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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
032h | Problem:
Let $M$ be the centroid of $\triangle ABC$. Prove that:
a) $\cot \angle AMB = \frac{BC^2 + CA^2 - 5 AB^2}{12 S_{ABC}}$.
b) $\cot \angle AMB + \cot \angle BMC + \cot \angle CMA \leq -\sqrt{3}$. | [
"Solution:\nWe shall use the standard notation for the elements of $\\triangle ABC$.\n\na) The Cosine theorem for $\\triangle AMB$ gives\n$$\n\\cos \\angle AMB = \\frac{AM^2 + BM^2 - AB^2}{2 \\cdot AM \\cdot MB}\n$$\nThis together with the equality $\\sin \\angle AMB = \\frac{2S}{3 AM \\cdot MB}$ and the median for... | Bulgaria | 53. Bulgarian 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 > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0fgt | Problem:
Sea $(x_{n}),\ n \in \mathbb{N}$, una sucesión de números enteros tal que
$$
\begin{aligned}
x_{1} & = 1 \\
x_{n+1} & > x_{n}, \text{ para } n \geq 1 \\
x_{n+1} & \leq 2n, \text{ para } n \geq 1
\end{aligned}
$$
Demostrar que para todo entero natural $k$ existen dos términos de la sucesión $x_{r}$ y $x_{s}$ ta... | [
"Solution:\nEn primer lugar $x_{1} = 1$, $x_{1} < x_{2} \\leq 2$ implica que $x_{2} = 2$ y $x_{2} - x_{1} = 1$.\n\nSea ahora $k$ cualquier número natural mayor que $1$; dispongamos los $2k$ números\n$$\n\\{1, 2, 3, \\ldots, 2k\\}\n$$\nagrupados en $k$ bloques de la siguiente forma:\n$$\n\\{1, 1 + k\\} \\quad \\{2, ... | Spain | OME 24 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
046u | Let $x_1, x_2, \dots, x_{22}$ be real numbers such that $2^{i-1} \le x_i \le 2^i$ holds for every $1 \le i \le 22$. Find the maximum value of
$$
(x_1 + x_2 + \dots + x_{22}) \left( \frac{1}{x_1} + \frac{1}{x_2} + \dots + \frac{1}{x_{22}} \right).
$$ | [
"Let $y_i = \\frac{x_i}{2^{i/2}}$, for $i = 1, 2, \\dots, 22$. It is well-known that $f(t) = t + \\frac{1}{t}$ is decreasing on $(0, 1]$ and increasing on $[1, +\\infty)$. For $1 \\le i \\le 11$, we have $\\frac{1}{2^{12-i}} \\le y_i \\le \\frac{1}{2^{11-i}}$, thus $y_i + \\frac{1}{y_i} \\le 2^{12-i} + \\frac{1}{2^... | China | 22nd Chinese Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | (2^{12} - 1 - 1/2^{11})^2 | |
0g3t | Problem:
Bestimme alle natürlichen Zahlen $n \geq 2$, sodass für jeden Teiler $d>1$ von $n$
$$
d^{2}+n \mid n^{2}+d
$$
gilt. | [
"Solution:\n\nAntwort: $n$ erfüllt die Bedingung genau dann, wenn es eine Primzahl ist.\n\nZuest bemerken wir, dass alle Primzahlen die Voraussetzung tatsächlich erfüllen: Wenn $n$ eine Primzahl ist, muss $d=n$ gelten. Für diese Wahl von $d$ gilt offensichtlich\n$$\nn^{2}+n \\mid n^{2}+n\n$$\n\nWenn nun $n$ keine P... | Switzerland | Zweite Runde 2021 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | All prime numbers | |
0fm6 | Let $ABC$ be a triangle, $P$ an interior point, and points $H_A$, $H_B$ and $H_C$ the orthocenter of triangles $PBC$, $PAC$ and $PAB$ respectively. Prove that triangles $H_AH_BH_C$ and $ABC$ have the same area. | [] | Spain | Spanija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geome... | English | proof only | null | |
014z | Problem:
Assume that $a$, $b$, $c$ and $d$ are the sides of a quadrilateral inscribed in a given circle. Prove that the product $(a b + c d)(a c + b d)(a d + b c)$ acquires its maximum when the quadrilateral is a square. | [
"Solution:\nLet $A B C D$ be the quadrilateral, and let $A B = a$, $B C = b$, $C D = c$, $A D = d$, $A C = e$, $B D = f$. Ptolemy's Theorem gives $a c + b d = e f$. Since the area of triangle $A B C$ is $a b e / 4 R$, where $R$ is the circumradius, and similarly the area of triangle $A C D$, the product $(a b + c d... | Baltic Way | Baltic Way 2008 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof only | null | |
07gh | Consider a circle $\omega_1$ centered at $O_1$ and a circle $\omega_2$ centered at $O_2$ passes through $O_1$ and intersects $\omega_1$ at $A$ and $B$. Let $\ell$ be the tangent line from $A$ to $\omega_1$. A circle passes through $O_1$ and $O_2$ and its center lies on $\ell$, intersects $\omega_2$ for the second time ... | [
"Let $Q$ and $C$ be the reflections of $P$ and $O_1$ with respect to $\\ell$, respectively. We have $AO_1 \\perp \\ell$ implying that $C$ lies on $AO_1$. Also note that $C$ lies on the circumcircle of $PO_1O_2$ because $\\ell$ crosses its center. It is enough to show that\n\n$CA = CP$ because this yields to the fac... | Iran | 38th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ftl | Problem:
Sei $\Gamma$ ein Kreis und $P$ ein Punkt ausserhalb von $\Gamma$. Eine Tangente von $P$ an den Kreis berühre ihn in $A$. Eine weitere Gerade durch $P$ schneide $\Gamma$ in den verschiedenen Punkten $B$ und $C$. Die Winkelhalbierende von $\Varangle A P B$ schneide $A B$ in $D$ und $A C$ in $E$. Beweise, dass d... | [
"Solution:\n\nWir nehmen an, $B$ liege näher bei $P$ als $C$. Da $P A$ eine Tangente an $\\Gamma$ ist, gilt $\\Varangle P A B=\\Varangle A C B$. Ausserdem ist $\\Varangle A P D=\\Varangle C P E$, also sind die Dreiecke $P A D$ und $P E C$ ähnlich. Daraus folgt $\\Varangle P D A=\\Varangle P E C$ und somit auch $\\V... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ioe | Problem:
Find the positive real number(s) $x$ such that
$$
\frac{1}{2}\left(3 x^{2}-1\right)=\left(x^{2}-50 x-10\right)\left(x^{2}+25 x+5\right).
$$ | [
"Solution:\n\nAnswer: $25+2 \\sqrt{159}$.\n\nWrite $a = x^{2} - 50x - 10$ and $b = x^{2} + 25x + 5$; the given becomes\n$$\n\\frac{a + 2b - 1}{2} = ab,\n$$\nso\n$$\n0 = 2ab - a - 2b + 1 = (a - 1)(2b - 1).\n$$\nThen $a - 1 = x^{2} - 50x - 11 = 0$ or $2b - 1 = 2x^{2} + 50x + 9 = 0$.\n\nThe former has a positive root,... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 25 + 2 sqrt(159) | |
0diu | Does there exist $P(x, y)$ such that $P(x, y)^2 + 2022$ is divisible by $x^2 + y^2 + 2023$? | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | No | |
079r | A subset $B$ of natural numbers is called *loyal* if there exist positive integers $i \le j$ such that $B = \{i, i+1, \dots, j\}$. $Q$ is the collection of *loyal* subsets of natural numbers. For every subset $A = \{a_1 < a_2 < \dots < a_k\}$ of $\{1, 2, \dots, n\}$ we define:
$$
g(A) = \max_{B \subset A, B \in Q} |B|... | [
"We divide subgroups into three groups\n\n1) Subset $\\varnothing$ such that $f(\\varnothing) = g(\\varnothing) = 0$.\n\n2) One element subsets. Of course we know that for an arbitrary element of this group like $X$, $f(X) = 0$ and $g(X) = 1$.\nSo for one element subsets we have $\\sum_{|A|=1} (f(A) - g(A)) = -n$.\... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
09vy | For a given value $t$, we consider number sequences $a_1, a_2, a_3, \dots$ such that $a_{n+1} = \frac{a_n + t}{a_n + 1}$ for all $n \ge 1$.
