So if Phil ensures tha... | United States | Berkeley Math Circle Monthly Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | a. Phil wins. b. Ellie wins. | |
0jcz | Problem:
Let $O, O_{1}, O_{2}, O_{3}, O_{4}$ be points such that $O_{1}, O, O_{3}$ and $O_{2}, O, O_{4}$ are collinear in that order, $O O_{1}=1$, $O O_{2}=2$, $O O_{3}=\sqrt{2}$, $O O_{4}=2$, and $\measuredangle O_{1} O O_{2}=45^{\circ}$. Let $\omega_{1}, \omega_{2}, \omega_{3}, \omega_{4}$ be the circles with respec... | [
"Solution:\n\nAnswer: $8+4 \\sqrt{2}$\n\nWe first maximize the area of triangle $P_{1} O P_{2}$, noting that the sum of the area of $P_{1} O P_{2}$ and the three other analogous triangles is the area of $P_{1} P_{2} P_{3} P_{4}$. Note that if $A \\neq P_{1}, P_{2}$, without loss of generality say $\\angle O A P_{1}... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 8 + 4√2 | |
06qw | Let $a$ and $b$ be distinct integers greater than $1$. Prove that there exists a positive integer $n$ such that $(a^{n}-1)(b^{n}-1)$ is not a perfect square. | [
"At first we notice that\n$$\n\\begin{align*}\n(1-\\alpha)^{\\frac{1}{2}}(1-\\beta)^{\\frac{1}{2}} & =\\left(1-\\frac{1}{2} \\cdot \\alpha-\\frac{1}{8} \\cdot \\alpha^{2}-\\cdots\\right)\\left(1-\\frac{1}{2} \\cdot \\beta-\\frac{1}{8} \\cdot \\beta^{2}-\\cdots\\right) \\\\\n& =\\sum_{k, \\ell \\geq 0} c_{k, \\ell} ... | IMO | IMO Problem Shortlist | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
02bl | Problem:
Uma urna tem 6 bolas numeradas de 1 a 6. Se duas bolas são extraídas, qual é a probabilidade da diferença entre os números dessas 2 bolas ser 1? | [
"Solution:\n\nObservemos que se extraímos a primeira bola com um número entre 2 e 5, então dentre as 5 bolas que ficam na urna temos duas possíveis bolas que cumprem a condição do problema, logo neste caso a probabilidade que a segunda bola cumpra a condição é $\\frac{2}{5}$ e a probabilidade que a primeira bola te... | Brazil | Lista 6 | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 1/3 | |
04tj | There is $|BC| = 1$ in a triangle $ABC$ and there is a unique point $D$ on $BC$ such that $|DA|^2 = |DB| \cdot |DC|$. Find all possible values of the perimeter of $ABC$. | [
"Let us denote by $E$ the second intersection of $AD$ with the circumcircle $k$. The power of $D$ with respect to $k$ gives $|DB| \\cdot |DC| = |DA| \\cdot |DE|$, which together with the given condition $|DA|^2 = |DB| \\cdot |DC|$ yields $|DA| = |DE|$. That is $E$ lies on the image $p$ of the line $BC$ in the homot... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof and answer | 1 + sqrt(2) | |
0krw | Problem:
Find all positive integers $n$ such that $n^{4}-27 n^{2}+121$ is a prime positive integer. | [
"Solution:\nWe can rewrite $n^{4}-27 n^{2}+121$ as\n$$\n\\left(n^{4}+22 n^{2}-121\\right)-49 n^{2}=\\left(n^{2}+11\\right)^{2}-(7 n)^{2}=\\left(n^{2}+7 n+11\\right)\\left(n^{2}-7 n+11\\right) .\n$$\nFor this to be prime, we would need $n^{2}-7 n+11=1$. Rearranging and factoring gives $(n-2)(n-5)=0$, so $n=2$ or $n=... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | n = 2, 5 | |
0gdj | 設三角形 $ABC$ 的內心為 $I$, 角 $A$ 內的旁心為 $J$。令 $\overline{AA'}$ 為 $\triangle ABC$ 外接圓的直徑, 點 $H_1, H_2$ 分別為 $\triangle BIA', \triangle CJA'$ 的垂心。證明: $H_1H_2$ 平行於 $BC$。 | [
"(∠ 代表有向角, $\\pm$ 代表正向相似。)\n\n平移 $\\triangle CJA' \\cup H_2$ 至 $\\triangle C_1BA'_1 \\cup H'_2$ (即使得 $J$ 平移至與 $B$ 重合)。由 $\\overline{IJ}$ 中點位於 $\\overline{BC}$ 的中垂線上, 可得 $C_1I \\perp BC$。令 $I'$ 為 $I$ 關於 $\\odot(BIA')$ 的對徑點、$C'_1$ 為 $C_1$ 關於 $\\odot(C_1BA'_1)$ 的對徑點。則由 $\\overline{BI'} = \\overline{H_1A'}$, $\\overlin... | Taiwan | 二〇一九數學奧林匹亞競賽第二階段選訓營,模擬競賽(二) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0gk9 | Consider the sequence $\{a_n\}_{n \ge 1}$ of positive integers with $a_n a_{n+3} = a_{n+2} a_{n+5}$ for all positive integer $n$. Determine the largest integer that always divides $\sum_{k=1}^{2550} a_{2k} a_{2k-1}$. | [
"From $a_n a_{n+3} = a_{n+2} a_{n+5}$ for every positive integer $n$, we have\n$a_{n+1} a_{n+4} = a_{n+3} a_{n+6}$ and $a_{n+2} a_{n+5} = a_{n+4} a_{n+7}$. Then\n$$\na_n a_{n+3} \\cdot a_{n+1} a_{n+4} \\cdot a_{n+2} a_{n+5} = a_{n+2} a_{n+5} \\cdot a_{n+3} a_{n+6} \\cdot a_{n+4} a_{n+7}.\n$$\nTherefore, $a_n a_{n+1... | Thailand | Thai Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 850 | |
