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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0k6s | Problem:
You are trying to cross a 6 foot wide river. You can jump at most 4 feet, but you have one stone you can throw into the river; after it is placed, you may jump to that stone and, if possible, from there to the other side of the river. However, you are not very accurate and the stone ends up landing uniformly a... | [
"Solution:\nTo be able to cross, the stone must land between 2 and 4 feet from the river bank you are standing on. Therefore the probability is $\\frac{2}{6} = \\frac{1}{3}$."
] | United States | HMMT November 2019 | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 1/3 | |
0enh | In the plane there are $2012$ congruent triangles of area $1$ which all have the same orientation. Each of these triangles contains the centroids of all the other triangles.
a. Show that the intersection of the $2012$ triangles is a triangle similar to them with area at least $\frac{1}{9}$.
b. Show that the area of t... | [
"a.\nLet $A_sB_sC_s$ be the given triangles, with $s \\in \\{1, 2, \\dots, 2012\\}$ and let $\\{XYZ\\}$ stand for the half-plane with the boundary line $XY$ and interior point $Z$. Each of the triangles is the intersection of the three half-planes $\\{A_sB_sC_s\\}$, $\\{B_sC_sA_s\\}$ and $\\{C_sA_sB_s\\}$, hence th... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0iib | Problem:
Let $a, b, c, d$ be real numbers so that $c, d$ are not both $0$. Define the function
$$
m(x) = \frac{a x + b}{c x + d}
$$
on all real numbers $x$ except possibly $-d / c$, in the event that $c \neq 0$. Suppose that the equation $x = m(m(x))$ has at least one solution that is not a solution of $x = m(x)$. Find... | [
"Solution:\nThat $0$ is a possible value of $a + d$ can be seen by taking $m(x) = -x$, i.e., $a = -d = 1$, $b = c = 0$. We will now show that $0$ is the only possible value of $a + d$.\n\nThe equation $x = m(m(x))$ implies\n$$\nx = \\frac{\\left(a^{2} + b c\\right) x + (a + d) b}{(a + d) c x + \\left(b c + d^{2}\\r... | United States | Harvard-MIT Mathematics Tournament, Team Round B | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | 0 | |
0424 | Given $X = \{1, 2, \dots, 100\}$, consider function $f: X \to X$ satisfying both the following conditions:
(1) $f(x) \neq x$ for all $x \in X$;
(2) $A \cap f(A) \neq \emptyset$ for all $A \subseteq X$ with $|A| = 40$.
Find the smallest positive integer $k$, such that for any such function $f$ there exists a set $B \su... | [
"First, we define a function $f: X \\to X$ with\n$$\n\\begin{aligned}\nf(3i - 2) &= 3i - 1, \\quad f(3i - 1) = 3i, \\quad f(3i) = 3i - 2, \\\\\ni &= 1, 2, \\dots, 30, \\\\\nf(j) &= 100, \\quad 91 \\le j \\le 99, \\quad f(100) = 99.\\end{aligned}\n$$\nObviously, $f$ satisfies condition (1). For any $A \\subseteq X$ ... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 69 | |
0ff6 | Problem:
El impuesto sobre el Rendimiento del Trabajo Personal es una función $f(x)$ del total $x$ de las retribuciones anuales (en pesetas). Sabiendo que
a) $f(x)$ es una función continua.
b) La derivada $d f(x) / d x$ en el intervalo $0 \leq x<60000$ es constante e igual a cero; en el intervalo $60000<x<P$ es cons... | [
"Solution:\n\nEn los tres intervalos la derivada es constante, luego la función es lineal y podemos expresarla\n$$\nf(x)= \\begin{cases}0 & 0 \\leq x<60000 \\\\ x-60000 & 60000 \\leq x<P \\\\ P-60000+0.14(x-P) & x \\geq P\\end{cases}\n$$\nImponiendo la condición $f(140000)=14000$, resulta\n$$\n14000 = P-60000 + 0.1... | Spain | Olimpiadas Matemáticas Españolas | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Functions"
] | null | proof and answer | P = 63255.81 pesetas (exact value 2720000/43). The function is: f(x) = 0 for 0 ≤ x < 60000; f(x) = x − 60000 for 60000 ≤ x < P; f(x) = (P − 60000) + 0.14 (x − P) for x ≥ P. | |
0bc0 | The parallelogram $ABCD$ has center $O$, $AD = DB$ and $BA > BC$. The point $V$ is outside the plane $(ABC)$ such that $VD \perp (ABC)$. Point $T$ is the foot of the perpendicular from $D$ onto $AB$ and $E, F, G$ are the feet of the bisectors of the angles $\angle VDA, \angle VDB, \angle VDC$ respectively (where $E \in... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 62nd NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0l2t | A data set containing 20 numbers, some of which are $6$, has mean $45$. When all the $6$s are removed, the data set has mean $66$. How many $6$s were in the original data set?
(A) 4 (B) 5 (C) 6 (D) 7 (E) 8 | [
"Let the number of $6$s in the original data set be $k$. Numbers in the original data set have sum $20 \\cdot 45 = 900$, and the data set with the $6$s removed has sum $900 - 6k$. Thus\n$$\n\\frac{900 - 6k}{20 - k} = 66.\n$$\nSolving this equation for $k$ yields $k = 7$. For example, the data set could consist of $... | United States | AMC 12 A | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0ebt | Find all pairs of positive integers $a$ and $b$ such that $a - b = 101$ holds and $ab$ is a perfect square. | [
"Let $d$ be the greatest common divisor of the numbers $a$ and $b$. Then we have $a = dm$ and $b = dn$, where $m$ and $n$ are co-prime positive integers. Since $ab = d^2mn$ is a perfect square and integers $m$ and $n$ are co-prime, they are also perfect squares. Let $m = x^2$ and $n = y^2$, where $x$ and $y$ are po... | Slovenia | National Math Olympiad 2015 – Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | a=2601, b=2500 | |
09nh | Find all sequences $a_1, a_2, \ldots$ of positive integers such that the expression $n a_n - m a_m + 2 a_m - 1$ is divisible by $a_n + a_m - 1$ for all $n, m \ge 1$.
(Unubold Munkhbat) | [
"Answer: $a_n = 1 + c(n - 1)$ for fixed $c \\ge 0$.\nIt is clear that the above is a solution, so we prove there are no other solutions.\nThe expression\n$$\nn - \\frac{n a_n - m a_m + 2 a_m - 1}{a_n + a_m - 1} = \\frac{n(a_m - 1) + (m - 2)a_m + 1}{a_n + a_m - 1}\n$$\nis an integer for all $n, m \\ge 1$. Let $c = a... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof and answer | a_n = 1 + c(n - 1) for a fixed nonnegative integer c | |
03eh | The first, seventh, and seventeenth terms of an arithmetic progression are distinct and consecutive terms of a geometric progression. To find the difference of the arithmetic progression if its first term is a solution of the equation
$$
x^2 - 9x + x\sqrt{12-x} - 9\sqrt{12-x} = 0.
$$ | [
"Let $a_1$ and $d$ be the first term and the difference of the arithmetic progression, respectively. From the condition $a_1$, $a_1 + 6d$ and $a_1 + 16d$ are consecutive members of a geometric progression, i.e.\n$$\n(a_1 + 6d)^2 = a_1 \\cdot (a_1 + 16d) \\iff d \\cdot (a_1 - 9d) = 0.\n$$\nSince $d \\neq 0$, we get ... | Bulgaria | 2 Bulgarian Winter Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | 1 or -4/9 | |
0dy1 | Find the smallest possible value of the expression $|5^{4m+3} - n^2|$, given that $m$ and $n$ are non-negative integers. | [
"Assume that $n^2 - 5^{4m+3} > 0$. If $n^2 - 5^{4m+3} = 1$, then $5^{4m+3} = n^2 - 1 = (n-1)(n+1)$. Since $n$ is obviously even the numbers $n-1$ and $n+1$ must be coprime, which implies $n-1 = 1$ and $n+1 = 5^{4m+3}$. This is not possible. If $n$ is divisible by $5$, then $n^2 - 5^{4m+3} \\ge 5$, if $n$ is of the ... | Slovenia | Slovenija 2008 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 4 | |
02m2 | Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that
(i) $f(f(n)) = f(n+1)$ for all $n \in \mathbb{Z}$;
(ii) $f(2009n + 2008) = 2009 \cdot f(n)$ for all $n \in \mathbb{Z}$. | [
"If $f$ is injective, then $f(f(n)) = f(n+1)$ reduces to $f(n) = n+1$, one of the solutions, another solution is $f(n) = 0$ for all $n \\in \\mathbb{Z}$. So suppose $f$ is neither injective nor identically zero. So there are two integers $a < b$ such that $f(a) = f(b)$. By applying (i) we obtain $f(a+1) = f(f(a)) =... | Brazil | XXXI Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Functional equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All solutions are: (1) f(n) = n + 1 for all integers n; (2) f(n) = 0 for all integers n; (3) f(n) = n + 1 for n < 0 and f(n) = 0 for n ≥ 0. | |
0erm | $\left(\frac{1}{8}\right)^2 \times 2^8$ is equal to
(A) 2
(B) 4
(C) 8
(D) 16
(E) 32 | [
"$\\left(\\frac{1}{8}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 256 = 4$\n\nOR\n\n$\\left(\\frac{1}{2^3}\\right)^2 \\times 2^8 = \\frac{1}{64} \\times 2^8 = 2^2 = 4$"
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | MCQ | B | |
07bn | a) Consider $n$ coprime natural numbers greater than $1$ like $d_1, d_2, \dots, d_n$ and arbitrary natural numbers $r_1, r_2, \dots, r_n$. Prove that there exists a natural number $x$, $1 \le x \le 3^n$, that satisfies the following system of modular inequalities:
$$
\begin{array}{l}
x \not\equiv r_1 \pmod{d_1} \\
x \n... | [
"a) Without loss of generality, we may suppose that $1 < d_1 < \\dots < d_n$. First, suppose that $d_1 > 2$. For each subset $I = \\{i_1 < \\dots < i_k\\} \\subseteq \\{1, 2, \\dots, n\\}$, we define $d_I$ to be equal to the product $d_{i_1} d_{i_2} \\dots d_{i_k}$. By the **Chinese Remainder Theorem**, we know tha... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0clm | Let $ABC$ and $DEF$ be triangles with the same circumcircle centred at $O$ and the same orthocentre $H \neq O$. The Simson lines of $D, E, F$ with respect to triangle $ABC$ form a non-degenerate triangle $\triangle$. Prove that the orthocentre of $\triangle$ lies on the circle on diameter $OH$.