a. Suppose that $t = 2$. Determine all starting values $a_1 > 0$ such that $\frac{4}{3} \le a_n \le \frac{3}{2}$ holds for all $n \ge 2$.
b. Suppose that $t = -3$. Investigate wh... | [
"a. First, we determine for what starting values $a_1 > 0$ the inequalities $\\frac{4}{3} \\le a_2 \\le \\frac{3}{2}$ hold. Then, we will prove that for those starting values, the inequalities $\\frac{4}{3} \\le a_n \\le \\frac{3}{2}$ are also valid for all $n \\ge 2$.\n\nFirst, we observe that $a_2 = \\frac{a_1+2}... | Netherlands | Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) Exactly those starting values with 1 ≤ a1 ≤ 2. b) Yes; for all starting values not equal to −1 and 1, the sequence is periodic with period 3, so a2020 = a1. | |
05qx | Problem:
Soit $n$ un entier positif. Montrer qu'il existe un entier positif $m$ tel que $n! = \varphi(m)$, où $\varphi$ est la fonction indicatrice d'Euler. (On rappelle que si $m = p_{1}^{\alpha_{1}} \cdots p_{k}^{\alpha_{k}}$ avec $p_{1}, \ldots, p_{k}$ des nombres premiers deux à deux distincts, $\varphi(m) = m \le... | [
"Solution:\n\nOn procède par construction en partant des plus grands nombres premiers au plus petit.\nL'idée est de contraindre la valeur de $\\varphi(m)$. La construction peut paraître un peu lourde mais il ne faut pas se laisser impressionner et comprendre l'idée qui est derrière.\n\nSoit $p_{1}, p_{2}, \\ldots$ ... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0ev2 | Find all functions $f : \mathbb{R} \to [0, \infty)$ such that, for all real $a, b, c, d$ that satisfy $ab + bc + cd = 0$, the following equality holds:
$$
f(a - b) + f(c - d) = f(a) + f(b + c) + f(d)
$$
(Here $\mathbb{R}$ is the set of real numbers and $[0, \infty)$ is the set of nonnegative real numbers.) | [
"We prove the following\n**Lemma** For all real $p, q, r$ that satisfy $p^2 + q^2 = r^2$, the following equality holds.\n$$\nf(p) + f(q) = f(r)\n$$\n**Proof** Put $a = \\frac{p-q+r}{2}$, $b = \\frac{p-q-r}{2}$, $c = \\frac{p+q+r}{2}$. Then we have $ab+bc+cd = \\frac{1}{2}(p^2+q^2-r^2)$. So if $p^2+q^2 = r^2$ then $... | South Korea | The 26th Korean Mathematical Olympiad Final Round | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | f(x) = λ x^2 for λ ≥ 0 | |
0hxs | Problem:
Bobbo starts swimming at 2 feet/s across a 100 foot wide river with a current of 5 feet/s. Bobbo doesn't know that there is a waterfall 175 feet from where he entered the river. He realizes his predicament midway across the river. What is the minimum speed that Bobbo must increase to make it to the other side... | [
"Solution:\n\nWhen Bobbo is midway across the river, he has travelled 50 feet. Going at a speed of 2 feet/s, this means that Bobbo has already been in the river for $\\frac{50 \\text{ feet}}{2 \\text{ feet} / \\mathrm{s}} = 25~\\mathrm{s}$. Then he has traveled $5 \\text{ feet} / \\mathrm{s} \\cdot 25~\\mathrm{s} =... | United States | HMMT 1998 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 3 feet per second | |
0cii | Determine the sets $A$ of positive natural numbers, with at least four elements $a_1 < a_2 < a_3 < \dots$, such that:
$$
\bullet \ a_1 = 1, a_2 = 2;
$$
$$
\bullet \ a_i a_{i+3} + 1 = a_{i+1} a_{i+2}, \text{ for every positive integer } i;
$$
$$
\bullet \ A \text{ contains exactly four prime numbers.}
$$ | [] | Romania | 75th NMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | {1, 2, 3, 5, 7} | |
0a21 | Lucas paints the entire outside of a cube blue. He then saws the cube into $27$ equally sized cubes. He neatly stacks these $27$ cubes so that he gets a tower of $27 \times 1 \times 1$ cubes.
At most how many of the $110$ side faces of cubes on the outside of his tower are blue? | [] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 48 | |
0hxb | Problem:
Let $p$ be a prime number, and $f\left(x_{1}, \ldots, x_{n}\right)$ be a polynomial with integer coefficients of total degree less than $n$. Prove that the number of ordered $n$-tuples $\left(x_{1}, \ldots, x_{n}\right)$ with $0 \leq x_{i}<p$ such that $f\left(x_{1}, \ldots, x_{n}\right)$ is an integer multip... | [
"Solution:\n\nBy Fermat's little theorem, the expression\n$$\nf\\left(x_{1}, \\ldots, x_{n}\\right)^{p-1}\n$$\nequals $0 \\bmod p$ if $f\\left(x_{1}, \\ldots, x_{n}\\right)$ is $0 \\bmod p$ and $1 \\bmod p$ otherwise. So\n$$\nS=\\sum_{0 \\leq x_{1}, \\ldots, x_{n}<p} f\\left(x_{1}, \\ldots, x_{n}\\right)^{p-1} \\eq... | United States | Berkeley Math Circle | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
061i | Problem:
In ein spitzwinkliges Dreieck $A B C$ wird ein Quadrat mit Mittelpunkt $A_{1}$ so einbeschrieben, dass zwei Ecken auf $B C$ und je eine auf $A B$ bzw. $A C$ liegen. Analog sind die Quadrate mit den Mittelpunkten $B_{1}$ bzw. $C_{1}$ definiert.
Man beweise, dass die Geraden $A A_{1}, B B_{1}$ und $C C_{1}$ ein... | [
"Solution:\n\nZusätzlich zur gegebenen Figur betrachten wir das nach außen errichtete Quadrat über der Seite $A B$ mit dem Mittelpunkt $C_{2}$. Da es eine zentrische Streckung gibt, welche das einbeschriebene Quadrat mit Mittelpunkt $C_{1}$ in das nach außen errichtete Quadrat überführt, liegen $C, C_{1}$ und $C_{2... | Germany | Auswahlwettbewerb zur IMO 2002 | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > D... | null | proof only | null | |
0fd9 | Problem:
Hallar los valores enteros positivos de $m$ para los que existe una función $f$ del conjunto de los números enteros en sí mismo tal que $f^{(m)}(n)=n+2017$, donde $f^{(m)}$ consiste en aplicar la función $f$ $m$ veces. | [
"Solution:\n\nLa primera observación es que $f$ debe ser biyectiva.\n\nLema 1. Si $f^{(a)}(n)=f^{(b)}(n)$, entonces tenemos $a=b$.\n\nDemostración: En otro caso, sin pérdida de generalidad asumamos $a>b$. Tenemos $f^{(a-b)}(n)=n$, lo que implica que $f^{(m(a-b))}(n)=n$, que es imposible, pues sería tanto como decir... | Spain | null | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"... | null | proof and answer | m = 1 or m = 2017 | |
09s6 | Problem:
Voor vierhoek $A B C D$ geldt $\angle A = \angle C = 90^{\circ}$. Zij $E$ een punt binnen de vierhoek. Zij $M$ het midden van $B E$. Bewijs dat $\angle A D B = \angle E D C$ dan en slechts dan als $|M A| = |M C|$. | [
"Solution:\n\nZij $N$ het midden van $B D$. Wegens Thales gaat de cirkel met middellijn $B D$ ook door $A$ en $C$ en punt $N$ is hiervan het middelpunt. Verder geldt dat $M N \\parallel D E$: als $E$ niet op $B D$ ligt, is $M N$ een middenparallel in driehoek $B D E$, en als $E$ wel op $B D$ ligt, dan zijn $M N$ en... | Netherlands | IMO-selectietoets | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
06kk | Given that $a$, $b$ and $c$ are positive real numbers such that $ab + bc + ca \ge 1$, prove that
$$
\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2} \ge \frac{\sqrt{3}}{abc}.