03pn | Given a positive integer $n$, find the least positive number $\lambda$ such that $\cos \theta_1 + \cos \theta_2 + \cdots + \cos \theta_n$ is not greater than $\lambda$ provided $\tan \theta_1 \cdot \tan \theta_2 \cdot \cdots \cdot \tan \theta_n = 2^{\frac{n}{2}}$ for any $\theta_i \in (0, \frac{\pi}{2})$ ($i=1, 2, \cdo... | [
"When $n=1$, $\\cos\\theta_1 = (1+\\tan^2\\theta_1)^{-\\frac{1}{2}} = \\frac{\\sqrt{3}}{3}$. Hence $\\lambda = \\frac{\\sqrt{3}}{3}$.\n\nWhen $n=2$, we can prove\n$$\n\\cos \\theta_1 + \\cos \\theta_2 \\le \\frac{2\\sqrt{3}}{3}, \\qquad \\textcircled{1}\n$$\nand when $\\theta_1 = \\theta_2 = \\arctan\\sqrt{2}$, the... | China | China Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | λ = √3/3 for n = 1; λ = 2√3/3 for n = 2; λ = n − 1 for n ≥ 3 | |
07qe | Let $ABC$ be a triangle with $|AC| \neq |BC|$. Let $P$ and $Q$ be the intersection points of the line $AB$ with the internal and external angle bisectors at $C$, so that $P$ is between $A$ and $B$. Prove that if $M$ is any point on the circle with diameter $PQ$, then $\angle AMP = \angle BMP$. | [
"The internal and external angle bisectors divide the segment $AB$ internally and externally in the same ratio $|AC| : |BC|$. This can be seen, for example, with the aid of the Sine Rule, applied to the triangles $ACP$ and $BCP$ for the internal bisector and to the triangles $ACQ$ and $BCQ$ for the external angle b... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0gqw | Let a line $l$ intersect the line $AB$ at $F$, the sides $AC$ and $BC$ of a triangle $ABC$ at $D$ and $E$, respectively and the internal bisector of the angle $BAC$ at $P$. Suppose that $F$ is at the opposite side of $A$ with respect to the line $BC$, $CD = CE$ and $P$ is in the interior the triangle $ABC$. Prove that
... | [
"Let $DP = a$, $PE = b$, $EF = c$, $CD = CE = x$. By the Stewart Theorem $CP^2 = x^2 - ab$ and $CF^2(a+b) + x^2c = (a+b+c)((a+b)c + x^2)$, hence $CF^2 - CP^2 = (a+c)(b+c)$. Let $Q$ be a point on the line segment $DE$ satisfying $EQ = a$. Then $CF^2 - CP^2 = (a+c)(b+c) = FP \\cdot FQ$. If $CF^2 - CP^2 = FB \\cdot FA... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04n8 | Determine all pairs $(x, y)$ of real numbers such that $x, y \in [0, \frac{\pi}{2}]$ for which the following holds:
$$
\frac{2 \sin^2 x + 2}{\sin x + 1} = 3 + \cos (x + y). \quad (\text{Petar Bakić})
$$ | [] | Croatia | Croatia_2018 | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | (π/2, π/2) | |
00rx | Let $n$, $a$, $b$, $c$ be natural numbers. Every point on the coordinate plane with integer coordinates is colored in one of $n$ colors. Prove there exists $c$ triangles whose vertices are colored in the same color, which are pairwise congruent, and which have a side whose length is divisible by $a$ and a side whose le... | [
"Let the colors be $d_1$, $d_2$, $d_3$, $\\dots$, $d_n$. Look at the coordinates\n$$\n(k, 0 + (n+1)abr),\\ (k, ab + (n+1)abr),\\ (k, 2ab + (n+1)abr),\\ \\dots,\\ (k, nab + (n+1)abr)\n$$\nfor integers $k$ and $r$. By the pigeonhole principle there are two points of the same color. For every pair $(k, r)$ we say the ... | Balkan Mathematical Olympiad | BMO 2017 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0dz9 | Let $ABC$ be a right triangle with the right angle at $C$. On the segment $BC$ choose a point $D$ different from $B$ and $C$. Denote the circumcircle of the triangle $ABD$ by $\mathcal{K}$. Let $T$ be a point on the side $AB$ such that $DT$ is perpendicular to $AB$. Let $E$ be the second intersection of the line $DT$ w... | [
"The sum of the two opposite angles $\\angle ATD$ and $\\angle ACD$ in the quadrilateral $ATDC$ is $\\pi$, so this quadrilateral is cyclic. Set $\\angle TCD = \\alpha$. Then\n\n$\\angle TAD = \\angle TCD = \\alpha$. Points $A$, $B$, $E$, $D$ are concyclic, so $\\angle BED = \\angle BAD = \\angle TAD = \\alpha$. We ... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08sn | What is the maximum number of times that you can divide by $2$ the number
$$
1004 \times 1005 \times 1006 \times \dots \times 2008?