**Note.** Assume that $... | [
"Let $\\ell_D, \\ell_E, \\ell_F$ be the Simson lines of $D, E, F$ with respect to triangle $ABC$ and let $X = \\ell_E \\cap \\ell_F$, $Y = \\ell_F \\cap \\ell_D$ and $Z = \\ell_D \\cap \\ell_E$. Let $D', E', F'$ be the midpoints of $HD, HE, HF$, respectively, and let $H'$ be the orthocentre of $\\triangle$.\n\nNote... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations... | English | proof only | null | |
09vk | Problem:
Gegeven is een positief geheel getal $n$. Bepaal de maximale waarde van $\operatorname{ggd}(a, b)+\operatorname{ggd}(b, c)+\operatorname{ggd}(c, a)$, onder de voorwaarde dat $a, b$ en $c$ positieve gehele getallen zijn met $a+b+c=5 n$. | [
"Solution:\n\nOplossing I. Schrijf $G=\\operatorname{ggd}(a, b)+\\operatorname{ggd}(b, c)+\\operatorname{ggd}(c, a)$. Zonder verlies van algemeenheid nemen we aan dat $a \\leq b \\leq c$. Nu geldt $\\operatorname{ggd}(a, b) \\leq a$, $\\operatorname{ggd}(b, c) \\leq b$ en $\\operatorname{ggd}(c, a) \\leq a$. Dus\n$... | Netherlands | IMO-selectietoets | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Maximum equals 5n if 3 divides n, and 4n otherwise. | |
0i0p | Problem:
Let $x_{1} = x_{2} = 1$ and $x_{n+2} = 2 x_{n+1} + 8 x_{n} - 1$ for $n \geq 1$. Prove that $x_{n}$ is a square for all $n$. | [
"Solution:\n\nLet $y_{1} = y_{2} = 1$ and $y_{n+2} = y_{n+1} + 2 y_{n}$ for $n \\geq 1$. Clearly $y_{n}$ is always an integer; we shall show inductively that $x_{n} = y_{n}^{2}$.\n\nFirst we demonstrate that $y_{n+1} - 2 y_{n} = \\pm 1$ for all $n$ by induction. This holds for $n = 1$, and if it holds for some $n$,... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
08jp | Problem:
Să se demonstreze inegalitatea
$$
\frac{x+y}{x^{2}-x y+y^{2}} \leq \frac{2 \sqrt{2}}{\sqrt{x^{2}+y^{2}}}
$$
pentru orice numere reale $x$ şi $y$, nu simultan nule. | [] | JBMO | Junior Balkan Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0agq | Let $a$, $b$, $c \in \mathbb{R}$ are such that $abc = 1$. Prove the inequality
$$
a^4 + b^4 + c^4 \geq a + b + c.
$$ | [
"By the trivial inequality $(x-y)^2 + (y-z)^2 + (z-x)^2 \\geq 0$ we have that for every $x$, $y$, $z \\in \\mathbb{R}$ it holds\n$$\nx^2 + y^2 + z^2 \\geq xy + yz + zx.\n$$\nBy applying this inequality, we get\n$$\n\\begin{align*}\na^4 + b^4 + c^4 &\\geq a^2 b^2 + b^2 c^2 + c^2 a^2 = (ab)^2 + (bc)^2 + (ca)^2 \\\\\n... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
098d | Problem:
Determinați toate valorile parametrului real $a$, pentru care ecuația $\sqrt{x-9}=a x+7 a-3$ are o soluție reală unică. | [
"Solution:\n\nRezolvare. $\\sqrt{x-9}=a x+7 a-3 \\Leftrightarrow \\sqrt{x-9}=a(\\sqrt{x-9})^{2}+16 a-3 \\Leftrightarrow a(\\sqrt{x-9})^{2}-\\sqrt{x-9}+16 a-3=0$.\n\n1. Dacă $a=0$, obținem $\\sqrt{x-9}=-3 \\Leftrightarrow x \\in \\varnothing$.\n\n2. Fie $a \\in \\mathbb{R}^{*}$. Notăm $\\sqrt{x-9}=t,\\ t \\geq 0$. O... | Moldova | Olimpiada Republicană la Matematică | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (0, 3/16) ∪ {1/4} | |
0i2f | Problem:
The manager of Chez Gastropod wants to write a menu consisting of 15 dishes. A "meal" is defined to be a subset of this menu (possibly empty), but some meals are legal and others are not. The manager may choose which meals are legal, but there is a requirement that the intersection of any two legal meals shou... | [
"Solution:\n\nThe answer is yes. Take any arbitrary 15-element set (menu), and call a collection of subsets (meals) \"valid\" if the intersection of any two sets in the collection is again in the collection. Thus, the objective is to show that there exists a valid collection containing exactly 2001 sets. We will sh... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Yes | |
069r | Let $P$ be an interior point to an equilateral triangle of altitude $1$. If $x$, $y$, $z$ are the distances from $P$ to the sides of the triangle, then prove that:
$$
x^2 + y^2 + z^2 \geq x^3 + y^3 + z^3 + 6xyz.
$$ | [
"It is well-known that in an equilateral triangle the sum of the distances from an interior point $P$ to its sides equals the altitude of the triangle, as can be easily proven. On account of the preceding, we have to prove that if $x + y + z = 1$ then it holds that\n$$\nx^2 + y^2 + z^2 \\geq x^3 + y^3 + z^3 + 6xyz\... | Greece | 21st Mediterranean Mathematical Competition | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0di6 | Let $ABCD$ be a trapezoid with $AB \parallel CD$ and $AB + CD = AD$. Its diagonals intersect at $E$. A line passing through the point $E$ and parallel to the bases of the trapezoid cuts $AD$ at $F$. Prove that $\angle BFC = 90^\circ$. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
040x | Let $a_1 \le a_2 \le \dots$ be a sequence of positive integers such that $\frac{r}{a_r} = k+1$ for some positive integers $k$ and $r$. Prove that there exists a positive integer $s$ such that $\frac{s}{a_s} = k$. | [] | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
0dgu | Let $x$, $y$, $z \ge 0$ and $x + y + z = 3$. Prove that
$$
\sqrt{x} + \sqrt{y} + \sqrt{z} \ge xy + xz + yz.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
03wp | The inequality
$$
\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n+1} < a - 2007 \frac{1}{3}
$$
holds for every positive integer $n$. Then the least positive integer of $a$ is $\underline{\hspace{2cm}}$. | [
"Obviously,\n$$\nf(n) = \\frac{1}{n+1} + \\frac{1}{n+2} + \\dots + \\frac{1}{2n+1}\n$$\nis monotonically decreasing. Therefore, $f(1)$ reaches the maximum of $f(n)$. From\n$$\nf(1) = \\frac{1}{2} + \\frac{1}{3} < a - 2007 \\frac{1}{3},\n$$\nwe have $a > 2008$. Therefore, the least positive integer of $a$ is $2009$.... | China | China Mathematical Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 2009 | |
01nk | Given cyclic pentagon $ABCDE$ with $AB = BC$, $CD = DE$. Segments $AD$ and $BE$ intersect at $P$, segment $BD$ intersects $CA$ and $CE$ at $Q$ and $T$, respectively.