$$ | [
"By the AM-GM inequality, we have\n$$\n\\begin{aligned}\n\\frac{bc}{a} + \\frac{ca}{b} &= c\\left(\\frac{b}{a} + \\frac{a}{b}\\right) \\ge 2c, \\\\\n\\frac{ca}{b} + \\frac{ab}{c} &= a\\left(\\frac{c}{b} + \\frac{b}{c}\\right) \\ge 2a, \\\\\n\\frac{ab}{c} + \\frac{bc}{a} &= b\\left(\\frac{a}{c} + \\frac{c}{a}\\right... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0eip | Problem:
Poišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb
$$
\begin{gathered}
\sqrt{x}-\sqrt{y}=1 \\
\sqrt{8x+7}-\sqrt{8y+7}=2
\end{gathered}
$$ | [
"Solution:\n\nEnačbi preoblikujemo v ekvivalentni enačbi $\\sqrt{x}=1+\\sqrt{y}$ in $\\sqrt{8x+7}=2+\\sqrt{8y+7}$ ter ju nato kvadriramo, da dobimo\n$$\n\\begin{aligned}\nx & = 1 + 2\\sqrt{y} + y \\, \\text{ in } \\\\\n8x + 7 & = 4 + 4\\sqrt{8y+7} + 8y + 7\n\\end{aligned}\n$$\nOd spodnje enačbe odštejemo 8-kratnik ... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Other"
] | null | proof and answer | x = 9/4, y = 1/4 | |
03tw | There are 63 points on a circle $C$ with radius 10. Let $S$ be the number of triangles whose sides are longer than 9 and whose vertices are chosen from the 63 points. Find the maximum value of $S$. | [
"Let $O$ be the center of circle $C$, $a_n$ is the length of a regular $n$-gon $A_1A_2\\cdots A_n$ inscribed in $\\odot O$. Then $a_6 = 10 > 9$, $a_7 < 10 \\times \\frac{2\\pi}{7} < 10 \\times \\frac{2 \\times 3.15}{7} < 9$.\n\n(1) Let $A_1A_2\\cdots A_6$ be a regular 6-gon inscribed in $\\odot O$, then $A_iA_{i+1}... | China | China National Team Selection Test | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics >... | English | proof and answer | 23121 | |
09tl | Problem:
Zij $p$ een priemgetal. Bewijs dat het mogelijk is om een permutatie $a_{1}, a_{2}, \ldots, a_{p}$ van $1,2, \ldots, p$ te kiezen zodat de getallen $a_{1}, a_{1} a_{2}, a_{1} a_{2} a_{3}, \ldots, a_{1} a_{2} a_{3} \cdots a_{p}$ allemaal verschillende resten geven na deling door $p$. | [
"Solution:\n\nNoem $b_{i}=a_{1} a_{2} \\cdots a_{i}$, voor $1 \\leq i \\leq p$. We bewijzen dat het mogelijk is de permutatie zo te kiezen dat $b_{i} \\equiv i \\bmod p$ voor alle $i$. Voor $i \\geq 2$ is $a_{i} \\equiv b_{i} \\cdot b_{i-1}^{-1} \\bmod p$ als $b_{i-1} \\not \\equiv 0 \\bmod p$. We kiezen nu dus $a_... | Netherlands | Selectietoets | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0hnz | Problem:
A king is placed in the left bottom corner of the $6 \times 6$ chessboard. At each step it can either move one square up, or one square to the right, or diagonally - one up and one to the right. How many ways are there for the king to reach the top right corner of the board? | [
"Solution:\nWe shall make a $6 \\times 6$ table. In each cell of the table we will write a number of ways in which the king can reach that cell. We will fill it out gradually starting with a row of ones at the bottom and a column of ones at the left. To fill out the rest we use the following rule: the number in eac... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 1683 | |
09fo | 51 distinct integers are placed on a circle in such a way that each number is greater than the sum of the next three numbers in clockwise direction. What is the maximal number of numbers greater than or equal to 1? | [
"If there are three consecutive positive numbers $a_i$, $a_{i+1}$ and $a_{i+2}$, then $a_{i-1} > 0$. Hence we conclude that all the numbers are positive. But for the smallest number on the circle it is impossible to be greater than the sum of the next three numbers. Therefore for any three consecutive numbers, at l... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 33 | |
0a5o | Problem:
Triangle $ABC$ is right-angled at $B$ and has incentre $I$. Points $D$, $E$ and $F$ are the points where the incircle of the triangle touches the sides $BC$, $AC$ and $AB$ respectively. Lines $CI$ and $EF$ intersect at point $P$. Lines $DP$ and $AB$ intersect at point $Q$. Prove that $AQ = BF$. | [
"Solution:\n\nFirst note that $ID = IE = IF$ because they are all radii of the incircle, and $\\angle BFI = \\angle BDI = 90^{\\circ}$ because tangents are perpendicular to radii. Since $\\angle ABC = 90^{\\circ}$ we have $BFID$ a square and so $BD = BF = ID$ too. Thus $\\triangle EIF$ is isosceles and so $\\angle ... | New Zealand | New Zealand 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... | null | proof only | null | |
0137 | Problem:
Let $X$ be a subset of $\{1,2,3, \ldots, 10000\}$ with the following property: If $a, b \in X, a \neq b$, then $a \cdot b \notin X$. What is the maximal number of elements in $X$? | [
"Solution:\nIf $X=\\{100,101,102, \\ldots, 9999,10000\\}$, then for any two selected $a$ and $b$, $a \\neq b$, $a \\cdot b \\geq 100 \\cdot 101 > 10000$, so $a \\cdot b \\notin X$. So $X$ may have 9901 elements.\n\nSuppose that $x_{1} < x_{2} < \\cdots < x_{k}$ are all elements of $X$ that are less than $100$. If t... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 9901 | |
0dac | Let $ABC$ be a triangle inscribed in circle $(O)$ with incenter $I$. The lines $IB$ and $IC$ intersect $(O)$ again at $J$ and $L$. Circumcircle $(\omega)$ of triangle $IBC$ meets $CA$, $AB$ again at $E$, $F$. Prove that $EL$ and $FJ$ intersect on $(\omega)$. | [
"Denote $X$ as the other intersection of $(IJL)$ and $(\\omega)$. We shall prove that $X$ is the intersection of $EL$ and $FJ$.\n\nBy the cyclic quadrilateral, we have\n$$\n\\angle LXI = \\angle LJI = \\angle LJB = \\angle LCB = \\angle LCA = 180^\\circ - \\angle EXI.\n$$\nHence $L$, $X$, $E$ are collinear, which m... | Saudi Arabia | Team selection tests for JBMO 2018 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
04e9 | Let $ABCD$ be a parallelogram and let $S$ be the intersection of its diagonals. The bisector of the angle $\angle ADC$ bisects the line segment $\overline{AS}$ and intersects with the line $BC$ at point $E$. Determine the ratios $|BE| : |BC|$ and $|AB| : |BC|$. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | BE:BC = 2:3, AB:BC = 3:1 | |
0g21 | Problem:
Entlang der Küste einer kreisrunden Insel befinden sich 20 verschiedene Dörfer. Jedes dieser Dörfer hat 20 Kämpfer, wobei alle 400 Kämpfer unterschiedlich stark sind.
Jeweils zwei benachbarte Dörfer $A$ und $B$ machen nun einen Wettkampf, indem sich jeder der 20 Kämpfer des Dorfs $A$ mit jedem der 20 Kämpfer... | [
"Solution:\n\nWir geben zuerst eine Konstruktion und beweisen dann, dass mehr als 290 nicht möglich ist. O.B.d.A können wir annehmen, dass die Kämpfer Stärken 1 bis 400 haben, und wir nummerieren die Dörfer im Uhrzeigersinn von eins bis zwanzig.\n\nKonstruktion: Im ersten Dorf leben die Kämpfer der Stärke 400, 190,... | Switzerland | SMO-Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 290 | |
02cs | Problem:
O colar - Um colar é composto de pérolas grandes e pérolas pequenas, num total de menos do que 500 pérolas.
i. Se substituirmos $70\%$ das pérolas grandes por pequenas, o peso do colar diminui de $60\%$.
ii. Se substituirmos $60\%$ das pérolas pequenas por grandes, o peso do colar aumenta de $70\%$.