$$ | [
"For a real number $x$, denote by $\\lfloor x \\rfloor$ the greatest integer $\\le x$. Then, for every positive integer $n$, there are exactly $\\lfloor \\frac{n}{2} \\rfloor$ even numbers among $1, 2, \\dots, n$ and therefore, $1 \\times 2 \\times 3 \\times \\dots \\times n$ can be divided by $2$ at least $\\lfloo... | Japan | Japan Junior Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1006 | |
0ioz | Problem:
Joe B. first places the black king in one corner of the board. In how many of the 35 remaining squares can he place a white bishop so that it does not check the black king? | [
"Solution:\n\nAny square not on the diagonal containing the corner is a possible location, and there are $36 - 6 = 30$ such squares."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Other"
] | null | final answer only | 30 | |
0cnc | Is it possible to replace the stars in the equality
$$ \{\text{l.c.m.}\}(*, *, *) - \{\text{l.c.m.}\}(*, *, *) = 2009 $$
by six consecutive positive integers (but not necessarily in the successive order) so that the equality would be valid? (R. Zhenodarov) | [
"It is not possible.\n\nSuppose such numbers exist. The least common multiple (LCM) of several numbers is divisible by each of them and, therefore, by each of their divisors. Thus, if among the numbers for which the LCM is taken there is an even number, then the LCM will also be even. Since $2009$ is an odd number,... | Russia | Euler olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof and answer | No | |
049q | Determine the minimal value of $\sin(x + 3) - \sin(x + 1) - 2\cos(x + 2)$ if $x \in \mathbb{R}$. | [] | Croatia | Hrvatska 2011 | [
"Precalculus > Trigonometric functions"
] | English | proof and answer | 2(sin 1 - 1) | |
0bi3 | Let $n$ be a natural number. Find the integers $x$, $y$, $z$ such that $x^2 + y^2 + z^2 = 2^n (x + y + z)$. | [
"If $n = 0$, using the inequalities $x^2 \\ge x$ and its analogues, we deduce that $x$, $y$, $z \\in \\{0, 1\\}$.\n\nIf $n \\ge 1$, then $2$ divides $x^2 + y^2 + z^2$, and hence, either the three numbers are even, or one is even and the others are odd. In the former case, if we take $x = 2x_1 + 1$, $y = 2y_1 + 1$, ... | Romania | 65th Romanian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | For n = 0: all triples with x, y, z in {0, 1}. For n ≥ 1: all triples with x, y, z in {0, 2^n}. | |
09uh | Problem:
Bestaan er een positief geheel getal $k$ en een niet-constante rij $a_{1}, a_{2}, a_{3}, \ldots$ van positieve gehele getallen zodat $a_{n}=\operatorname{ggd}\left(a_{n+k}, a_{n+k+1}\right)$ voor alle positieve gehele getallen $n$? | [
"Solution:\n\nOplossing I. Zo'n $k$ en bijbehorende rij bestaan niet. We bewijzen dit uit het ongerijmde, dus stel dat ze wel bestaan. Merk op dat $a_{n} \\mid a_{n+k}$ en $a_{n} \\mid a_{n+k+1}$ voor alle $n \\geq 1$. Met een eenvoudige inductie volgt ook dat $a_{n} \\mid a_{n+\\ell k}$ en $a_{n} \\mid a_{n+\\ell ... | Netherlands | Selectietoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
0isr | Problem:
Suppose that at some point Joe B. has placed 2 black knights on the original board, but gets bored of chess. He now decides to cover the 34 remaining squares with 17 dominos so that no two overlap and the dominos cover the entire rest of the board. For how many initial arrangements of the two pieces is this po... | [
"Solution:\nAnswer: 324 Color the squares of the board red and blue in a checkerboard pattern, and observe that any domino will cover exactly one red square and one blue square. Therefore, if the two knights cover squares of the same color, this is impossible. We now claim that it is always possible if they cover s... | United States | 1st Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 324 | |
0kub | Problem:
Let $ABCD$ and $WXYZ$ be two squares that share the same center such that $WX \parallel AB$ and $WX < AB$. Lines $CX$ and $AB$ intersect at $P$, and lines $CZ$ and $AD$ intersect at $Q$. If points $P$, $W$, and $Q$ are collinear, compute the ratio $AB / WX$. | [
"Solution:\n\n\n\nWithout loss of generality, let $AB = 1$. Let $x = WX$. Then, since $BPWX$ is a parallelogram, we have $BP = x$. Moreover, if $T = XY \\cap AB$, then we have $BT = \\frac{1 - x}{2}$, so $PT = x - \\frac{1 - x}{2} = \\frac{3x - 1}{2}$. Then, from $\\triangle PXT \\sim \\tri... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof and answer | sqrt(2) + 1 | |
0dek | We have $n > 2$ nonzero integers such that every one of them is divisible by the sum of the other $n-1$ numbers. Show that the sum of the $n$ numbers is precisely $0$. | [
"Let these numbers be $a_1, a_2, \\dots, a_n$ and $S$ is the sum of these numbers. For every $i \\in \\{1, 2, \\dots, n\\}$, we have $S - a_i \\mid a_i$, which means $S - a_i \\nmid S$. Suppose that $S \\neq 0$, without the loss of generality, $S > 0$. We investigate two cases\n\n* If $\\exists i \\in \\{1, 2, \\do... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization"
] | null | proof only | null | |
0inw | Problem:
Kevin has four red marbles and eight blue marbles. He arranges these twelve marbles randomly, in a ring. Determine the probability that no two red marbles are adjacent. | [
"Solution:\n\nAnswer: $\\frac{7}{33}$.\n\nSelect any blue marble and consider the remaining eleven marbles, arranged in a line. The proportion of arrangements for which no two red marbles are adjacent will be the same as for the original twelve marbles, arranged in a ring. The total number of ways of arranging $4$ ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 7/33 | |
0h52 | Let us call a year *colored* if the decimal representation of its number has no repeating digits. For example, all years from $2013$ to $2019$ are colored, unlike $2020$.
a) Find the nearest chain of seven consecutive colored years in the future.
b) Can a chain of more than seven consecutive years happen in the futur... | [
"a) Let us show that the nearest sequence of $7$ colored years is $2103, \\ldots, 2109$. First, we prove that in this century no sequence of more than six colored years can happen any longer. We see that digits $0$ and $2$ cannot represent units or tens. Therefore, a chain is broken at each year ending in $0$ or $2... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Third Round | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | a) 2103, 2104, 2105, 2106, 2107, 2108, 2109; b) No, a chain longer than seven consecutive colored years cannot occur. | |
0hnt | Problem:
Ten cups lie upside down in a line. It is known that pennies lie under two of the cups which are consecutive in the line. Choosing several of the cups, you may ask for the total number of coins under them. Is it possible to determine the positions of the pennies by asking two such questions, without knowing t... | [
"Solution:\n\nThe answer is yes; here is one strategy that works. On the first turn choose the cups $1, 4, 5, 6, 7$, and on the second turn choose $1, 6, 7, 8, 9$. Depending on the locations of the coins, the answers will be:\n\n| Coins | $1,4,5,6,7$ | $1,6,7,8,9$ |\n| :---: | :---: | :---: |\n| 1,2 | 1 | 1 |\n| 2,... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic"
] | null | proof and answer | Yes | |
05ha | Problem:
Soit $ABC$ un triangle, $D$ est le milieu de l'arc $BC$ du cercle $ABC$ ne contenant pas $A$, $Z$ est l'unique point sur la bissectrice extérieure de $\widehat{BAC}$ tel que $ZA = ZC$. Montrer que le cercle $ADZ$ passe par le milieu du côté $[AB]$. | [
"Solution:\n\nOn note $\\alpha = \\widehat{BAC}$, on pose $M_B$ le milieu de $[AB]$ et $M_C$ le milieu de $[AC]$. On note $X$ l'intersection des droites $(M_BM_C)$ et $(DC)$.\n\nAlors comme $(M_BM_C) \\parallel BC$, on trouve $\\widehat{M_BXC} = \\widehat{BCD} = \\widehat{BAD} = \\frac{\\alpha}{2}$, l'avant-dernièr... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0j3e | Problem:
What is the sum of all four-digit numbers that are equal to the cube of the sum of their digits (leading zeros are not allowed)? | [
"Solution:\n\nWe want to find all integers $x$ between $1000$ and $9999$ that are the cube of the sum of their digits. Of course, our search is only restricted to perfect cubes. The smallest such cube is $10^{3} = 1000$ and the largest such cube is $21^{3} = 9261$. This means we only have to check $12$ different cu... | United States | Harvard-MIT November Tournament | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 10745 | |
0k2c | Problem:
Kevin starts with the vectors $(1,0)$ and $(0,1)$ and at each time step, he replaces one of the vectors with their sum. Find the cotangent of the minimum possible angle between the vectors after 8 time steps. | [
"Solution:\n\nSay that the vectors Kevin has at some step are $(a, b)$ and $(c, d)$. Notice that regardless of which vector he replaces with $(a+c, b+d)$, the area of the triangle with vertices $(0,0)$, $(a, b)$, and $(c, d)$ is preserved with the new coordinates. We can see this geometrically: the parallelogram wi... | United States | HMMT February 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 987 | |
0hd3 | Prove that for any real numbers $x, y, z$ the following inequality is true:
$$
x^2(3y^2 + 3z^2 - 2yz) \geq yz(2xy + 2xz - yz).