Prove that triangle $PQT$ is isosceles. | [
"Since $AB = BC$, $CD = DE$ we have $\\sim AB = \\sim BC$, $\\sim CD = \\sim DE$.\n\n\n\nTherefore,\n$$\n\\angle BPA = \\angle DPE = \\frac{1}{2}(\\sim AB + \\sim DE) = \\frac{1}{2}(\\sim AB + \\sim CD) = \\angle BQA. \\quad (1)\n$$\n\nSo, points $A$, $B$, $Q$, $P$ belong to the same circle... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fe1 | Problem:
Encontrad todas las funciones $f: \mathbb{N} \rightarrow \mathbb{N}$ tales que $f(f(n))=n+2$ para todo número natural $n$. | [
"Solution:\n\nSea $f$ una función que cumple las condiciones del enunciado.\nSea $f(1)=a$. Reiterando obtenemos $f(a)=3$, $f(3)=a+2$, $\\ldots$, $f(n)=n+a-1$ si $n$ es impar.\nSea $f(2)=b$. De igual manera, $f(b)=4$, $f(n)=n+b-2$ si $n$ es par.\nDe hecho, las condiciones $f(a)=3$, $f(b)=4$, $f(n)=n+a-1$ si $n$ es i... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Two functions:
1) f(n) = n + 1 for all n ∈ ℕ.
2) f(n) = { n + 3 if n is odd; n − 1 if n is even }. | |
0j44 | Problem:
A classroom has 30 students and 30 desks arranged in 5 rows of 6. If the class has 15 boys and 15 girls, in how many ways can the students be placed in the chairs such that no boy is sitting in front of, behind, or next to another boy, and no girl is sitting in front of, behind, or next to another girl? | [
"Solution:\n\n$2 \\cdot 15!^{2}$\n\nIf we color the desks of the class in a checkerboard pattern, we notice that all of one gender must go in the squares colored black, and the other gender must go in the squares colored white. There are 2 ways to pick which gender goes in which color, $15!$ ways to put the boys in... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | final answer only | 2 * 15!^2 | |
07wy | Let $H$ denote the orthocentre of $\triangle ABC$, with circumcircle $\Gamma$. The altitudes $AH$, $BH$, $CH$ intersect $\Gamma$ for the second time at the points $A'$, $B'$ and $C'$, respectively. The circle of diameter $AH$ intersects $\Gamma$ at the points $A$ and $D$. Lines $DC$ and $A'B'$ intersect at $E$, while l... | [
"The idea of the proof is to show that *EF* is tangent at *H* to the circle which has *AH* as diameter. This will imply the result since *AH* is perpendicular to *BC*.\n\n$$\n\\angle HDE = \\angle MDC = \\angle MAC = 90^\\circ - \\angle AMC = 90^\\circ - \\angle B.\n$$\n\nBecause $AA'$ is p... | Ireland | IRL_ABooklet_2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Colline... | null | proof only | null | |
0iyy | Let $P(x)$ be a polynomial with integer coefficients such that $P(0) = 0$ and
$$
\text{gcd}(P(0), P(1), P(2), \dots) = 1.
$$
Prove that there are infinitely many positive integers $n$ such that
$$
\text{gcd}(P(n) - P(0), P(n+1) - P(1), P(n+2) - P(2), \dots) = n.
$$ | [
"Write\n$$\nP(x) = a_{r}x^{r} + a_{r-1}x^{r-1} + \\dots + a_{1}x,\n$$\nand consider its formal derivative\n$$\nQ(x) = r a_{r} x^{r-1} + (r-1) a_{r-1} x^{r-2} + \\dots + a_{1}.\n$$\nSince $P$ is not identically zero, neither is $Q$, so we may choose some positive integer $m$ such that $Q(m) \\neq 0$. We claim that w... | United States | Team Selection Test 2010 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
00dc | Find three consecutive odd numbers such that the sum of its squares is a 4-digit number whose digits are all equal. | [
"Let us compute the sum $S$ of the squares of three consecutive odd numbers $x$, $x+2$, $x+4$:\n$$\nS = x^2 + (x+2)^2 + (x+4)^2 = 3x^2 + 12x + 20.\n$$\nWe can see that $S$ is odd: indeed, $x$ is odd, hence $3x^2$ is odd, and $12x$ and $20$ are both even. Furthermore, $S$ has a remainder of 2 upon division by 3 (bec... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 41, 43, 45 | |
08j9 | Problem:
Is there a triangle whose area is $12~\mathrm{cm}^2$ and whose perimeter is $12~\mathrm{cm}$.
Problem:
Există un triunghi cu aria $12~\mathrm{cm}^2$ şi perimetrul $12$? | [
"Solution:\nNo. Let $r$ be the radius of the inscribed circle. Then $12 = 6r$, i.e. $r = 2~\\mathrm{cm}$. But the area of the inscribed circle is $4\\pi > 12$, and it is known that the area of any triangle is bigger than the area of its inscribed circle."
] | JBMO | 7th JBMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | No | |
08b9 | Problem:
Ada e Charles fanno un gioco. All'inizio un numero intero $n>1$ è scritto sulla lavagna. A turno, Ada e Charles cancellano il numero $k$ che trovano sulla lavagna e lo rimpiazzano
1 - o con un divisore positivo di $k$ diverso da 1 e da $k$ stesso
2 - oppure con $k+1$.
Inizialmente ciascuno dei giocatori po... | [] | Italy | XXXI Olimpiade Italiana di Matematica | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Exactly when the starting number is prime | |
08dv | Problem:
Dato il polinomio
$$
\left(x^{2}+5 x-19\right)^{50}=a_{0}+a_{1} x+a_{2} x^{2}+\cdots+a_{100} x^{100},
$$
si consideri la quantità
$$
M=a_{0}-a_{1}+4 a_{2}-9 a_{3}+16 a_{4}-81 a_{5}+64 a_{6}-729 a_{7}+\cdots+2^{98} a_{98}-3^{98} a_{99}+2^{100} a_{100}
$$
in cui il coefficiente di $a_{2 k}$ è $2^{2 k}$ e il coe... | [
"Solution:\n\nLa risposta è $(\\mathbf{D})$. Detto $p(x)$ il polinomio, $p(1)$ è la somma di tutti i coefficienti e $p(-1)$ è la somma a segni alterni dei coefficienti, quelli di indice pari con segno positivo e quelli di indice dispari con segno negativo. Di conseguenza, $p(1)+p(-1)$ è il doppio della somma dei co... | Italy | Olimpiadi della Matematica | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization"
] | null | MCQ | D | |
06g7 | Let $I$ be the incentre of a triangle $ABC$ with $AB > AC$. A line through $I$ perpendicular to $AI$ cuts $BC$ produced at $P$. Show that $IP^2 = BP \cdot CP$. | [
"We have\n$$\n\\angle CIP = \\angle CIA - 90^\\circ = \\angle CBI.\n$$\nThis shows $IP$ is tangent to $(BCI)$. Hence, $IP^2 = BP \\cdot CP$ by considering the power of $P$ with respect to $(BCI)$.\n\n"
] | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
09ii | Can the expression $\frac{a}{a+1} + \frac{a}{a+2} + \frac{a}{a+3} + \frac{a}{a+4}$ be an integer for some positive rational number $a$? | [
"Let $A = \\frac{a}{a+1} + \\frac{a}{a+2} + \\frac{a}{a+3} + \\frac{a}{a+4}$. Clearly, we have $0 < A < 4$.\n\n**Solution using GCD:** Write $a = \\frac{p}{q}$ with $p, q$ relatively prime positive integers. Then can write\n$$\nA = \\frac{p}{p+q} + \\frac{p}{p+2q} + \\frac{p}{p+3q} + \\frac{p}{p+4q} = \\frac{2p(2p+... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof and answer | No | |
0e1c | Problem:
Naj bosta $E$ in $F$ taki točki na stranicah $AB$ in $AD$ konveksnega štirikotnika $ABCD$, da je $EF \parallel BD$. Daljica $CE$ seka diagonalo $BD$ v točki $G$, daljica $CF$ pa seka diagonalo $BD$ v točki $H$. Dokaži: če je $AGCH$ paralelogram, je tudi $ABCD$ paralelogram. | [
"Solution:\n\nPresečišče premic $EF$ in $AG$ označimo z $I$, presečišče premic $EF$ in $AH$ pa z $J$. Ker je $EF \\parallel BD$ in $AH \\parallel CE$, je štirikotnik $EGHJ$ paralelogram. Zaradi vzporednosti $AG$ in $CF$ pa je tudi $FGIH$ paralelogram. Zato je $|FH| = |IG|$ in $|HJ| = |GE|$ ter velja $\\angle FHJ = ... | Slovenia | Slovenian Secondary School Mathematical Competition | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04mn | Let $ABC$ be a triangle such that $|AB| = 6$ and $|BC| = 9$, and let the measure of the angle $\angle ABC$ be $120^\circ$. Let $P$ and $Q$ be points on the side $\overline{AC}$ such that the triangle $BPQ$ is equilateral. Determine the length of the segment $\overline{PQ}$. | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English | proof and answer | 18√19/19 | |
04iu | For any two positive real numbers $x$ and $y$ prove
$$
\log^2(xy) \ge \log(x^2)\log(y^2).