Quanta... | [
"Solution:\n\nSejam $n$ o número de pérolas grandes, $p$ o número de pérolas pequenas, $a$ o peso de uma pérola grande e $b$ o de uma pérola pequena. Com essa notação temos:\n\n- número total de pérolas no colar $= p + n$. Logo: $n + p < 500$\n- peso das pérolas grandes $= n \\times a$\n- peso das pérolas pequenas ... | Brazil | Lista 4 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 85, 170, 255, 340, 425 | |
079d | Do there exist two functions $f, g: \mathbb{R} \to \mathbb{R}$ such that for all $x \neq y$ the following inequality holds?
$$
|f(x) - f(y)| + |g(x) - g(y)| > 1
$$ | [
"For all $x \\neq y$ we have:\n$$\n(f(x) - f(y))^2 + (g(x) - g(y))^2 \\geq \\frac{1}{2} (|f(x) - f(y)| + |g(x) - g(y)|)^2 > \\frac{1}{2}\n$$\nFor every $x$, consider the point $(f(x), g(x))$ in the $\\mathbb{R}^2$ plane. The above inequality shows that the distance between every two of these points is more than $\\... | Iran | 27th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No, such functions do not exist. | |
03ig | Problem:
The numbers from $1$ to $50$ are printed on cards. The cards are shuffled and then laid out face up in $5$ rows of $10$ cards each. The cards in each row are rearranged to make them increase from left to right. The cards in each column are then rearranged to make them increase from top to bottom. In the final ... | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | null | proof and answer | Yes | |
05ol | Problem:
Soient deux cercles $\omega_{1}$, $\omega_{2}$ tangents l'un à l'autre en un point $T$, tels que $\omega_{1}$ soit à l'intérieur de $\omega_{2}$. Soient $M$ et $N$ deux points distincts sur $\omega_{1}$, différents de $T$. Soient $[AB]$ et $[CD]$ deux cordes du cercle $\omega_{2}$ passant respectivement par $... | [
"Solution:\n\nSoit $E$ et $F$ les points d'intersection respectifs de $(TM)$ et $(TN)$ avec $\\omega_{2}$, autres que $T$. Puisque $\\omega_{1}$ et $\\omega_{2}$ sont tangents en $T$, l'homothétie de centre $T$ et qui envoie $\\omega_{1}$ sur $\\omega_{2}$ envoie également $M$ sur $E$ et $N$ sur $F$. On en déduit q... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06xv | Let $a_{1} < a_{2} < a_{3} < \cdots$ be positive integers such that $a_{k+1}$ divides $2(a_{1} + a_{2} + \cdots + a_{k})$ for every $k \geqslant 1$. Suppose that for infinitely many primes $p$, there exists $k$ such that $p$ divides $a_{k}$. Prove that for every positive integer $n$, there exists $k$ such that $n$ divi... | [
"For every $k \\geqslant 2$ define the quotient $b_{k} = 2(a_{1} + \\cdots + a_{k-1}) / a_{k}$, which must be a positive integer. We first prove the following properties of the sequence $(b_{k})$:\n\nClaim 1. We have $b_{k+1} \\leqslant b_{k} + 1$ for all $k \\geqslant 2$.\n\nProof. By subtracting $b_{k} a_{k} = 2(... | IMO | International Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
08h2 | Problem:
Find all positive integers $n$ such that there exists a prime number $p$, such that
$$
p^{n}-(p-1)^{n}
$$
is a power of $3$. | [
"Solution:\nSuppose that the positive integer $n$ is such that\n$$\np^{n}-(p-1)^{n}=3^{a}\n$$\nfor some prime $p$ and positive integer $a$.\n\nIf $p=2$, then $2^{n}-1=3^{a}$ by $(1)$, whence $(-1)^{n}-1 \\equiv 0\\pmod{3}$, so $n$ should be even. Setting $n=2s$ we obtain $\\left(2^{s}-1\\right)\\left(2^{s}+1\\right... | JBMO | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | n = 2 | |
080b | Problem:
La tela di un dipinto rettangolare è circondata da un passepartout (cioè un riquadro) largo $10~\mathrm{cm}$. Attorno a quest'ultimo vi è poi una cornice, anch'essa larga $10~\mathrm{cm}$ (nella figura, il rettangolo bianco rappresenta la tela, la superficie tratteggiata il passepartout, la superficie nera la... | [
"Solution:\n\nLa risposta è (D). Siano infatti $a$ e $b$ le dimensioni della tela.\nQuelle del passepartout sono allora $(a+20)$ e $(b+20)$ e quelle della cornice sono $(a+40)$ e $(b+40)$. L'ipotesi dice che si ha $(a+40)(b+40)=2(a+20)(b+20)$ da cui $a b=800$, condizione ovviamente equivalente all'ipotesi stessa.\n... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | D | |
0hsg | Problem:
If $\left(a+\frac{1}{a}\right)^{2}=3$, find $\left(a+\frac{1}{a}\right)^{3}$ in terms of $a$. | [
"Solution:\n0 ."
] | United States | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 0 | |
0g6g | 令 $T = \{1, 2, \dots, n\}$, 對於一個子集合 $S = \{s_1, s_2, \dots, s_k\} \subseteq T$ 其中 $s_1 > s_2 > \dots > s_k$, 定義
$$
f(S) = s_1 - s_2 + s_3 - s_4 + \dots;
$$
$$
g(S) = s_1 - s_3 + s_5 - s_7 + \dots,
$$
也就是 $g$ 是依序且循環地給 $1, 0, -1, 0$ 加權之和, 而 $f$ 是依序且循環地給 $1, -1$ 加權之和; 明顯地, $f(\emptyset) = g(\emptyset) = 0$.
a. 試求 $\sum_{... | [
"(a) (2分) 令 $F_n = \\sum_{S \\subseteq T} f(S)$ 對於 $n = 0, 1, 2, \\dots$ 考慮數字 $n$ 是“在”或“不在”子集合 $S$ 之中, 我們得到以下的等式：\n$$\n\\begin{aligned}\nF_n &= \\sum_{S \\subseteq T - \\{n\\}} (n - f(S)) + F_{n-1} \\\\\n&= n \\times 2^{n-1} - F_{n-1} + F_{n-1} \\\\\n&= n \\times 2^{n-1}.\n\\end{aligned}\n$$\n\n(b) (5分) 令 $G_n = \\... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | (a) Sum over all subsets of f(S) equals n · 2^{n−1}.
(b) Sum over all subsets of g(S) equals [(-1+i)/4]·(1+i)^n + [(-1−i)/4]·(1−i)^n + (n+1)·2^{n−1}.
(Equivalently: G_n = (n+1)·2^{n−1} − 2^{(n−1)/2}·cos((n−1)·π/4).)
(c) For n = 8, the value is 1144. | |
0i7f | Problem:
An integer is a perfect number if and only if it is equal to the sum of all of its divisors except itself. For example, $28$ is a perfect number since $28 = 1 + 2 + 4 + 7 + 14$.
Let $n!$ denote the product $1 \cdot 2 \cdot 3 \cdots n$, where $n$ is a positive integer. An integer is a factorial if and only if... | [
"Solution:\n\nThe only perfect factorial is $6 = 3!$. Certainly, $2! = 2$ is not perfect. For $n > 3$, note that $n! = 6k$, where $k > 1$, and thus the factors of $n!$ will include $1, k, 2k, 3k$. This sums to $6k + 1$, showing that $n!$ is not perfect."