$$
Find the triples that turn it into equality. | [
"**Answer:** $(x, y, z) = (t, 2t, 2t), (t, 0, 0), (0, t, 0), (0, 0, t)$.\n\nConsider the given inequality as quadratic relative to the variable $x$:\n$$\nx^2(3y^2 + 3z^2 - 2yz) - 2x(y^2z + z^2y) + y^2z^2 \\geq 0.\n$$\nSince $3y^2 + 3z^2 \\geq 6|yz| \\geq 2yz$, consider the case that $3y^2 + 3z^2 = 2yz$. Then both v... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | (x, y, z) = (t, 2t, 2t), (t, 0, 0), (0, t, 0), (0, 0, t) for any real t | |
0a0t | At the beginning of the day, a toy shop has 20 sticker sheets (0.30 euro each), 18 footballs (3 euros each), 5 teddy bears (5 euros each) and 8 water guns (15 euros each) in stock. The new cashier makes a mess of the records and reports at the end of the day about the sale of these four items only that the total amount... | [
"D) 4"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | MCQ | D | |
08iq | Problem:
Let $x$, $y$, $z$ be real numbers greater than $-1$. Prove that
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
$$
Problem:
Fie $x$, $y$, $z$ numere mai mari ca $-1$. Demonstrați că
$$
\frac{1+x^{2}}{1+y+z^{2}}+\frac{1+y^{2}}{1+z+x^{2}}+\frac{1+z^{2}}{1+x+y^{2}} \geq 2
... | [
"Solution:\nWe have $y \\leq \\frac{1+y^{2}}{2}$, hence\n$$\n\\frac{1+x^{2}}{1+y+z^{2}} \\geq \\frac{1+x^{2}}{1+z^{2}+\\frac{1+y^{2}}{2}}\n$$\nand the similar inequalities.\nSetting $a=1+x^{2}$, $b=1+y^{2}$, $c=1+z^{2}$, it suffices to prove that\n$$\n\\frac{a}{2c+b}+\\frac{b}{2a+c}+\\frac{c}{2b+a} \\geq 1\n$$\nfor... | JBMO | 7th JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kli | Call a fraction $\frac{a}{b}$, not necessarily in simplest form, *special* if $a$ and $b$ are positive integers whose sum is $15$. How many distinct integers can be written as the sum of two, not necessarily different, special fractions?
(A) 9 (B) 10 (C) 11 (D) 12 (E) 13 | [] | United States | AMC 12 B | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | C | |
0knu | Quadratic polynomials $P(x)$ and $Q(x)$ have leading coefficients of $2$ and $-2$, respectively. The graphs of both polynomials pass through the two points $(16, 54)$ and $(20, 53)$. Find $P(0) + Q(0)$. | [
"Because the leading coefficients of $P(x)$ and $Q(x)$ are negatives of each other, the polynomial $R(x) = P(x) + Q(x)$ is linear. Furthermore, $R(16) = 54 + 54 = 108$ and $R(20) = 53 + 53 = 106$. It follows that $R(x) = 116 - 0.5x$, so $P(0) + Q(0) = R(0) = 116$.\n\nNote that\n$$\nP(x) = 2x^2 - \\frac{289}{4}x + 6... | United States | 2022 AIME I | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 116 | |
0hx6 | Problem:
Let $n > 1$ be an integer. Three complex numbers have the property that their sum is $0$ and the sum of their $n$th powers is also $0$. Prove that two of the three numbers have the same absolute value. | [
"Solution:\nGiven that\n$$\na + b + c = 0 \\quad \\text{and} \\quad a^{n} + b^{n} + c^{n} = 0,\n$$\nlet\n$$\nt = ab + bc + ca \\quad \\text{and} \\quad u = abc.\n$$\nThen $a$, $b$, and $c$ are the roots of the polynomial\n$$\nf(z) = (z - a)(z - b)(z - c) = z^{3} + t z - u.\n$$\nIf $t = 0$, then $a$, $b$, and $c$ ar... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
04s4 | There are 234 visitors in a cinema auditorium. The visitors are sitting in $n$ rows, where $n \ge 4$, so that each visitor in the $i$-th row has exactly $j$ friends in the $j$-th row, for any $i, j \in \{1, 2, ..., n\}, i \neq j$. Find all the possible values of $n$. (Friendship is supposed to be a symmetric relation.) | [
"For any $k \\in \\{1, 2, ..., n\\}$ denote by $p_k$ the number of visitors in the $k$-th row. The stated condition on given $i$ and $j$ implies that the number of friendly pairs $(A, B)$, where $A$ and $B$ are from the $i$-th row and from $j$-th row respectively, is equal to the product $j p_i$. Interchanging the ... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 12 | |
00u7 | Let $a$, $b$, $n$ be positive integers such that:
(i) $a^{2021} \mid n$ and $b^{2021} \mid n$
(ii) $2022 \mid a - b$ and $a > b$.