$$ | [
"**3.1.** Let $D$ be the point on the side $\\overline{BC}$ such that $\\angle DAC = 90^\\circ$. Let us denote $\\varphi = \\angle CDA$ and $x = |CD|$. Then $\\cos \\angle ACB = \\sin \\varphi$.\n\nThen we have $|AC| = x \\sin \\varphi$ and $|BD| = |AD| = x \\cos \\varphi$. We also have $\\... | Croatia | First round – City competition | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0gek | 設三角形 $ABC$ 的內心為 $I$, $X \neq A$ 在外接圓 $\Omega$ 使得 $AI = XI$, 內切圓分別和 $AC, AB$ 切於 $E, F$, $M_a, M_b, M_c$ 分別為三邊 $BC, CA, AB$ 的中點, 再設 $T$ 為 $M_bF$ 和 $M_cE$ 的交點, $S$ 為 $AT$ 和 $\Omega$ 的另一個交點。
試證：$X, M_a, S, T$ 共圓。
Let $ABC$ be a triangle with incenter $I$ and circumcircle $\Omega$. A point $X$ on $\Omega$ which is differen... | [
"設 $O$ 為 $\\triangle ABC$ 的外心, $J$ 為 $EF$ 和 $OI$ 的交點, $L$ 為 $AJ$ 和 $\\Omega$ 異於 $A$ 的交點。\n\n**Claim.** $O, M_a, X, J, L$ 五點共圓。\n\n*Proof.* 由於 $JO$ 為 $\\angle XJA$ 的角平分線且 $XO = LO$, 因此 $J, L, O, X$ 共圓。另一方面, 設 $M$ 在 $OI$ 上為 $AX$ 中點, $Y$ 在 $\\Omega$ 上使得 $XY \\parallel BC$, 那麼由 $A, F, I, M, E$ 共圓且 $AFIE$ 為調和四邊形, 可知\n$$... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0l8l | Let be given a positive integer $n$ and two coprime integers $a, b$ greater than $1$. Let $p, q$ be two odd divisors greater than $1$ of $a^{6n} + b^{6n}$. Find the remainder of the division of $p^{6n} + q^{6n}$ by $6 \cdot (12)^n$. | [
"The answer follows easily from the following remarks:\n\na) If $a, b$ are coprime integers greater than $1$ and $p$ is an odd prime divisor of $a^{6n} + b^{6n}$ then $p \\equiv 1 \\pmod{2^{n+1}}$.\n\nb) If $x \\equiv 1 \\pmod{c^k}$ then $x^{c^m} \\equiv 1 \\pmod{2^{m+k}}$."
] | Vietnam | VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 2 | |
0hku | Problem:
Which is larger,
$$
3^{3^{3^{3}}} \text{ or } 2^{2^{2^{2^{2}}}} ?
$$ | [
"Solution:\nStarting to evaluate, the first expression becomes\n$$\nA = 3^{3^{3^{3}}} = 3^{3^{27}}.\n$$\nThe second expression is\n$$\nB = 2^{2^{2^{2^{2}}}} = 2^{2^{2^{4}}} = 2^{2^{16}}.\n$$\nIt is now obvious that\n$$\nA = 3^{3^{27}} > 3^{3^{16}} > 3^{2^{16}} > 2^{2^{16}} = B\n$$\nso the first expression is the la... | United States | Berkeley Math Circle | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 3^{3^{3^{3}}} | |
04t9 | Let $r$ and $r_a$ be the radii of the inscribed circle and the excircle opposite $A$ of the triangle $ABC$. Show that if
$$
r + r_a = |BC|,
$$
then the triangle is right-angled. | [
"Let us use the standard notation of the inner angles of the triangle $ABC$, further let $I$ be the incenter and $I_a$ be the excenter (of the excircle opposite $A$), and let $D$ and $E$ be in order the touching points of the thought circles. Since the bisectors $BI$ and $BI_a$ of the supplementary angles are perpe... | Czech Republic | 65th Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
03d0 | The lengths of the sides and the diagonal $AC$ of inscribed quadrilateral $ABCD$ are positive integers. If $\cos \angle ABC = \frac{1}{4}$, $2CD = AD + AC$ and $2AB = CA + CB + CD$ find the smallest possible value of the perimeter of $ABCD$. | [
"Let $\\alpha = \\angle ABC$, $AB = a$, $BC = b$, $CD = c$, $DA = d$ and $AC = e$. Then $\\angle ADC = 180^\\circ - \\alpha$, $2c = d + e$, $2a = b + e + c$ and it follows from the cosine theorem for $\\triangle ADC$ that\n$$\nd^2 + c^2 + \\frac{dc}{2} = e^2 = (2c - d)^2 \\iff 2c = 3d.\n$$\nSince $2c = d + e$ we ob... | Bulgaria | Bulgaria 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 83 | |
02hw | Problem:
Quantos são os números inteiros $x$ tais que $-5 < x-1 \leq 5$?
(A)
(B) 9
(C) 10
(D) 11
(E) 12 | [
"Solution:\n\nSomando 1 a todos os membros das duas desigualdades temos $-5+1 < x-1+1 \\leq 5+1 \\Rightarrow -4 < x \\leq 6$.\n\nOs valores inteiros de $x$ que satisfazem as duas desigualdades são: $-3, -2, -1, 0, 1, 2, 3, 4, 5, 6$.\n\n"
] | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | C | |
04yd | Let $ABC$ be a triangle and $D$ a point on its side $BC$. Points $E, F$ lie on the lines $AB, AC$ beyond vertices $B, C$, respectively, such that $BE = BD$ and $CF = CD$. Let $P$ be a point such that $D$ is the incenter of triangle $PEF$. Prove that $P$ lies inside the circumcircle $Ω$ of triangle $ABC$ or on it. (Jose... | [
"Let $ω$ be the circumcircle of triangle $AEF$ and let $I_A$ be the A-excenter of triangle $ABC$. First, we prove that $I_A$ is the midpoint of the arc $EF$ of $ω$ that does not contain point $A$ (see the left figure).\nTo that end, note that since $I_A$ lies on the external angle bisector of $∠B$ and $BE = BD$, tr... | Czech-Polish-Slovak Mathematical Match | CAPS Match 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals ... | English | proof only | null | |
0duf | Problem:
Dana je daljica $AB$. Na tej daljici izberimo poljubno točko $P$, različno od $A$ in $B$, ter konstruirajmo enakokraka pravokotna trikotnika $APQ$ in $PBR$, katerih vrhova $Q$ in $R$ ležita na istem bregu premice skozi $A$ in $B$. Naj bo točka $M$ razpolovišče daljice $QR$. Dokaži, da oddaljenost točke $M$ od... | [
"Solution:\n\nNalogo bomo rešili nekoliko splošneje. Privzeli bomo, da sta $APQ$ in $PBR$ podobna enakokraka trikotnika (z vrhoma $Q$ oz. $R$).\n\nOznačimo s $C$ presečišče premic $AQ$ in $BR$. Potem je $QPRC$ paralelogram in $M$ razpolovišče njegovih diagonal.\n\nČe sedaj označimo z $A'$ in $B'$ razpolovišči dalji... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0kj2 | Problem:
On a chessboard, a queen attacks every square it can reach by moving from its current square along a row, column, or diagonal without passing through a different square that is occupied by a chess piece. Find the number of ways in which three indistinguishable queens can be placed on an $8 \times 8$ chess boar... | [
"Solution:\n\nThe configuration of three cells must come in a $45$-$45$-$90$ triangle. There are two cases, both shown above: the triangle has legs parallel to the axes, or it has its hypotenuse parallel to an axis.\n\nThe first case can be solved by noticing that each selection of four cel... | United States | HMMT November 2021 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 864 | |
0bdw | We say that a group $(G, \cdot)$ has the property $(P)$ if, for any automorphism $f$ of $G$, there exist two automorphisms $g$ and $h$ of $G$ such that $f(x) = g(x) \cdot h(x)$, for any $x \in G$. Prove that:
a. Any group with the property (P) is commutative.
b. Any finite group of odd order has the property (P).
c.... | [
"a. Suppose $g, h \\in \\text{Aut}(G)$ with $x = g(x)h(x)$ for all $x \\in G$. Let $a$ and $b$ be elements of $G$. Then $a = g(a)h(a)$, $b = g(b)h(b)$ and $ab = g(ab)h(ab)$.\n\nIt follows that $g(a)h(a)g(b)h(b) = ab = g(ab)h(ab) = g(a)g(b)h(a)h(b)$, hence $h(a)g(b) = g(b)h(a)$.\nSince $g$ and $h$ are onto, the clai... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
048k | One of the midlines in a triangle is longer than one of its medians. Prove that the triangle is obtuse. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0dog | Problem:
Дат је природан број $k$. Посматрајмо следећу игру на бесконачној једнодимензионалној табли. На почетку игре, на поља постављамо укупно $n$ жетона, при чему може бити више жетона на истом пољу. Након тога, у сваком потезу извршавамо једну од следећих операција:
$(1^{\circ})$ бирамо два суседна поља која су об... | [
"Solution:\n\n(a) Потезе врсте $(1^{\\circ})$ зваћемо скупљањем, а потезе врсте $(2^{\\circ})$ сејањем. Поља нумеришемо редом бројевима $\\ldots, -1, 0, 1, 2, \\ldots$. Поља се могу поделити на $k$ класа по модулу $k$, тако да при сејању жетони не мењају своју класу.\n\nПретпоставимо прво да у једној од класа нема ... | Serbia | 14. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | For all k, there exists a choice of n and an initial placement that allows an infinite sequence of moves. | |
09y7 | Problem:
Stekel en Prik spelen een spel op een $m \times n$-bord, waarbij $m$ en $n$ positieve gehele getallen zijn. Ze zijn afwisselend aan de beurt, waarbij Stekel begint. Stekel zet in zijn beurt steeds een pion op een vakje waar nog geen pion staat. Prik doet in zijn beurt hetzelfde, maar zijn pion moet altijd kom... | [
"Solution:\n\nAls $m$ (het aantal rijen) even is, dan koppelen we de vakjes van het bord in tweetallen aan elkaar: in elke kolom vormen de bovenste twee vakjes een tweetal, en het derde en vierde vakje, enzovoorts. Omdat er een even aantal rijen is, lukt dit. Prik kan nu de volgende strategie hanteren: als Stekel e... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Stekel (first player) wins if and only if m and n are both odd and either m = n = 3 or at least one of m or n is at least 5 (i.e., all odd-by-odd boards except 1×1 and 1×3/3×1). Prik (second player) wins in all other cases: if m is even or n is even, or (m, n) ∈ {(1,1), (1,3), (3,1)}. | |
07xj | Suppose $a, b, c$ are real numbers such that $a + b + c = 1$ and $a^2 + b^2 + c^2 = 1$. Prove that
$$
-\frac{4}{27} \le abc \le 0.