] | United States | 5th Bay Area Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6 | |
06gq | Prove that for every positive integer $n$ and every group of real numbers $a_1, a_2, \dots, a_n > 0$,
$$
\sum_{k=1}^{n} \frac{k}{a_{1}^{-1} + a_{2}^{-1} + \dots + a_{k}^{-1}} \le 2 \sum_{k=1}^{n} a_{k}.
$$
Can “2” immediately to the right of the inequality be replaced by a smaller positive number? | [
"By the Cauchy-Schwarz inequality, we have\n$$\n(1^2 a_1 + 2^2 a_2 + \\dots + k^2 a_k)(a_1^{-1} + a_2^{-1} + \\dots + a_k^{-1}) \\ge (1 + 2 + \\dots + k)^2 = \\frac{k^2 (k+1)^2}{4}.\n$$\nThis implies\n$$\n\\frac{k}{a_1^{-1} + a_2^{-1} + \\dots + a_k^{-1}} \\le \\frac{4}{k(k+1)^2} \\sum_{j=1}^{k} j^2 a_j.\n$$\n\nAls... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No. The constant two is optimal and cannot be replaced by any smaller positive number. | |
0051 | Encuentre todas las funciones $f: \mathbb{Z} \rightarrow \mathbb{Z}$ con la siguiente propiedad:
$$
\text{Si } x + y + z = 0, \text{ entonces } f(x) + f(y) + f(z) = xyz .
$$ | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Divisibility / Factorization"
] | Spanish | proof and answer | All functions of the form f(n) = (n^3 - n)/3 + k n for any integer k. | |
0b8u | Two circles in the plane, $\gamma_1$ and $\gamma_2$, meet at points $M$ and $N$. Let $A$ be a point on $\gamma_1$, and let $D$ be a point on $\gamma_2$. The lines $AM$ and $AN$ meet again $\gamma_2$ at points $B$ and $C$, respectively, and the lines $DM$ and $DN$ meet again $\gamma_1$ at points $E$ and $F$, respectivel... | [
"Since $AB = DE$, the triangles $NAB$ and $NED$ are congruent, so $NA = NE$ and $NB = ND$. Let $K$ and $L$ be the antipodes of $N$ in $\\gamma_1$ and $\\gamma_2$, respectively, and notice that they are the midpoints of the arcs $AE$ and $BD$, respectively. Notice further that the angles $KME$ and $LMD$ have equal m... | Romania | NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0jwc | Problem:
Let $A$, $B$, $C$, $D$ be four points on a circle in that order. Also, $AB = 3$, $BC = 5$, $CD = 6$, and $DA = 4$. Let diagonals $AC$ and $BD$ intersect at $P$. Compute $\frac{AP}{CP}$. | [
"Solution:\n\nNote that $\\triangle APB \\sim \\triangle DPC$ so $\\frac{AP}{AB} = \\frac{DP}{CD}$. Similarly, $\\triangle BPC \\sim \\triangle APD$ so $\\frac{CP}{BC} = \\frac{DP}{DA}$. Dividing these two equations yields\n$$\n\\frac{AP}{CP} = \\frac{AB \\cdot DA}{BC \\cdot CD} = \\frac{3 \\cdot 4}{5 \\cdot 6} = \... | United States | February 2017 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 2/5 | |
064s | Let $ABCD$ be a convex quadrilateral with $AB = BC = CD$, $AC \neq BD$ and let $E$ be the intersection point of its diagonals. Prove that $AE = DE$ if and only if $\angle BAD + \angle ADC = 120^{\circ}$. | [
"Let $\\angle BAC = \\angle BCA = \\alpha$, $\\angle CBD = \\angle CDB = \\beta$, and $S$ the point of intersection of the lines $AB$ and $DC$.\nIf $AE=DE$, then $\\frac{AE}{\\sin(2\\alpha + \\beta)} = \\frac{AB}{\\sin(\\alpha + \\beta)} = \\frac{CD}{\\sin(\\alpha + \\beta)} = \\frac{DE}{\\sin(\\alpha + 2\\beta)}$.... | Greece | 24th Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0f8v | Problem:
$ABC$ is a triangle. Points $D$, $E$, $F$ are chosen on $BC$, $CA$, $AB$ such that $B$ is equidistant from $D$ and $F$, and $C$ is equidistant from $D$ and $E$. Show that the circumcenter of $AEF$ lies on the bisector of $EDF$. | [] | Soviet Union | 23rd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
00st | Let $S \subset \{1, \dots, n\}$ be a nonempty set, where $n$ is a positive integer. We denote by $s$ the greatest common divisor of the elements of the set $S$. We assume that $s \neq 1$ and let $d$ be its smallest divisor greater than $1$. Let $T \subset \{1, \dots, n\}$ be a set such that $S \subset T$ and $|T| \geq ... | [
"Let $t$ be the greatest common divisor of the elements in $T$. Due to the fact that $S \\subset T$, we immediately get that $t/s$. Let us assume for the sake of contradiction that $t \\neq 1$. From the previous observation we get that $t \\geq d$.\nBy taking into account that $|T| \\geq 1 + \\lfloor \\frac{n}{d} \... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | 1 + floor(n/d) | |
01f1 | Let $m \ge 2$ be a positive integer. Consider the $2m$ numbers
$$
1 \cdot 2, 2 \cdot 3, 3 \cdot 4, \dots, 2m(2m + 1).
$$
A move consists of choosing three numbers $a, b, c$, and replacing them with the single number
$$
\frac{abc}{ab + bc + ca}.
$$
After $m-1$ such moves, only two numbers will remain. Supposing one of t... | [
"Denoting the new number\n$$\ng = \\frac{abc}{ab + bc + ca},\n$$\nits reciprocal is\n$$\n\\frac{1}{g} = \\frac{ab + bc + ca}{abc} = \\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c}.\n$$\nThe sum of all reciprocals is therefore invariantly\n$$\n\\sum_{k=1}^{2m} \\frac{1}{k(k+1)} = \\sum_{k=1}^{2m} \\left( \\frac{1}{k} - \... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof only | null | |
067a | Let $AB\Gamma$ an acute angled scalene triangle with $AB < A\Gamma < B\Gamma$. Let $\Delta, E, Z$ be the midpoints of the sides $B\Gamma, A\Gamma, AB$, respectively, and let $BK, G\Lambda$ be altitudes. At the extension of $\Delta Z$, to the part of $Z$, we consider a point $M$, such that the parallel from $M$ to $K\La... | [
"Since $\\Delta, E, Z$ are the midpoints of the sides $B\\Gamma, A\\Gamma, AB$, respectively, the quadrilaterals $AE\\Delta Z, ZE\\Delta B$ and $ZE\\Gamma\\Delta$ are parallelograms.\nFrom $\\Delta M \\parallel \\Gamma \\Sigma$ and $MN \\parallel K\\Lambda$ we conclude that: $\\hat{M} = \\hat{\\Sigma}_1$ and $\\hat... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | English | proof only | null | |
0cig | Prove that if $x, y, z \in \mathbb{R}$, then the following inequality holds:
$$
\sum_{\text{cyc}} \sqrt{1 + (x\sqrt{1+y^2} + y\sqrt{1+x^2})^2} \geq xy + yz + zx + 2(x+y+z).
$$ | [] | Romania | 75th NMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0hyk | Problem:
Prove that $1993^{1993} + 1994^{1994} + 1995^{1995} + 1996^{1996}$ is divisible by $10$. | [
"Solution:\nSince $5^{2} = 25$ ends in $5$ again, and $6^{2} = 36$ ends in $6$ again, no matter to what power we raise $5$ or $6$, the resulting numbers will end in $5$ or $6$, respectively. Thus, $1995^{1995}$ ends in $5$ and $1996^{1996}$ ends in $6$.\n\nFurther, $4^{1} = 4$, $4^{2} = 16$, $4^{3} = 64$, $4^{4} = ... | United States | BAMO | [
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0cgr | a) Prove that $12n+13$ and $13n+14$ are coprime, for every natural $n$.