Prove that there is a subset of the divisors of the number $n$ having sum of elements divisible by $2022$ but not by $2022^2$. | [
"**Solution 1.** Write $a = dr$ and $b = ds$ where $d = \\gcd(a, b)$ and $(r, s) = 1$. Then $d^{2021} r^{2021} s^{2021}$ divides $n$. Furthermore $2 \\cdot 3 \\cdot 337 \\mid d(r - s)$.\n\n**Case 1:** Assume $337 \\mid d$. Since $(r, s) = 1$, we may assume that $r$ is odd. Then\n$$\n\\{337r^2, 337r^4, \\dots, 337r^... | Balkan Mathematical Olympiad | BMO 2022 shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0fq9 | Problem:
Sean $a, b, c$ números naturales primos, distintos dos a dos. Demostrar que el número
$$
(a b)^{c-1} + (b c)^{a-1} + (c a)^{b-1} - 1
$$
es un múltiplo del producto $a b c$. | [
"Solution:\nAl ser $a, b, c$ primos, el producto $a b$ no es divisible por $c$ (si $c \\mid a b$, sería $c \\mid a$ o $c \\mid b$; habida cuenta de que $c > 1, a \\neq c, b \\neq c$, tendríamos una contradicción con el hecho de ser $a$ y $b$ primos).\nEntonces, por la congruencia de Fermat $(\\dagger)$,\n$$\n(a b)^... | Spain | null | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0bmv | Determine $n \in \mathbb{N}$ such that the numbers $n+8$, $2n+1$, $4n+1$ are all perfect cubes. | [
"We will show that the only such integer is $1$. We start noticing that the product $(n+8)(4n+1)(2n+1) = 8n^3 + 70n^2 + 49n + 8$ must be also a cube.\n\nFor $n \\in \\{1, 2, \\dots, 18\\}$ the number $n+8$ cannot be a cube. For $n \\ge 19$, remark that $(2n+2)^3 \\le 8n^3 + 70n^2 + 49n + 8 < (2n+6)^3$. We have to c... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | n = 0 | |
0alx | Problem:
Find all real solutions to the system of equations
$$
\begin{cases}
x(y-1) + y(x+1) = 6 \\
(x-1)(y+1) = 1
\end{cases}
$$ | [
"Solution:\nLet us first expand and simplify the equations.\n\nFrom the first equation:\n$$\nx(y-1) + y(x+1) = 6\n$$\nExpand:\n$$\nxy - x + yx + y = 6\n$$\nCombine like terms:\n$$\nxy - x + xy + y = 6\n$$\n$$\n2xy - x + y = 6\n$$\n\nFrom the second equation:\n$$\n(x-1)(y+1) = 1\n$$\nExpand:\n$$\nxy + x - y - 1 = 1\... | Philippines | 18th PMO Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | ((4/3), 2) and (-2, -4/3) | |
03a8 | Let $a$ be a real number such that the quadratic equation $x^2 - x + a = 0$ has two real distinct roots $x_1$ and $x_2$. Prove that $|x_1^2 - x_2^2| = 1$ if and only if $|x_1^3 - x_2^3| = 1$. | [] | Bulgaria | Winter Mathematical Competition | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
096s | Problem:
Să se afle toate valorile parametrului real $a$, pentru care toate soluţiile ecuaţiei
$$
x^{4}-2 x^{3}-3 x^{2}-4 a x-a^{2}=0
$$
sunt reale. Pentru valorile aflate, să se rezolve ecuaţia. | [
"Solution:\nEcuaţia se ordonează ca o ecuaţie de gradul 2 în raport cu necunoscuta $a$ :\n$$\na^{2}+4 x \\cdot a+\\left(3 x^{2}+2 x^{3}-x^{4}\\right)=0\n$$\nDiscriminantul ei, $\\Delta=16 x^{2}-12 x^{2}-8 x^{3}+4 x^{4}=4 x^{2}\\left(x^{2}-2 x+1\\right)=4 x^{2}(x-1)^{2}$. Se află soluţiile $a_{1}=-x^{2}-x$ şi $a_{2}... | Moldova | A 63-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Parameter range: a ∈ [-9/4, 1/4]. For such a, the roots are x = (-1 ± sqrt(1 − 4a))/2 and x = (3 ± sqrt(4a + 9))/2. | |
045w | Let $m$ and $n$ be two positive integers with $m \ge n \ge 2022$. Let $a_1, a_2, \dots, a_n, b_1, b_2, \dots, b_n$ be $2n$ real numbers. Prove that the numbers of ordered pairs $(i, j)$ ($1 \le i, j \le n$) such that
$$
|a_i + b_j - ij| \le m
$$
is less than or equal to $3n\sqrt{m \log n}$. | [
"**Proof:** Mark red all points $(i, j)$ for which $|a_i + b_j - ij| \\le m$.\n\n*Lemma:* If $(i_1, j_1)$, $(i_1, j_2)$, $(i_2, j_1)$, $(i_2, j_2)$ are all red, then $|(i_2 - i_1)(j_2 - j_1)| \\le 4m$.\n\n*Proof of the lemma:* By the definition of red points, we know\n$$\n|a_{i_1} + b_{j_1} - i_1 j_1| \\le m, \\qua... | China | 2022 China Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
008g | a) Is it possible to divide a square with side length $1$ into $30$ rectangles, each with a perimeter equal to $2$?
b) Let us assume that a square with side length $1$ is divided into $25$ rectangles with a perimeter equal to $p$. Find the minimum and maximum value of $p$. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | a) No. b) Minimum p is 4/5 and maximum p is 52/25. | |
0inn | Problem:
The function $f: \mathbb{R} \rightarrow \mathbb{R}$ satisfies $f\left(x^{2}\right) f^{\prime\prime}(x) = f^{\prime}(x) f^{\prime}\left(x^{2}\right)$ for all real $x$. Given that $f(1) = 1$ and $f^{\prime\prime\prime}(1) = 8$, determine $f^{\prime}(1) + f^{\prime\prime}(1)$. | [
"Solution:\n\nLet $f^{\\prime}(1) = a$ and $f^{\\prime\\prime}(1) = b$. Then setting $x = 1$ in the given equation, $b = a^{2}$.\n\nDifferentiating the given yields\n$$\n2x f^{\\prime}\\left(x^{2}\\right) f^{\\prime\\prime}(x) + f\\left(x^{2}\\right) f^{\\prime\\prime\\prime}(x) = f^{\\prime\\prime}(x) f^{\\prime}\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 6 | |
0egj | Problem:
Za $3~\mathrm{kg}$ pomaranč in $5~\mathrm{kg}$ limon plačamo skupaj $8,40~€$. Za $5~\mathrm{kg}$ pomaranč in $4~\mathrm{kg}$ limon plačamo skupaj $8,80~€$. Koliko skupno plačamo za $2~\mathrm{kg}$ pomaranč in $3~\mathrm{kg}$ limon?