$$ | [
"To begin with,\n$$\n1 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca) = 1 + 2(ab + bc + ca)\n$$\nso that $ab+bc+ca = 0$. Now we can prove the upper bound of zero. Suppose (for a contradiction) that a real triple $a, b, c$ satisfies the given constraints and $abc > 0$. Then either\n\n(i) all three of $a, b, c$ ... | Ireland | IRL_ABooklet_2025 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof only | null | |
07yp | Problem:
Una macchina misteriosa contiene una combinazione segreta costituita da 2016 numeri interi $x_{1}, \ldots, x_{2016}$. Sappiamo che tutti i numeri della combinazione sono uguali salvo uno. È possibile interrogare la macchina sottoponendo ad essa una sequenza di 2016 numeri interi $y_{1}, \ldots, y_{2016}$. La ... | [
"Solution:\n\nLa risposta ad entrambe le richieste dell'esercizio è due domande. Per risolvere completamente l'esercizio basta esibire un metodo per determinare la combinazione in due domande, e mostrare che, anche sapendo che il numero diverso è zero, una domanda non è sufficiente. A scopo illustrativo, daremo anc... | Italy | null | [
"Algebra > Linear Algebra > Vectors",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | (a) 2, (b) 2 | |
0dvm | Problem:
V letalu, ki ima 62 vrst s po 6 sedeži v vsaki vrsti, so se potniki posedli tako, da v nobenih 2 vrstah niso zasedeni sedeži na istih mestih. Največ koliko potnikov je lahko v letalu? | [
"Solution:\n\nKer je v vsaki vrsti 6 sedežev, imamo $2^{6} = 64$ različnih možnosti posedanja potnikov v posamezni vrsti. Ker lahko za vsakega izmed 64 načinov posedanja najdemo njemu komplementarnega (tj. takega, ki ima zasedene natanko tiste sedeže, ki so v dani vrsti nezasedeni), lahko iz 32 komplementarnih paro... | Slovenia | 47. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 191 | |
0872 | Problem:
Determinare tutti gli interi positivi $m$ per i quali sia $\frac{2 \cdot 5^{m}+10}{3^{m}+1}$ che $\frac{9^{m}+1}{5^{m}+5}$ sono interi. | [
"Solution:\n\nNotiamo che se $a=\\frac{2 \\cdot 5^{m}+10}{3^{m}+1}$ e $b=\\frac{9^{m}+1}{5^{m}+5}$ sono entrambi interi, allora anche il loro prodotto,\n$$\na b=\\frac{\\left(2 \\cdot 5^{m}+10\\right)\\left(9^{m}+1\\right)}{\\left(3^{m}+1\\right)\\left(5^{m}+5\\right)}\n$$\nè ancora intero.\nRaccogliendo ora un fat... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m = 1 | |
041n | In a group of $m$ girls and $n$ boys, any two of them either know each other, or do not know each other. For any two boys and two girls, at least one boy and one girl do not know each other. Prove that the number of boy-girl pairs that know each other is at most $m + \frac{n(n-1)}{2}$. | [
"From the hypothesis, for any two boys, there is at most one girl that knows both of them. Let $x_i$ be the number of girls that know exactly $i$ boys, $1 \\le i \\le n$. So $\\sum_{i=1}^{n} x_i = m$. By counting the number of the above two boys-one girl combinations, we have\n$$\n\\sum_{i \\ge 2} \\frac{i(i-1)}{2}... | China | China Girls' Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0gdh | 設三角形 $ABC$ 的外接圓為 $\Omega$, 內心為 $I$, $A$-旁心為 $J$。令 $T$ 為 $J$ 對 $BC$ 的對稱點, $P$ 為 $BC$ 與 $AT$ 的交點。若 $\triangle AIP$ 的外接圓交 $BC$ 於 $X \neq P$, 點 $Y \neq A$ 位於 $\Omega$ 上使得 $\overline{IA} = \overline{YI}$, 證明: $\triangle IXY$ 的外接圓與直線 $AI$ 相切。 | [
"(∠代表有向角。)\n\n\n\n令 $M$, $N$ 分別為 $\\overline{BC}$, $\\overline{IJ}$ 中點, $N'$ 為 $N$ 對 $M$ 的對稱點, $D$ 為 $J$ 關於 $BC$ 的垂足。熟知 $IM \\parallel AD$, 又 $MN \\perp BC \\perp DJ$, 所以 $\\triangle ADJ$ 與 $\\triangle IMN$ 位似, 因此由 $T$ 為 $J$ 對 $D$ 的對稱點, 知 $AT \\parallel IN'$. 設 $O$ 為 $\\triangle ABC$ 的外心, 由... | Taiwan | 二〇一九數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
09a1 | Let $I = \{1, 2, 3, ..., 2n\}$. Is it possible to set $I$ divide into $A, B$ two subsets with following properties?
(i) $A = \{a_1, a_2, ..., a_n\}, B = \{b_1, b_2, ..., b_n\}$
(ii) $A \cup B = I, A \cap B = \emptyset$
(iii) The number of divisors of $(a_1+b_1)(a_2+b_2)...(a_n+b_n)$ is less than or equal to $2^n$.
(pro... | [
"By the induction method we will show that divide in to two subset $A, B$ such that $a_i + b_i$ is prime number for $i|1, n$.\nIf $n = 1$ then $\\{1, 2\\} = \\{1\\} \\cup \\{2\\}$, $1 + 2 = 3$ is prime number and $\\{1\\} \\cap \\{2\\} = \\emptyset$. Assume now that the result is proved for $1, 2, 3, ..., n - 1$. N... | Mongolia | 45th Mongolian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0875 | Problem:
A un tavolo, vi sono quattro persone: Luca, Maria, Nicola e Paola. Ognuno dei quattro mente sempre, oppure non mente mai. Inoltre non amano parlare di loro stessi, ma piuttosto dei loro amici; tant'è che quando gli viene chiesto chi di loro menta sempre, le loro risposte sono:
Luca: "ogni ragazza è sempre si... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Certamente Luca non può essere sincero, in quanto le due ragazze si contraddicono, quindi non possono dire entrambe il vero. Se Maria dicesse il vero, Nicola mentirebbe; perciò, non è vero che c'è una ragazza sincera, e in particolare Maria non lo è, contraddizione. Quin... | Italy | Progetto Olimpiadi di Matematica GARA di SECONDO LIVELLO | [
"Discrete Mathematics > Logic"
] | null | MCQ | C | |
05sv | Problem:
On considère une grille $2019 \times 2019$. On note $C_{i, j}$ la case sur la $i$-ième colonne et la $j$-ième ligne pour $1 \leqslant i, j \leqslant 2019$. Un coloriage de la grille est formidable s'il n'existe pas $1 \leqslant i < \ell \leqslant 2019$ et $1 \leqslant j < k \leqslant 2019$ tels que les cases ... | [
"Solution:\n\nDans ce problème, on cherche le plus petit entier satisfaisant une certaine propriété. Supposons que l'on veuille montrer que le plus petit entier recherché est l'entier $c$. Pour montrer que $c$ est bien le plus petit entier recherché, on va d'une part montrer que si un entier $n$ satisfait la propri... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1010 | |
0khg | Let $c = \frac{2\pi}{11}$. What is the value of
$$
\frac{\sin 3c \cdot \sin 6c \cdot \sin 9c \cdot \sin 12c \cdot \sin 15c}{\sin c \cdot \sin 2c \cdot \sin 3c \cdot \sin 4c \cdot \sin 5c} ?