b) Find the number of the pairs $(a, b)$ of natural numbers for which there exists a natural number $n$ so that $\frac{a}{b} = \frac{12n+13}{13n+14}$ and $17a + 19b < 2024$. | [
"a) If $d$ is a common divisor of the numbers $12n + 13$ and $13n + 14$, then $d$ divides the numbers $13(12n + 13)$ and $12(13n + 14)$. Then $d$ divides $13(12n + 13) - 12(13n + 14)$, that is $d \\mid 1$. So $d = 1$, hence the numbers $12n + 13$ and $13n + 14$ are coprime.\n\nb) The relation $\\frac{a}{b} = \\frac... | Romania | 74th Romanian Mathematical Olympiad | [
"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"
] | English | proof and answer | 8 | |
09wi | The *digit sum* of a number is obtained by adding all digits of the number. For example, the digit sum of $1303$ is $1 + 3 + 0 + 3 = 7$. Find the smallest positive integer $n$ for which both the digit sum of $n$ and the digit sum of $n + 1$ are divisible by $5$. | [
"$49999$"
] | Netherlands | Second Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 49999 | |
0hip | Problem:
Prove that
$$
n(n+1)(2 n+1)
$$
is always divisible by $6$, for $n$ a positive integer. | [
"Solution:\nThe number is even, because either $n$ or $n+1$ is even.\nNow we show it is always divisible by three. Assume for contradiction that it isn't. Then neither $n$ nor $n+1$ is divisible by three, so $n+2$ must be. However, $2 n+1 = 2(n+2) - 3$ is then also a multiple of three, which is a contradiction.\n\n... | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
00wm | Problem:
Express the value of $\sin 3^{\circ}$ in radicals. | [
"Solution:\n\nWe use the equality\n$$\n\\sin 3^{\\circ} = \\sin (18^{\\circ} - 15^{\\circ}) = \\sin 18^{\\circ} \\cos 15^{\\circ} + \\cos 18^{\\circ} \\sin 15^{\\circ}\n$$\nwhere\n$$\n\\sin 15^{\\circ} = \\sin \\frac{30^{\\circ}}{2} = \\sqrt{\\frac{1 - \\cos 30^{\\circ}}{2}} = \\frac{\\sqrt{6} - \\sqrt{2}}{4}\n$$\n... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | sin 3° = [ (√5 − 1)(√6 + √2) − √(10 + 2√5)(√6 − √2) ] / 16 | |
09xx | Peter gets bored during the lockdown, so he decides to write numbers the whole day. He makes a sequence of numbers starting with $0$, $1$ and $-1$, and then going on indefinitely. On the next line he writes the same sequence of numbers, but shifted one place to the right. On the third line he writes again the same sequ... | [
"2020"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | final answer only | 2020 | |
07ez | Hossna is playing with an $m \times n$ grid of points. She can draw segments between some of the points with the following conditions:
a. No two segments intersect except in points of the grid.
b. Each segment is drawn between two consecutive rows.
c. There is at most one segment between any two points.
Find the maximu... | [
"We claim the answer is $mn - n$.\n\nFirst note that by Pick's theorem, the area of a region is at least $\\frac{b}{2} - 1$, where $b$ is the number of the lattice points on its boundary. Obviously, for any region, $b \\ge 3$ and if $b = 3$, the region must be a triangle. But each of the edges of this triangle lie ... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Pick's theorem",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | mn - n | |
05fg | Problem:
Soit $m, n \geqslant 2$ des entiers vérifiant la propriété suivante:
$$
a^{n} \equiv 1 \quad(\bmod m) \quad a=1, \ldots, n
$$
Prouver que $m$ est un nombre premier et que $n = m - 1$. | [
"Solution:\n\nCommençons par supposer que $m = p$ est un nombre premier. On doit donc montrer $n = p - 1$. Si $n \\geqslant p$, on a $p^{n} \\equiv 1 (\\bmod p)$, une évidente contradiction. Donc $n < p$.\n\nConsidérons $T = X^{n} - 1$ le polynôme de $\\mathbb{Z} / p \\mathbb{Z}[X]$. Il a pour racines par hypothèse... | France | ENVOi 3 : ARITHMÉTIQUE | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof only | null | |
0bz2 | In triangle $ABC$, we have $m(\angle A) < m(\angle C)$. Let $E$ be a point on the bisector of angle $B$ such that $\angle EAB = \angle ACB$. Let $D$ be a point on line $BC$ such that $B \in (CD)$ and $[BD] = [AB]$. Prove that the midpoint $M$ of the line segment $[AC]$ belongs to the line $DE$.
Bogdan Antohe
![](atta... | [
"Let $N$ be the midpoint of $[AD]$; it follows that $BN \\perp AD$. If $AP \\perp BE$, $P \\in BE$; then the quadrilateral $BNAP$ is a rectangle.\n\nSince the points $N$ and $P$ are the projections of $A$ on the interior and exterior bisectors of the angle $ABC$, respectively, it follows that the line $NP$ contains... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Concurrency and Collinearity"
] | English | proof only | null | |
0gof | For an interior point $D$ of a triangle $ABC$, let $\Gamma_D$ denote the circle passing through the points $A$, $E$, $D$, $F$ if these points are concyclic where $BD \cap AC = \{E\}$ and $CD \cap AB = \{F\}$. Show that all circles $\Gamma_D$ pass through a second common point different from $A$ as $D$ varies. | [
"Let $A'$ be the midpoint of the side $BC$, and let $A''$ be the second point of intersection of the line $AA'$ and the circumcircle $\\Gamma$ of the triangle $BDC$. Since both $D, E, A, F$ and $D, C, A'', B$ are concyclic, $\\angle BA''C = \\angle BAC$. Hence $A'$ is the midpoint of $AA''$. In particular, the circ... | Turkey | Team Selection Test for IMO 2010 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fdk | Problem:
Decimos que un conjunto $E$ de números naturales es especial cuando al tomar dos elementos cualesquiera distintos $a, b \in E$ se tiene que $(a-b)^2$ divide al producto $ab$.
a. Encuentra un conjunto especial formado por tres elementos.
b. ¿Existe un conjunto especial formado por cuatro números naturales qu... | [
"Solution:\n\na. Un conjunto especial de tres elementos es $\\{2, 3, 4\\}$.\n\nb. Supongamos que $\\{x, x+y, x+2y, x+3y\\}$ forman un conjunto especial.\nPodemos suponer que $x$ e $y$ son primos relativos, pues si llamamos $d = \\operatorname{mcd}\\{x, y\\}$ y $d \\neq 1$, tomando $x' = x/d$ e $y' = y/d$ tenemos un... | Spain | null | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) {2, 3, 4}. b) No; no four-term arithmetic progression can be special. | |
03y7 | Function $f(x) = a^{2x} + 3a^x - 2$ ($a > 0, a \neq 1$) reaches the maximum value $8$ on interval $[-1, 1]$. Then its minimum value on this interval is ________. | [
"Let $a^x = y$. The original function is then changed to\n$$g(y) = y^2 + 3y - 2,$$ which is increasing over $\\left(-\\frac{3}{2}, +\\infty\\right)$.\n\nWhen $0 < a < 1$, we have $y \\in [a, a^{-1}]$ and\n$$\ng(y)_{\\max} = a^{-2} + 3a^{-1} - 2 = 8 \\Rightarrow a^{-1} = 2 \\Rightarrow a = \\frac{1}{2}.\n$$\nThen\n$... | China | China Mathematical Competition | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | -1/4 | |
08do | Problem:
Sia $ABCD$ un trapezio di basi $AB$ e $CD$ inscritto in una circonferenza $\Gamma$, tale che le diagonali $AC$ e $BD$ siano perpendicolari. Detto $P$ il punto d'incontro delle diagonali $AC$ e $BD$, quanto vale il rapporto fra l'area di $\Gamma$ e la somma delle aree dei triangoli $APB$ e $CPD$?
(A) 1
(B) $\... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Siano $O$ e $r$ rispettivamente il centro e il raggio della circonferenza $\\Gamma$. Osserviamo che necessariamente il trapezio è isoscele (perché inscrivibile in una circonferenza), quindi le diagonali sono uguali e $AP = PB$ (quindi anche $PC = PD$). Quindi i triangoli... | Italy | Progetto Olimpiadi della Matematica | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
0l6q | Problem:
Compute the largest possible radius of a circle contained in the region defined by $|x + |y|| \leq 1$ in the coordinate plane. | [
"Solution:\n\n\nAfter drawing the graph, it's clear that the circle should pass through $(-1, 0)$ and be tangent to $y = x - 1$ and $y = -x + 1$. Letting the radius of this circle be $r$, we have $r\\sqrt{2} + r = 2$, so $\\boxed{r = 2\\sqrt{2} - 2}$."
] | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 2√2 - 2 | |
0a9w | Problem:
A game is played on an $n \times n$ chessboard. At the beginning there are 99 stones on each square. Two players $A$ and $B$ take turns, where in each turn the player chooses either a row or a column and removes one stone from each square in the chosen row or column. They are only allowed to choose a row or a ... | [
"Solution:\nPlayer $A$ has a winning strategy if and only if $n$ is odd.\n\nFirst we prove that no matter how they play, the play will not end before the board is empty. Let $(i, j)$ denote the square in row $i$ and column $j$, let $r_{i}$ denote the number of times row $i$ has been chosen when the game ends, and l... | Nordic Mathematical Olympiad | The 28th Nordic Mathematical Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | all odd n | |
04dd | Prove that every positive integer $d$ has a positive multiple $n$, such that one can delete one of the non-zero digits of $n$ to obtain another multiple of $d$. | [
"We will find a number $n$ in the form $10^k(10a + b) + c$ where $0 \\le c < 10^k$ and $a, b$ are digits. Deleting the digit $b$ gives the number $n_1 = 10^k a + c$. Since $n - n_1 = 10^k(9a+b)$, to satisfy the conditions of the problem it is sufficient that $d$ divides $9a+b$ and $10^k a + c$.\n\nThe digit $b$ can... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof only | null | |
032s | Problem:
The lines through the vertices $A$ and $B$ that are tangent to the circumcircle of an acute $\triangle ABC$ meet at a point $D$. If $M$ is the midpoint of the side $AB$, prove that $\Varangle ACM = \Varangle BCD$. | [
"Solution:\nThe Sine theorem for $\\triangle AMC$ and $\\triangle BMC$ gives\n$$\n\\frac{AM}{CM} = \\frac{\\sin \\Varangle ACM}{\\sin \\alpha}, \\quad \\frac{BM}{CM} = \\frac{\\sin (\\gamma - \\Varangle ACM)}{\\sin \\beta}\n$$\nSince $AM = BM$, we get that $\\frac{\\sin \\Varangle ACM}{\\sin \\alpha} = \\frac{\\sin... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof only | null | |
04mb | Marko has drawn a rectangle with two blue sides of length $24$ and two red sides of length $36$. He coloured each point in the interior of the rectangle in the colour of the side closest to that point. He also coloured in black all points equally distant from a blue and red side. Determine the area of the red part of t... | [] | Croatia | Croatia_2018 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 576 | |
01mk | Do there exist integers $x$ and $y$ satisfying the equality $2x^2 - 5y^3 = 2011$? | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | No | |
06if | Two parallel chords of a circle have lengths $24$ and $32$ respectively, and the distance between them is $14$. What is the length of another parallel chord midway between the two chords? | [
"Let $r$ be the radius of the circle. Denote by $x$ and $y$ the distances from the two chords to the centre of the circle (see the figure). Since the perpendicular from the centre to a chord bisects this chord, by Pythagoras' Theorem we see that $\\left(\\frac{24}{2}\\right)^2 + x^2 = r^2$, i.e. $12^2 + x^2 = r^2$.... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2√249 | |
0eyw | Problem:
20 teams compete in a competition. What is the smallest number of games that must be played to ensure that given any three teams at least two play each other? | [] | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 90 | |
0kfd | Problem:
A small village has $n$ people. During their yearly elections, groups of three people come up to a stage and vote for someone in the village to be the new leader. After every possible group of three people has voted for someone, the person with the most votes wins.
This year, it turned out that everyone in the... | [
"Solution:\nThe problem asks for the number of $n$ that divide $\\binom{n}{3}$, which happens exactly when $\\frac{(n-1)(n-2)}{2 \\cdot 3}$ is an integer. Regardless of the parity of $n$, $(n-1)(n-2)$ is always divisible by $2$. Also, $(n-1)(n-2)$ is divisible by $3$ if and only if $n$ is not a multiple of $3$. Of ... | United States | HMMO 2020 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization"
] | null | final answer only | 61 | |
05ly | Problem:
On considère un quadrillage formé de tous les petits carrés de côté $1$ entièrement inclus dans le disque donné par l'inéquation $x^{2}+y^{2} \leq 2014^{2}$ dans le plan muni d'un repère orthonormé $Oxy$. À chaque étape du jeu, chaque carré contient le nombre $1$ ou le nombre $-1$. On suppose que chaque case ... | [
"Solution:\n\nOn raisonne par l'absurde en supposant qu'on peut arriver à avoir un seul $-1$. On appelle $C_{1}$ le petit carré qui contiendra ce $-1$. Par symétrie de la figure, on peut supposer que le centre de ce petit carré a pour coordonnées $\\left(a-\\frac{1}{2}, b-\\frac{1}{2}\\right)$ avec $a \\geq b \\geq... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Number Theory > Other"
] | null | proof and answer | No | |
0f9t | Problem:
Find three non-zero reals such that all quadratics with those numbers as coefficients have two distinct rational roots. | [
"Solution:\n\nAnswer $1$, $2$, $-3$\n\nIf $a + b + c = 0$, then $1$ is a root of $a x^2 + b x + c$, and so the other root is $-\\frac{b}{a} - 1$, which is rational."
] | Soviet Union | 24th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | 1, 2, -3 | |
083m | Problem:
Il professor Abacus ha scritto sulla lavagna due numeri naturali, risultato di parecchie ore di lavoro. Il figlio dispettoso cancella i due numeri e li sostituisce con il loro prodotto meno $1$ e la loro somma. Non soddisfatto cancella anche questi e li sostituisce di nuovo con il loro prodotto meno $1$ e la ... | [] | Italy | UNIONE MATEMATICA ITALIANA Progetto Olimpiadi di Matematica 2004 - GARA di SECONDO LIVELLO TRIENNIO | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 22 | |
09hi | Let $ABCD$ be an isosceles trapezoid with $AD = BC$ and $AB \parallel CD$. Let $O$ be the intersection of the diagonals and let $M$ be the midpoint of $AD$. Circumcircle of $BCM$ intersects $AD$ again at $K$. Prove that $OK$ is parallel to $AB$.
(Proposed by B. Bat-Od) | [
"Let $N$ be a middle point of $CB$ and $AB \\cap CK = E$. Since $MBCK$ inscribed in a circle, we have $\\angle AMB = \\angle BCE$ and since $ABCD$ is equilateral, we have $\\angle AMB = \\angle ANB$. Thus $AN \\parallel EC$. By Thale's theorem $1 = \\frac{BN}{NC} = \\frac{BA}{AE}$, thus $BA = AE$ and $\\frac{DK}{KA... | Mongolia | Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gdk | 是否存在正整數 $m$, 可以讓等式
$$
(a^3 - a)(b^3 - b) = mc^2.
$$
有無窮多組滿足 $a \neq b$ 的正整數解 $(a, b, c)$?
Does there exist a positive integer $m$ for which the equation
$$
(a^3 - a)(b^3 - b) = mc^2.