(A) $7,20~€$
(B) $5,20~€$
(C) $3,60~€$
(D) $5,60~€$
(E) $4,80~€$ | [
"Solution:\n\nZapišemo sistem enačb $3P + 5L = 8,40$ in $5P + 4L = 8,80$. Rešimo sistem enačb in dobimo $P = 0,80~€$, $L = 1,20~€$. Izračunamo $2P + 3L = 5,20~€$."
] | Slovenia | Državno tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | B | |
0c25 | Find all the real numbers $k$ such that
$$
\sqrt{3(a^2 + b^2 + c^2)} \leq a + b + c + k\sqrt{a^2 + b^2 + c^2 - ab - bc - ca},
$$
for every positive real numbers $a$, $b$, $c$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | k ≥ sqrt(3) - 1 | |
013g | Problem:
Through a point $P$ exterior to a given circle pass a secant and a tangent to the circle. The secant intersects the circle at $A$ and $B$, and the tangent touches the circle at $C$ on the same side of the diameter through $P$ as $A$ and $B$. The projection of $C$ on the diameter is $Q$. Prove that $Q C$ bisec... | [
"Solution:\n\nDenoting the centre of the circle by $O$, we have $O Q \\cdot O P = O A^{2} = O B^{2}$. Hence $\\triangle O A Q \\sim \\triangle O P A$ and $\\triangle O B Q \\sim \\triangle O P B$. Since $\\triangle A O B$ is isosceles, we have $\\angle O A P + \\angle O B P = 180^{\\circ}$, and therefore\n$$\n\\beg... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06g9 | Given a triangle $ABC$, let $M$ be the midpoint of $BC$. The circle passing through $A$ and tangent to $BC$ at $M$ cuts $AB$ and $AC$ at $D$ and $E$ respectively. Suppose $B, C, E$ and $D$ are concyclic. Show that $AB = AC$. | [
"Firstly, consider the power of $B$ with respect to $(ADME)$. This gives\n$$\nBD \\times BA = BM^2.\n$$\nSimilarly, we get\n$$\nCE \\times CA = CM^2\n$$\nby considering the power of $C$ with respect to the same circle. As $M$ is the midpoint of $BC$, the two expressions are equal, so that\n$$\nBD \\times BA = CE \\... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | null | proof only | null | |
0drp | Let $D$ be the point in the interior of a triangle $ABC$ such that $AB = ab$, $AC = ac$, $AD = ad$, $BC = bc$, $BD = bd$ and $CD = cd$. Prove that $\angle ABD + \angle ACD = 60^\circ$. | [
"Let $A_1$, $B_1$, $C_1$ be the feet of the perpendiculars from $D$ onto the sides $BC$, $CA$, $AB$, respectively.\n\nThen, $B_1C_1 = DA \\sin A$, $C_1A_1 = DB \\sin B$, $A_1B_1 = DC \\sin C$. Thus, $B_1C_1 : C_1A_1 : A_1B_1 = (ad)(bc) : (bd)(ac) : (cd)(ab) = 1 : 1 : 1$. Therefore, $A_1B_1C... | Singapore | Singapur | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00m1 | Es seien $x_1, x_2, \dots, x_9$ nicht negative reelle Zahlen, für die gilt:
$$
x_1^2 + x_2^2 + \dots + x_9^2 \geq 25
$$
*Man beweise, dass es drei dieser Zahlen gibt, deren Summe mindestens 5 ist.* | [
"Es sei o. B. d. A. $x_1 \\ge x_2 \\ge \\dots \\ge x_9 \\ge 0$. Annahme: $x_1 + x_2 + x_3 < 5$. Dann gilt\n$$\n25 \\le x_1^2 + x_2^2 + \\dots + x_9^2 \\le x_1(x_1 + x_2 + x_3) + x_2(x_4 + x_5 + x_6) + x_3(x_7 + x_8 + x_9) \\\\ \\le 5x_1 + 5x_2 + 5x_3 < 25.\n$$\nDas ist ein Widerspruch und daher gibt es drei Zahlen,... | Austria | 48. Österreichische Mathematik-Olympiade | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | German | proof only | null | |
0418 | Given $A = \{2, 0, 1, 3\}$, let $B = \{x \mid -x \in A, 2 - x^2 \notin A\}$. Then the sum of elements in $B$ is ______. | [
"It is easy to find that $B \\subseteq \\{-2, 0, -1, -3\\}$. We have $2 - x^2 \\notin A$ when $x = -2, -3$, and $2 - x^2 \\in A$ when $x = 0, -1$. Therefore, $B = \\{-2, -3\\}$, the sum of whose elements is $-5$.\nThe answer is $-5$."
] | China | China Mathematical Competition | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | final answer only | -5 | |
06jd | Let $f(x) = \frac{15}{x+1} + \frac{16}{x^2+1} - \frac{17}{x^3+1}$. Find the value of
$f(\tan 15°) + f(\tan 30°) + f(\tan 45°) + f(\tan 60°) + f(\tan 75°)$. | [
"Note that $f\\left(\\frac{1}{x}\\right) = \\frac{15}{\\frac{1}{x}+1} + \\frac{16}{\\frac{1}{x^2}+1} - \\frac{17}{\\frac{1}{x^3}+1} = \\frac{15x}{x+1} + \\frac{16x^2}{x^2+1} - \\frac{17x^3}{x^3+1}$.\nHence we have\n$$\n\\begin{aligned}\nf(x) + f\\left(\\frac{1}{x}\\right) &= 15 \\left( \\frac{1}{x+1} + \\frac{x}{x+... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | 35 | |
0buv | Problem:
a) Fie $(G, )$ un grup abelian cu $2016$ elemente. Demonstrați că există $x \in G \setminus \{e\}$ astfel încât $x^{2} = e$ ($e$ este elementul neutru din $G$).