$$
(A) -1 (B) $-\frac{\sqrt{11}}{5}$ (C) $\frac{\sqrt{11}}{5}$ (D) $\frac{10}{11}$ (E) 1 | [] | United States | AMC 12 B | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | E | |
0dxd | Problem:
Poišči vse pare kompleksnih števil $z$ in $w$, ki rešijo sistem enačb
$$
\begin{aligned}
w^{2} & =63+16 i \\
2 z-i|z| & =w
\end{aligned}
$$ | [
"Solution:\n\nNaj bo $z=a+b i$ in $w=c+d i$. Iz prve enačbe sledi\n$$\nc^{2}+2 i c d-d^{2}=63+16 i\n$$\nS primerjavo realnih in imaginarnih delov dobimo sistem enačb\n$$\n\\begin{aligned}\nc^{2}-d^{2} & =63 \\\\\n2 c d & =16\n\\end{aligned}\n$$\nIz druge enačbe izrazimo $d=\\frac{8}{c}$ in vstavimo v prvo enačbo, d... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Two solutions: w = 8 + i with z = 4 + 3i; and w = −8 − i with z = −4 + (5/3)i. | |
0haf | Find the biggest three-digit number $n$, for which the following condition is true: there exist exactly 16 pairs of natural numbers $(a, b)$, where $a < b$, for which $n$ is the least common multiple. | [
"At first suppose, that number $n = p_1^{m_1} \\dots p_k^{m_k}$, where $p_i$, $i = 1, \\overline{k}$ – are distinct prime numbers and $m_i$ – are natural numbers, $i = 1, \\overline{k}$. Then let's choose all ordered pairs of numbers $(a, b)$, for which $[a, b] = n$. Here we do not have condition $a < b$, so pairs ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 992 | |
093m | Problem:
Let $ABC$ be a right-angled triangle with its right angle at $B$ and circumcircle $c$. Denote by $D$ the midpoint of the shorter arc $AB$ of $c$. Let $P$ be the point on the side $AB$ such that $CP = CD$ and let $X$ and $Y$ be two distinct points on $c$ satisfying $AX = AY = PD$. Prove that the points $X$, $Y... | [
"Solution:\n\nIt will be enough to prove that $PX \\perp AC$. Without loss of generality assume that $X$ lies in another half-plane with regard to the line $AC$ than the point $B$. Let us denote $K \\neq D$ the point of intersection of the line $DP$ and the circle $c$. Further, let $S$ be the point on the ray $AX$ ... | Middle European Mathematical Olympiad (MEMO) | MEMO Team Competition | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | null | proof only | null | |
0ff1 | Problem:
En el triángulo $ABC$, se trazan la bisectriz interior $AL$ ($L$ pertenece al lado $BC$), la altura $BH$ ($H$ pertenece al lado $AC$) y la mediana $CM$ ($M$ pertenece al lado $AB$).
Se sabe que los ángulos $\angle CAL$, $\angle ABH$ y $\angle BCM$ son iguales.
Determinar, razonadamente, las medidas de los áng... | [
"Solution:\n\nLlamemos $\\alpha = \\angle CAL = \\angle ABH = \\angle BCM$. Entonces $\\angle BAC = 2\\alpha$.\n\nSi el ángulo $BAC$ fuera obtuso, el punto $H$ estaría en la prolongación del lado $AC$ más allá de $A$, y por lo tanto $2\\alpha = \\alpha + 90^\\circ$, de donde $\\alpha = 90^\\circ$. Esto es imposible... | Spain | TANDA III | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | A = B = C = 60° or A = 60°, B = 30°, C = 90° | |
03j4 | Problem:
Let $P_{1}$ and $P_{2}$ be regular polygons of $1985$ sides and perimeters $x$ and $y$ respectively. Each side of $P_{1}$ is tangent to a given circle of circumference $c$ and this circle passes through each vertex of $P_{2}$. Prove $x + y \geq 2c$. (You may assume that $\tan \theta \geq \theta$ for $0 \leq \... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
09mx | Bobby constructed a rectangular parallelepiped frame using wire. He possessed a tape measure capable of measuring both the distance between any two points and the length of any curved line. Bobby boasted that he could determine the surface area of his parallelepiped using just two measurements with this tape measure. W... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Solid Geometry > 3D Shapes",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0jzw | Problem:
Let $\omega$ and $\Gamma$ be circles such that $\omega$ is internally tangent to $\Gamma$ at a point $P$. Let $AB$ be a chord of $\Gamma$ tangent to $\omega$ at a point $Q$. Let $R \neq P$ be the second intersection of line $PQ$ with $\Gamma$. If the radius of $\Gamma$ is $17$, the radius of $\omega$ is $7$, a... | [
"Solution:\nLet $r$ denote the circumradius of triangle $AQR$. By Archimedes Lemma, $R$ is the midpoint of arc $AB$ of $\\Gamma$. Therefore $\\angle RAQ = \\angle RPB = \\angle RPA$ so $\\triangle RAQ \\sim \\triangle RPA$. By looking at the similarity ratio between the two triangles we have\n$$\n\\frac{r}{17} = \\... | United States | February 2017 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(170) | |
0a8e | Problem:
The real-valued function $f$ is defined for $0 \leq x \leq 1$, $f(0)=0$, $f(1)=1$, and
$$
\frac{1}{2} \leq \frac{f(z)-f(y)}{f(y)-f(x)} \leq 2
$$
for all $0 \leq x < y < z \leq 1$ with $z-y = y-x$. Prove that
$$
\frac{1}{7} \leq f\left(\frac{1}{3}\right) \leq \frac{4}{7}
$$ | [
"Solution:\n\nWe set $f\\left(\\frac{1}{3}\\right)=a$ and $f\\left(\\frac{2}{3}\\right)=b$. Applying the inequality of the problem for $x=\\frac{1}{3}$, $y=\\frac{2}{3}$ and $z=1$, as well as for $x=0$, $y=\\frac{1}{3}$, and $z=\\frac{2}{3}$, we obtain\n$$\n\\frac{1}{2} \\leq \\frac{1-b}{b-a} \\leq 2, \\quad \\frac... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 14 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1/7 <= f(1/3) <= 4/7 | |
05r8 | Problem:
Soit $P$ un polynôme à coefficients rationnels de degré supérieur ou égal à $2$, et $\left(q_{n}\right)_{n \in \mathbb{N}}$ une suite de rationnels tels que pour tout $n \geqslant 0$, $q_{n} = P\left(q_{n+1}\right)$. Montrer que la suite $q_{n}$ est périodique à partir d'un certain rang. | [
"Solution:\n\nSoit $u$ entier tel que $u_{1}$ soit entier. Soit $\\tilde{P}(X) = u P\\left(\\frac{1}{u} X\\right)$. Soit $v$ entier tel que $v \\tilde{P}$ soit à coefficients entiers. Soit $a$ le coefficient dominant de $v \\tilde{P}$. On pose $Q(X) = a \\tilde{P}\\left(\\frac{1}{a} X\\right)$, $m = a^{\\operatorna... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0gb9 | 平面上有一個三角形 $ABC$，其外接圓為 $\Gamma$，設點 $A'$ 是點 $A$ 在 $\Gamma$ 上的對徑點。作正三角形 $BCD$，使得 $A, D$ 兩點位於 $BC$ 的異側。設過 $A'$ 且與 $A'D$ 垂直的直線分別與直線 $AC, AB$ 交於 $E, F$ 兩點。以 $EF$ 為底，作底角為 $30^\circ$ 的等腰三角形 $ETF$，並使 $A, T$ 兩點位於 $EF$ 的異側。證明：$AT$ 經過三角形 $ABC$ 的九點圓圓心 $N$。
註：三角形 $ABC$ 的九點圓，係指通過三角形 $ABC$ 三邊的中點、三高的垂足、與頂點到垂心的三條線段的中點這九個點的圓。 | [
"我們首先證明兩個引理。\n\n**引理 1.** 設 $P, Q$ 為 $\\triangle ABC$ 的一對等角共軛點。若 $O, O_a, O_b, O_c, T$ 分別為 $\\triangle ABC, \\triangle BPC, \\triangle CPA, \\triangle APB$ 及 $\\triangle O_aO_bO_c$ 的外心，則 $PQ$ 與 $OT$ 平行。\n\n**引理 1. 證明** 設 $\\triangle Q_aQ_bQ_c$ 為 $Q$ 關於 $\\triangle ABC$ 的垂足三角形，且設 $V$ 為 $\\triangle Q_aQ_bQ_c$ 的外心。因為 ... | Taiwan | 二〇一七數學奧林匹亞競賽第三階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"... | null | proof only | null | |
0c0o | If $a$, $b$, $c$ are positive real numbers, prove that
$$
\frac{a}{\sqrt{(a+2b)^3}} + \frac{b}{\sqrt{(b+2c)^3}} + \frac{c}{\sqrt{(c+2a)^3}} \ge \frac{1}{\sqrt{a+b+c}}.
$$ | [
"From Hölder's inequality we have:\n$$\n\\sum_{cycl} \\frac{a}{\\sqrt{(a+2b)^3}} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\cdot \\sum_{cycl} a\\sqrt{a+2b} \\ge (a+b+c)^4,\n$$\nand from Cauchy-Buniakowsky-Schwarz it follows that\n$$\n\\left(\\sum_{cycl} a\\sqrt{a+2b}\\right)^2 = \\left(\\... | Romania | 69th NMO Selection Tests for JBMO | [
"Algebra > Equations and Inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
09rh | Problem:
Vind alle drietallen $(x, n, p)$ van positieve gehele getallen $x$ en $n$ en priemgetallen $p$ waarvoor geldt
$$
x^{3}+3 x+14=2 \cdot p^{n} .