b) Calculați: $\int \frac{1}{x^{2016} + x} \, dx$, $x > 0$. | [] | Romania | Olimpiada de Matematică Etapa Locală | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | a) Such an element exists (there is x ≠ e with x^2 = e). b) ∫ dx/(x^{2016} + x) = ln x − (1/2015) ln(1 + x^{2015}) + C, for x > 0. | |
00zm | Problem:
Consider the sequence
$$
\begin{aligned}
x_{1} & = 19, \\
x_{2} & = 95, \\
x_{n+2} & = \operatorname{lcm}\left(x_{n+1}, x_{n}\right) + x_{n},
\end{aligned}
$$
for $n > 1$, where $\operatorname{lcm}(a, b)$ means the least common multiple of $a$ and $b$. Find the greatest common divisor of $x_{1995}$ and $x_{19... | [
"Solution:\n\nLet $d = \\operatorname{gcd}\\left(x_{k}, x_{k+1}\\right)$. Then $\\operatorname{lcm}\\left(x_{k}, x_{k+1}\\right) = x_{k} x_{k+1} / d$, and\n$$\n\\operatorname{gcd}\\left(x_{k+1}, x_{k+2}\\right) = \\operatorname{gcd}\\left(x_{k+1}, \\frac{x_{k} x_{k+1}}{d} + x_{k}\\right) = \\operatorname{gcd}\\left... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 19 | |
0373 | Problem:
Denote by $d(a, b)$ the number of the divisors of a positive integer $a$, which are greater than or equal to $b$. Find all positive integers $n$ such that
$$
d(3 n+1,1)+d(3 n+2,2)+\cdots+d(4 n, n)=2006
$$ | [
"Solution:\nDenote by $D(a, b)$ the set of the divisors of $a$, which are greater than or equal to $b$. Thus, $|D(a, b)|=d(a, b)$. Every integer $k$, $1 \\leq k \\leq 4$, belongs to at most one of the sets\n$$\nD(3 n+1,1), D(3 n+2,2), \\ldots, D(4 n, n)\n$$\nEvery integer $k$, $1 \\leq k \\leq n$, $3 n+1 \\leq k \\... | Bulgaria | Team selection test for 47. IMO | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 708 | |
03xr | Given integer $n > 2$, suppose positive real numbers $a_1, a_2, \dots, a_n$ satisfy $a_k \le 1$, $k = 1, 2, \dots, n$.
Let $A_k = \frac{a_1 + a_2 + \dots + a_k}{k}$, $k = 1, 2, \dots, n$.
Prove $\left| \sum_{k=1}^n a_k - \sum_{k=1}^n A_k \right| < \frac{n-1}{2}$. | [
"For $1 \\le k \\le n-1$, we have $0 < \\sum_{i=1}^k a_i \\le k$ and $0 < \\sum_{i=k+1}^n a_i \\le n-k$. By using the fact that $|x-y| < \\max\\{x, y\\}$ for $x, y > 0$, we get\n$$\n\\begin{align*}\n|A_n - A_k| &= \\left| \\left(\\frac{1}{n} - \\frac{1}{k}\\right) \\sum_{i=1}^{k} a_i + \\frac{1}{n} \\sum_{i=k+1}^{n... | China | China Mathematical Competition (Complementary Test) | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02wp | Problem:
Em um torneio com 5 times, não existem empates. De quantos modos podem ocorrer os $\frac{5 \cdot 4}{2}=10$ jogos do torneio de modo que, tanto não tenhamos um time que ganhou todas quanto um time que não perdeu todas as partidas? | [
"Solution:\n\nRepresente cada time por um vértice de um pentágono e o resultado de cada jogo por uma seta partindo do jogador que ganhou para o jogador que perdeu.\n\n\n\nComo cada partida possui 2 resultados possíveis, existem $2^{10}$ possíveis resultados para o torneio. Suponha que o jog... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 544 | |
05dn | Problem:
Let $D$ and $E$ be two points on the sides $AB$ and $AC$, respectively, of a triangle $ABC$, such that $DB = BC = CE$, and let $F$ be the point of intersection of the lines $CD$ and $BE$. Prove that the incenter $I$ of the triangle $ABC$, the orthocenter $H$ of the triangle $DEF$ and the midpoint $M$ of the $... | [
"Solution:\n\nAs $DB = BC = CE$ we have $BI \\perp CD$ and $CI \\perp BE$. Hence $I$ is orthocenter of triangle $BFC$. Let $K$ be the point of intersection of the lines $BI$ and $CD$, and let $L$ be the point of intersection of the lines $CI$ and $BE$. Then we have the power relation $IB \\cdot IK = IC \\cdot IL$. ... | European Girls' Mathematical Olympiad (EGMO) | European Girls' Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Ana... | null | proof only | null | |
073n | Suppose $d$, $u$, $v$, $w$ are positive integers such that $u$, $v$, $w$ are distinct and
$$
d^3 - d(uv + vw + wu) - 2uvw = 0.