$$ | [
"Solution:\nOmdat de rechterkant een product is van priemfactoren, zou het handig zijn als de linkerkant ook een product is. Het zou dan te factoriseren moeten zijn in een eerstegraads en een tweedegraads polynoom in $x$. Proberen geeft dan de volgende ontbinding. We kunnen de linkerkant ontbinden als $(x+2)\\left(... | Netherlands | Selectietoets | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (1, 2, 3) and (3, 2, 5) | |
07hw | For two rational numbers $r, s$ we say $r|s$ if there is $k \in \mathbb{Z}$ so that $s = kr$. The sequence $(a_n)_{n \in \mathbb{N}}$ is an increasing sequence of natural numbers such that for all $i, j \in \mathbb{N}$, $\gcd(a_i, a_j) = 1$ and $(b_n)_{n \in \mathbb{N}}$ is a sequence of distinct natural numbers. Assum... | [
"$$\n\\sum_{i=1}^{n} \\frac{1}{b_i} = k_n \\sum_{i=1}^{n} \\frac{1}{a_i}\n$$\nby subtracting these equalities for $n, n + 1$ we have:\n$$\n\\frac{1}{b_{n+1}} = k_{n+1} \\frac{1}{a_{n+1}} + (k_{n+1} - k_n) \\sum_{i=1}^{n} \\frac{1}{a_i}\n$$\nChoose a big enough $m$, such that for $n > m$ we have $b_n > a_0$, (this i... | Iran | 40th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
007j | Several white and black balls can be divided into pairs so that exactly $\frac{10}{11}$ of the white balls are in mixed pairs (with one white and one black ball), and the remaining ones are in pairs with the same color. Also the balls can be divided into pairs so that exactly $\frac{12}{13}$ of the black balls are in m... | [
"Let there be $x$ white and $y$ black balls. Since $\\frac{10}{11}x$ is an integer, $x$ is divisible by $11$. Next, the $x - \\frac{10}{11}x = \\frac{x}{11}$ balls not in a mixed pair in the first division must be paired up among themselves. Hence $\\frac{x}{11}$ is even, i.e. $x$ is even. Thus $x$ is divisible by ... | Argentina | National Olympiad of Argentina | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (154, 156), (176, 182), (198, 182), (198, 208) | |
090g | A pentagon **ABCDE** is inscribed in a circle $\Omega$, and it satisfies $AC = AD$ and $BC \parallel DE$. Take a point $P$ on the arc $CD$ of $\Omega$ which does not contain $A$. Let $P_1, P_2, P_3, P_4$, and $P_5$ be the points symmetric to $P$ with respect to the lines $AB, BC, CD, DE$, and $EA$, respectively. When $... | [
"$\\boxed{\\frac{\\sqrt{37}}{10}}$\n\nSince $C$ is the circumcenter of the triangle $PP_2P_3$, we have $P_2P_3 = 2PC \\sin \\angle P_2PP_3$ by the law of sines. Similarly, since $D$ is the circumcenter of the triangle $PP_4P_3$, we have\n\n$P_4P_3 = 2PD \\sin \\angle P_4PP_3$. Since the lines $BC$ and $DE$ are para... | Japan | The 35th Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Ge... | English | proof and answer | sqrt(37)/10 | |
0hoy | Problem:
Prove that for any positive integer $n$, we have
$$
\prod_{k=1}^{n} \operatorname{lcm}\left(1,2, \ldots,\left\lfloor\frac{n}{k}\right\rfloor\right)=n !
$$ | [
"Solution:\nWe show that the exponents of $p$ coincide for any prime $p$. Actually, we'll prove the stronger claim that for any prime power $q$, the number of terms on each side divisible by $q$ is exactly the same. By the fundamental theorem of arithmetic, that will imply the desired equality.\n\nClearly, the numb... | United States | Berkeley Math Circle: Monthly Contest 6 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0bza | A planar country has an odd number of cities separated by pairwise distinct distances. Some of these cities are connected by direct two-way flights. Each city is directly connected to exactly two other cities, and the latter are located farthest from it. Prove that, using these flights, one may go from any city to any ... | [
"Consider the graph $G$ whose vertices are the cities and whose edges are the direct two-way flights. We must show that $G$ is connected.\n\nAssuming $G$ has at least two components, we show that the cardinality of each component is even; this contradicts the fact that the total number of vertices is odd, whence th... | Romania | THE 68th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0ir3 | Let $n$ be a positive integer. Denote by $S_n$ the set of points $(x, y)$ with integer coordinates such that
$$
|x| + \left\lfloor y + \frac{1}{2} \right\rfloor < n.
$$
A *path* is a sequence of distinct points $(x_1, y_1), (x_2, y_2), \dots, (x_\ell, y_\ell)$ in $S_n$ such that, for $i = 2, \dots, \ell$, the distance ... | [
"Color the points in $S_n$ as follows (see Figure 1):\n- if $y \\ge 0$, color $(x, y)$ white if $x + y - n$ is even and black if $x + y - n$ is odd;\n- if $y < 0$, color $(x, y)$ white if $x + y - n$ is odd and black if $x + y - n$ is even.\n\n\nFigure 1: Coloring of $S_3$\n\nConsider a pat... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0csk | Точка $M$ — середина стороны $AC$ остроугольного треугольника $ABC$, в котором $AB > BC$. Окружность $\Omega$ описана около треугольника $ABC$. Касательные к $\Omega$, проведённые в точках $A$ и $C$, пересекаются в точке $P$. Отрезки $BP$ и $AC$ пересекаются в точке $S$. Пусть $AD$ — высота треугольника $ABP$. Окружнос... | [
"Поскольку $\\angle AMP = \\angle ADP = 90^\\circ$, точки $M$ и $D$ лежат на окружности $\\gamma$ с диаметром $AP$. Поскольку $PA$ — касательная к $\\Omega$, имеем $\\angle KAP = \\angle ACK$. Так как точки $C, K, D$ и $S$ лежат на окружности $\\omega$, имеем $\\angle ACK = \\angle KDP$. Значит, $\\angle KAP = \\an... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aps | Problem:
What is the least 6-digit natural number that is divisible by $198$? | [
"Solution:\n\n$100188$\n\nSince $198 \\times 500 = 99000$ and $198 \\times 5 = 990$, we have $198 \\times 505 = 99990$. Thus, the least six-digit natural number that is divisible by $198$ is $99990 + 198 = 100188$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 100188 | |
07w1 | Prove that
$$
B = \sum_{k=0}^{n} (-1)^k \binom{n}{k}^2 = \begin{cases} 0, & \text{if } n \text{ is odd,} \\ (-1)^{\frac{n}{2}} \binom{n}{\frac{n}{2}}, & \text{if } n \text{ is even.} \end{cases}
$$ | [
"If $p, q$ are polynomials of $\\deg \\le n$ with coefficients $a_k, b_k$ respectively, then the coefficients of the product $pq$ may be found in terms of those of $p$ and $q$ as follows:\n$$\n\\begin{aligned}\np(x)q(x) &= \\sum_{i=0}^{n} a_i x^i \\sum_{j=0}^{n} b_j x^j = \\sum_{i=0}^{n} a_i b_j x^{i+j} \\\\\n&= \\... | Ireland | IRL_ABooklet_2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Generating functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0h9a | Given quadrilateral $ABCD$, such that $AB = BC$. $K$ is the midpoint of $CD$, rays $BK$ and $AD$ intersect at $M$, the circumcircle of $\triangle ABM$ intersects line $AC$ for the second time at point $P$. Show that $\angle BKP = 90^\circ$.
(Anton Trygub) | [
"Let $N$ be the midpoint of diagonal $AC$, then $NK$ is the midline of $\\triangle ACD$ (Fig. 31). Thus, $\\angle (KN, NP) = \\angle (AM, AP) = \\angle (BM, BP) = \\angle (KB, BP)$, hence $BNKP$ is inscribed. Thus, since $\\triangle ABC$ is equilateral, $\\angle BNP = 90^\\circ$, therefore, $\\angle BKP = 90^\\circ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bof | Let $m$ and $n$ be positive integers, $n \ge 2$, let $R$ be an $n$-element ring, and let $x$ be an element of $R$ such that $1 - x^k$ is invertible for each $k \in \{m+1, m+2, \dots, m+n-1\}$. Show that $x$ is nilpotent (i.e., $x^p = 0$ for some positive integer $p$). | [
"Since $0$ is nilpotent, we will assume in the sequel that $x \\ne 0$.\nGiven a positive integer $k < n$, notice that $m+p$ is divisible by $k$ for some positive integer $p < n$; write $m + p = kl$. Since $1 - x^{m+p}$ is invertible, and $1 - x^{m+p} = (1 - x^k)(1 + x^k + \\dots + x^{k(\\ell-1)}) = (1 + x^k + \\dot... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Ring Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0dxf | Problem:
Kmet Martin je pri baliranju sena izdelal 3 enake bale sena. Prečni prerez vsake bale sena je krog s polmerom $r$. Bale sena je postavil v piramido, tako da so se dotikale druga druge, in čez njih napel vrv, ki jo je na obeh koncih pritrdil v tla. Zgornji del vrvi je napet po obodu zgornje bale sena, stranska... | [
"Solution:\n\n\n\nOznačimo nekaj točk kot je prikazano na sliki. Točke $A, B$ in $C$ so središča bal sena, točka $D$ je eden od koncev vrvi, točki $E$ in $F$ sta dotikališči leve bale sena s tlemi in z vrvjo, točki $G$ in $H$ pa sta skrajni točki, kjer se vrv še dotika zgornje bale sena. Tr... | Slovenia | 66. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | (2*sqrt(3)+4+2*pi/3)*r | |
0fdz | Problem:
Describir todas las soluciones enteras positivas $(m, n)$ de la ecuación $8m - 7 = n^2$ y dar el primer valor de $m$ (si existe) mayor que $1959$. | [
"Solution:\nSea $m = k + 1$, tenemos $8k + 1 = n^2$, o equivalentemente, $8k = (n - 1)(n + 1)$. Evidentemente, $n$ es impar y si consideramos el entero $s = \\frac{n - 1}{2}$, deducimos que $k = \\frac{1 + s}{2} s$ es un número triangular. De hecho, si $k$ es triangular (i.e. $m$ es el número posterior a un triangu... | Spain | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All solutions are (m, n) = (1 + s(s + 1)/2, 2s + 1) for integers s ≥ 0. The smallest m > 1959 is 2017. | |
014i | Problem:
Determine all positive integers $n$ such that $3^{n}+1$ is divisible by $n^{2}$. | [
"Solution:\n\nFirst observe that if $n^{2} \\mid 3^{n}+1$, then $n$ must be odd, because if $n$ is even, then $3^{n}$ is a square of an odd integer, hence $3^{n}+1 \\equiv 1+1=2\\pmod{4}$, so $3^{n}+1$ cannot be divisible by $n^{2}$ which is a multiple of $4$.\n\nAssume that for some $n>1$ we have $n^{2} \\mid 3^{n... | Baltic Way | Baltic Way | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | n = 1 | |
08z3 | For each positive integer $n$, define $f(n)$ by
$$
f(n) = \begin{cases}
n^{100} & \text{(if digit sum of $n$ is even)}, \\
-n^{100} & \text{(if digit sum of $n$ is odd)}.
\end{cases}
$$
Let $S = f(1) + f(2) + \dots + f(10^{100} - 1)$, and determine the largest non-negative integer $m$ such that $S$ is divid... | [
"In the following, we denote $d = 100$ and let $f(0) = 0$. Note that $S = f(0) + f(1) + \\dots + f(10^d - 1)$. For any integer $n$ such that $0 \\le n \\le 10^d$, there exists unique $(a_0, a_1, \\dots, a_{d-1})$ with $0 \\le a_i \\le 9$ ($i = 0, 1, \\dots, d-1$) such that\n$$\nn = a_0 + a_1 \\cdot 10 + \\dots + a_... | Japan | Japan 2022 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 5074 | |
04qr | Let $P(x)$ be a polynomial with integer coefficients. Prove that the polynomial $Q(x) = P(x^4)P(x^3)P(x^2)P(x) + 1$ has no integer roots. | [
"By the Little Fermat theorem $n^3 \\equiv n \\pmod{3}$ for every $n \\in \\mathbb{Z}$. Then also $n^4 \\equiv n^2 \\pmod{3}$. Since $P(x)$ is a polynomial with all integer coefficients, it follows that\n$$\nP(n^3) \\equiv P(n) \\pmod{3}, \\quad P(n^4) \\equiv P(n^2) \\pmod{3}\n$$\nPolynomial $Q(x)$ has integer coe... | Czech Republic | 6-th Czech-Slovak Match | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
04kz | A cutting of an $8 \times 8$ chessboard into rectangles is called diversified if the following two conditions hold:
* Each rectangle contains the same number of black and white cells.
* No two rectangles have the same total number of cells.
Find the largest $n$ such that a diversified cutting into $n$ rectangles exists... | [] | Croatia | Mathematical competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 7 | |
067x | (α) For every real number $x > 0$ prove that: $x^3 - 3x \ge -2$.
(β) For all real numbers $x, y, z > 0$, prove that:
$$
\frac{x^2 y}{z} + \frac{y^2 z}{x} + \frac{z^2 x}{y} + 2 \left( \frac{y}{xz} + \frac{z}{xy} + \frac{x}{yz} \right) \ge 9. \quad (1)
$$
When equality holds? | [
"(α) We have\n$$\n\\begin{align*} x^3 - 3x \\ge -2 &\\Leftrightarrow x^3 - 3x + 2 \\ge 0 \\\\ &\\Leftrightarrow x^3 - x - 2x + 2 \\ge 0 \\\\ &\\Leftrightarrow x(x-1)(x+1) - 2(x-1) \\ge 0 \\\\ &\\Leftrightarrow (x-1)(x^2 + x - 2) \\ge 0 \\\\ &\\Leftrightarrow (x+2)(x-1)^2 \\ge 0 \\end{align*}\n$$\nwhich is valid, be... | Greece | Selection Examination | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Equality holds when x = y = z = 1. | |
0l54 | Problem:
Holden has a collection of polygons. He writes down a list containing the measure of each interior angle of each of his polygons. He writes down the list $30^{\circ}$, $50^{\circ}$, $60^{\circ}$, $70^{\circ}$, $90^{\circ}$, $100^{\circ}$, $120^{\circ}$, $160^{\circ}$, and $x^{\circ}$, in some order. Compute $... | [
"Solution:\n\nWe work in degrees. The sum of all 9 angles is $680 + x$. The sum of the angles in a polygon with $n$ sides is $180(n - 2) \\equiv 180n$ mod $360$. Since there are 9 angles, the polygons have a total of 9 sides, so the sum of the 9 angles must be $9 \\cdot 180 \\equiv 180$ mod $360$. Thus $680 + x \\e... | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 220 | |
0axa | Problem:
The segments $A P$ and $A Q$ are tangent to circle $O$ at points $P$ and $Q$, respectively. Moreover, $Q E$ is perpendicular to diameter $P D$ of length $4$. If $P E = 3.6$ and $A P = 6$, what is the length of $Q E$? | [
"Solution:\n\nDraw $A O$, $P Q$ and $O Q$. We note that $A O \\perp P Q$ since $A Q = A P$ and $O Q = O P$.\n\n\n\nThis implies that $\\angle P A O \\cong \\angle E P Q$ (both complementary to $\\angle A P R$).\n\nHence $\\triangle P A O \\approx \\triangle E P Q$. Therefore\n$$\n\\frac{A P... | Philippines | Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1.2 | |
06yu | Problem:
Find $f(x)$ such that $f(x)^{2} f\left(\frac{1-x}{1+x}\right) = 64x$ for $x$ not $0, \pm 1$. | [
"Solution:\n\nPut $x = \\frac{1-y}{1+y}$, then $\\frac{1-x}{1+x} = y$, so $f\\left(\\frac{1-y}{1+y}\\right)^{2} f(y) = 64 \\frac{1-y}{1+y}$. Hence $f\\left(\\frac{1-x}{1+x}\\right)^{2} f(x) = 64 \\frac{1-x}{1+x}$.\n\nBut $f(x)^{4} f\\left(\\frac{1-x}{1+x}\\right)^{2} = 64^{2} x^{2}$, so $f(x)^{3} = 64 x^{2} \\frac{... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 4 ((x^2(1+x))/(1-x))^{1/3} | |
0c1q | Prove that a continuous function $f: \mathbb{R} \to \mathbb{R}$ is non-decreasing if and only if
$$
(c - b) \int_{a}^{b} f(x) \, dx \leq (b - a) \int_{b}^{c} f(x) \, dx,
$$
for all $a < b < c$. | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Applications",
"Precalculus > Functions"
] | null | proof only | null | |
0hgf | You are given a set of $n$ not necessarily distinct numbers $\{a_1, a_2, \dots, a_n\}$, (meaning that some of them can be equal). Consider all $2^n - 1$ nonempty subsets of this set, and for each such subset, find the sum of its elements. What largest number of these sums could turn out to be equal to $1$? For example,... | [
"Example, where we reach equality, is: $(1, 0, 0, \\dots, 0)$."
] | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 2^{n-1} | |
04ln | A sequence $a_0, a_1, a_2, \dots$ of positive real numbers is given such that
$$
a_1 = 1 - a_0, \quad a_{n+1} = 1 - a_n(1 - a_n) \quad \text{for all } n \ge 1.
$$
Prove that
$$
a_0 a_1 \cdots a_n \left( \frac{1}{a_0} + \frac{1}{a_1} + \cdots + \frac{1}{a_n} \right) = 1
$$
holds for every positive integer $n$. | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0fp7 | Hemos empezado la Olimpiada Matemática puntualmente a las 9:00, como he comprobado en mi reloj, que funcionaba en ese momento correctamente. Cando he terminado, a las 13:00, he vuelto a mirar el reloj y he visto que las manecillas se habían desprendido de su eje pero manteniendo la posición en la que estaban cuando el ... | [
"Si medimos el tiempo $t$ en segundos a partir de las 00:00 y los ángulos en grados, en sentido horario y a partir de la posición de las manecillas a las 00:00, tenemos que el ángulo barrido por la manecilla de las horas en el instante $t$ es $\\alpha_{hor}(t) = t/120$ y barrido por el minutero, $\\alpha_{min}(t) =... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Spanish | proof and answer | 9:49:05 |
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