$$
Prove that $d$ cannot be a prime. Find also the least possible value of $d$. | [
"Note that $u = 1$, $v = 2$, $w = 3$ gives $d^3 - 11d - 12 = 0$ which has no integer solutions. Thus $uvw > 8$ and $uv + vw + wu \\ge 3(uvw)^{2/3} > 12$. This shows that $d^3 > 12d + 16$ and we infer that $d \\ge 5$.\n\nIf $d$ is a prime, $d$ has to be an odd prime, say $d = p$. Then $p^3 = p(uv + vw + wu) + 2uvw$,... | India | Indija TS 2008 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | d is not prime; the least possible value of d is 8. | |
09kw | There are four pipes installed in the pool. The first two pipes are used to fill the pool with water, while the remaining two pipes are responsible for draining water from the pool. If all four pipes are operational simultaneously, the reservoir fills up in $2.5$ hours. When only pipes $1$, $2$, and $3$ are in use, the... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 3 hours | |
051z | Let $I$ be the incenter of triangle $ABC$. Let $R_A$, $R_B$ and $R_C$ be the radii of the circumcircles of triangles $BIC$, $CIA$ and $AIB$, respectively, and $R$ be the radius of the circumcircle of triangle $ABC$. Prove that $R_A + R_B + R_C \le 3R$. | [
"Let $|BC| = a$, $|CA| = b$ and $|AB| = c$ and the angles opposite to those sides be $\\alpha$, $\\beta$ and $\\gamma$, respectively (see fig. 24).\n\n\nFigure 24\n\nThe law of sines in triangles $ABC$ and $IBC$ gives $\\frac{a}{\\sin \\alpha} = 2R$ and $\\frac{a}{\\sin \\left(\\frac{\\beta... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Algebra > Equations and Inequalit... | null | proof only | null | |
037f | Problem:
The sequence $\{x_{n}\}_{n=1}^{\infty}$ is defined by $x_{1}=2$ and $x_{n+1}=1+a x_{n}$, $n \geq 1$, where $a$ is a real number. Find all values of $a$ for which the sequence is:
a) an arithmetic progression;
b) convergent and find its limit. | [
"Solution:\na) It follows by the recurrence relation that $x_{1}=2$, $x_{2}=1+2a$ and $x_{3}=1+a+2a^{2}$. Then $x_{1}+x_{3}=2x_{2} \\Longleftrightarrow 3+a+2a^{2}=2(1+2a)$ with solutions $a=1$ and $a=\\frac{1}{2}$. For $a=1$ we get $x_{n+1}=x_{n}+1$, i.e. the sequence is an arithmetic progression. For $a=\\frac{1}{... | Bulgaria | Spring Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) a = 1. b) Convergent for −1 < a < 1, with limit 1/(1 − a). | |
0ium | Problem:
Stan has a stack of 100 blocks and starts with a score of 0, and plays a game in which he iterates the following two-step procedure:
a. Stan picks a stack of blocks and splits it into 2 smaller stacks each with a positive number of blocks, say $a$ and $b$. (The order in which the new piles are placed does no... | [
"Solution:\n\nLet $E(n)$ be the expected value of the score for an $n$-block game. It suffices to show that the score is invariant regardless of how the game is played. We proceed by induction. We have $E(1) = 0$ and $E(2) = 1$. We require that $E(n) = E(n-k) + E(k) + (n-k)k$ for all $k$. Setting $k = 1$, we hypoth... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 4950 | |
00yr | Problem:
Consider the following two person game. A number of pebbles are situated on the table. Two players make their moves alternately. A move consists of taking off the table $x$ pebbles where $x$ is the square of any positive integer. The player who is unable to make a move loses. Prove that there are infinitely m... | [
"Solution:\n\nSuppose that there is an $n$ such that the first player always wins if there are initially more than $n$ pebbles. Consider the initial situation with $n^{2}+n+1$ pebbles. Since $(n+1)^{2}>n^{2}+n+1$, the first player can take at most $n^{2}$ pebbles, leaving at least $n+1$ pebbles on the table. By the... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
012r | Problem:
Two magicians show the following trick. The first magician goes out of the room. The second magician takes a deck of 100 cards labelled by numbers $1,2, \ldots, 100$ and asks three spectators to choose in turn one card each. The second magician sees what card each spectator has taken. Then he adds one more ca... | [
"Solution:\n\nWe will identify ourselves with the second magician. Then we need to choose a card in such a manner that another magician will be able to understand which of the 4 cards we have chosen and what information it gives about the order of the other cards. We will reach these two goals independently.\nLet $... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof only | null | |
08jz | Problem:
Let $ABC$ be a triangle with $\angle C = 90^{\circ}$ and $D \in CA$, $E \in CB$, and $k_1, k_2, k_3, k_4$ semicircles with diameters $CA$, $CB$, $CD$, $CE$ respectively, which have common part with the triangle $ABC$. Let also,
$$
k_1 \cap k_2 = \{C, K\},\quad k_3 \cap k_4 = \{C, M\},\quad k_2 \cap k_3 = \{C,... | [
"Solution:\n\nThe points $K, L, M, N$ belong to the segments $AB$, $BD$, $DE$, $EA$ respectively, where $CK \\perp AB$, $CL \\perp BD$, $CM \\perp DE$, $CN \\perp AE$. Then quadrilaterals $CDLM$ and $CENM$ are inscribed. Let $\\angle CAE = \\varphi$, $\\angle DCL = \\theta$. Then $\\angle EMN = \\angle ECN = \\varp... | JBMO | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07an | Let $n$ be a natural number. Permutation $a_1, a_2, \dots, a_n$ of numbers $1, 2, \dots, n$ is called square (cubic), if for each natural number $1 \le i \le n - 1$, $a_i a_{i+1} + 1$ is a perfect square (cube).
a) Prove that for infinitely many natural numbers $n$ there exists at least one square permutation of numbe... | [
"a) Let $a_1 = 2, a_2 = 4, \\dots, a_k = 2k, a_{k+1} = 1, a_{k+2} = 3, \\dots, a_{2k} = 2k-1$. We can easily check that $a_i a_{i+1} + 1$ is a perfect square for $1 \\le i \\le 2k$ except $i = k$, which can be repaired if $2k+1$ is a perfect square which is possible for infinitely many values of $k$.\n\nb) Let $a_1... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0ct0 | Determine which of two numbers $(100!)!$ and $99!^{100!} \cdot 100!^{99!}$ is greater than the other one. | [
"Let $a = 99!$. Then we need to compare the numbers $(100a)!$ and $a^{100a} \\cdot (100a)^a$.\n\nNote that\n$$\n\\begin{aligned}\n& 1 \\cdot 2 \\cdot 3 \\cdots a < a^a, \\\\\n& (a+1)(a+2)(a+3) \\cdots 2a < (2a)^a, \\\\\n& (2a+1)(2a+2)(2a+3) \\cdots 3a < (3a)^a, \\\\\n& \\vdots \\\\\n& (99a+1)(99a+2)(99a+3) \\cdots ... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 99!^{100!} · 100!^{99!} is greater than (100!)! | |
0ejs | Problem:
Dan je enakokrak trikotnik $ABC$ z vrhom pri $C$, v katerem je $\angle ACB < 90^\circ$. Naj bo $X$ od $C$ različna točka na stranici $AC$ in $Y$ od $C$ različna točka na stranici $BC$. Naj bo $D$ taka točka, da je premica $DX$ vzporedna premici $AB$, premica $AC$ pa je notranja simetrala kota $\angle BAD$. Po... | [
"Solution:\n\n\n\nOznačimo $\\angle BAC = \\angle CBA = \\alpha$. Ker je premica $AC$ simetrala kota $\\angle BAD$, je $\\angle XAD = \\alpha$. Zaradi vzporednosti premic $XD$ in $AB$ pa je tudi $\\angle DXA = \\alpha$. Trikotnik $AXD$ je torej enakokrak z vrhom pri $D$, zato velja $|XD| = ... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AB^2/BC |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.