$$
has infinitely many positive integer solutions $(a, b, c)$ in which $a \neq b$? | [
"We assert that $m = 13$ works.\nThere are infinitely many integers $b$ satisfying $4b^2 - 3 = 13k^2$ for some $k$, because $(b, k) = (2, 1)$ is a solution, and $18^2 - 13 \\cdot 5^2 = -1$, so one can generate arbitrarily many solutions using Pell equations. Specifically, we take $(2 + \\sqrt{13})(18 - 13\\sqrt{5})... | Taiwan | 2020 Taiwan IMO 2J | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof and answer | m = 13 | |
0iqk | Problem:
Let $ABC$ be a triangle with $AB = 5$, $BC = 4$ and $AC = 3$. Let $\mathcal{P}$ and $\mathcal{Q}$ be squares inside $ABC$ with disjoint interiors such that they both have one side lying on $AB$. Also, the two squares each have an edge lying on a common line perpendicular to $AB$, and $\mathcal{P}$ has one ver... | [
"Solution:\n\n$\\boxed{\\dfrac{144}{49}}$\n\nLet the side lengths of $\\mathcal{P}$ and $\\mathcal{Q}$ be $a$ and $b$, respectively. Label two of the vertices of $\\mathcal{P}$ as $D$ and $E$ so that $D$ lies on $AB$ and $E$ lies on $AC$, and so that $DE$ is perpendicular to $AB$. The triangle $ADE$ is similar to $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 144/49 | |
0ja4 | Problem:
Brian has a 20-sided die with faces numbered from 1 to 20, and George has three 6-sided dice with faces numbered from 1 to 6. Brian and George simultaneously roll all their dice. What is the probability that the number on Brian's die is larger than the sum of the numbers on George's dice? | [
"Solution:\n\nAnswer: $\\frac{19}{40}$\n\nLet Brian's roll be $d$ and let George's rolls be $x, y, z$. By pairing the situation $d, x, y, z$ with $21-d, 7-x, 7-y, 7-z$, we see that the probability that Brian rolls higher is the same as the probability that George rolls higher. Given any of George's rolls $x, y, z$,... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 19/40 | |
06io | There are 12 lamps, initially all off, each of which comes with a switch. When a switch is pressed, a lamp which is off will be turned on, and a lamp which is on will be turned off. Now one is allowed to press exactly 5 different switches in each round. What is the minimum number of rounds needed so that all lamps will... | [
"Suppose all lamps are turned on after $n$ rounds. Then we have pressed the switches $5n$ times in total. Note that each lamp should change state for an odd number of times. As there are 12 lamps, the total number of times the lamps have changed state is an even number, meaning that $n$ is even.\n\nClearly, $n \\ne... | Hong Kong | Hong Kong Preliminary Selection Contest | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 4 | |
0fxd | Problem:
Sei $n > 3$ eine natürliche Zahl. Beweise, dass $4^{n} + 1$ einen Primteiler $> 20$ besitzt. | [
"Solution:\n\nWir verwenden folgendes Resultat:\nLemma 1. Jeder Primteiler $p$ einer Zahl der Form $a^{2} + 1$ erfüllt $p \\equiv 1 \\pmod{4}$. Jeder Primteiler $p$ einer Zahl der Form $a^{4} + 1$ erfüllt $p \\equiv 1 \\pmod{8}$.\nBeweis. Es gilt $-1 \\equiv a^{2} \\pmod{p}$, und somit ist $d = 4$ die kleinste natü... | Switzerland | SMO Finalrunde | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0896 | Problem:
Alice, Berto e Carlo devono seppellire un tesoro e decidono di seppellirlo in un punto equidistante da tutti e tre. Sapendo che si trovano ai vertici di un triangolo rettangolo con un angolo di $30^{\circ}$ e di perimetro $6~\mathrm{m}$, quale sarà la distanza del tesoro da ciascuno?
(A) $1~\mathrm{m}$
(B) ... | [
"Solution:\n\nLa risposta è **(D)**. Il tesoro è sepolto nel circocentro $O$ del triangolo ai vertici del quale si trovano Alice, Berto e Carlo. Poiché tale triangolo è rettangolo, $O$ è il punto medio dell'ipotenusa. Detta $x$ la sua lunghezza espressa in metri, si ha $x+\\frac{x}{2}+\\frac{x}{2} \\sqrt{3}=6$, da ... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
0jdk | Problem:
Find the sum of all positive integers $n$ such that there exists an integer $b$ with $|b| \neq 4$ such that the base $-4$ representation of $n$ is the same as the base $b$ representation of $n$. | [
"Solution:\nAll 1-digit numbers, $1,2,3$, are solutions when, say, $b=5$. (Of course, $d \\in \\{0,1,2,3\\}$ works for any base $b$ of absolute value greater than $d$ but not equal to $4$.)\n\nConsider now positive integers $n=(a_{d} \\ldots a_{1} a_{0})_{4}$ with more than one digit, so $d \\geq 1$, $a_{d} \\neq 0... | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1026 | |
00vv | Let $\triangle ABC$ be an acute-angled triangle with $AB < AC$, which is inscribed in the circle $\omega_1$ centered at $O$. Denote by $H$ the orthocenter of $\triangle ABC$ and by $M$ the midpoint of side $BC$. Let $\omega_2$ be the circumcircle of triangle $\triangle BHC$, let the line $HO$ meet $\omega_2$ at $K \neq... | [
"Let $D$ be the point such that $ABDC$ is a parallelogram and $O_A$ be the center of $\\omega_2$. Let $AM$ meet $\\omega_1$ at $E \\neq A$, and $AO$ meet $\\omega_1$ at $X \\neq A$. Furthermore, let $L$ be the reflection of $K$ across $M$, and $P'$ be the reflection of $P$ across $M$.\n$LBKC$ is a parallelogram, so... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
02gt | A graph $G$ with $n$ vertices is called *great* if we can label each vertex with a different positive integer not exceeding $\left\lfloor \frac{n^2}{4} \right\rfloor$ and find a set of non-negative integers $D$ so that there is an edge between two vertices if and only if the difference between their labels is in $D$. S... | [
"Let's count the number of ordered pairs $(f, D)$, where $f$ is a labeling of vertices and $D$ a possible set of differences of labels of vertices joined by an edge.\nNote that an ordered pair $(f, D)$ determines at most one great graph, since the labeling $f$ determine the vertices and the set of differences $D$ d... | Brazil | XXV OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
09va | In a square $ABCD$ of side length $2$ we draw lines from each vertex to the midpoints of the two opposite sides. For example, we connect $A$ to the midpoint of $BC$ and to the midpoint of $CD$. The eight resulting lines together bound an octagon inside the square (see figure). What is the area of this octagon?
![](att... | [
"$\\frac{2}{3}$"
] | Netherlands | Second Round, March 2019 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 2/3 | |
0klu | Problem:
A perfect power is an integer $n$ that can be represented as $a^{k}$ for some positive integers $a \geq 1$ and $k \geq 2$. Find the sum of all prime numbers $0 < p < 50$ such that $p$ is 1 less than a perfect power. | [
"Solution:\n\nFirst, it is known that $a^{k}-1 = (a-1)\\left(a^{k-1} + a^{k-2} + \\ldots + 1\\right)$. This means either $a-1$ or $a^{k-1} + a^{k-2} + \\ldots + 1$ must be 1 in order for $a^{k}-1$ to be prime. But this only occurs when $a$ is $2$. Thus, the only possible primes are of the form $2^{k}-1$ for some in... | United States | HMMT November 2021 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 41 | |
0clg | We will call a positive integer *special* if the sum of its (decimal) digits and the sum of the digits of its successor are divisible by $11$.
a) Find the last five digits of a special number.
b) Prove that there are infinitely many special numbers. | [
"a) Denote $s(m)$ the sum of the digits of a positive integer $m$.\nLet $n = \\overline{a_k a_{k-1} \\dots a_2 a_1}$ be a special number with $k \\ge 1$ digits. The statement says that $11 \\mid s(n)$ and $11 \\mid s(n+1)$.\nIf $a_1 \\le 8$, then $n+1 = \\overline{a_k a_{k-1} \\dots a_2 (a_1+1)}$, hence $s(n+1) = s... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | a) 99999. b) There are infinitely many, for example all numbers of the form n = 2·10^(q+5) + 899999 for any integer q ≥ 1. | |
023a | Problem:
Seja $ABC$ um triângulo acutângulo com alturas $BD$ e $CE$. Os pontos $F$ e $G$ são os pés das perpendiculares $BF$ e $CG$ à reta $DE$. Prove que $EF = DG$. | [
"Solution:\n\nOs ângulos $FBE$ e $DEC$ possuem a mesma medida, pois ambos são o complemento do ângulo $FEB$.\n\nObserve que o quadrilátero $BCDE$ é inscritível. De fato, a circunferência de diâmetro $BC$ contém $E$ e $D$, pois $\\angle BEC = \\angle BDC = 90^\\circ$.\n\nSegue que $\\angle FBE = \\angle DEC = \\angl... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08rz | $AB$ is a segment on a plane with length $7$, and $P$ is a point such that the distance between $P$ and line $AB$ is $3$. Find the smallest possible value of $AP \times BP$. | [
"Take $\\angle APB = \\theta$ and let $S$ be the area of $APB$. Then $\\frac{1}{2} \\times AP \\times BP \\times \\sin \\theta = S = \\frac{3 \\times 7}{2} = \\frac{21}{2}$. Since $\\sin \\theta$ is positive, $AP \\times BP$ takes minimum value when $\\sin \\theta$ takes maximum. Since $\\frac{7}{2} > 3$, we can ta... | Japan | Japan 2007 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | 21 | |
0leg | Find all functions $f: \mathbb{R} \to \mathbb{R}$ such that
$$
f(x)f(y) = f(xy - 1) + x f(y) + y f(x), \quad \forall x, y \in \mathbb{R}.
$$ | [
"By letting $y = 0$ in (1), we have $f(x) \\cdot f(0) = f(-1) + x f(0)$.\nWe distinguish two cases regarding the value of $f(0)$.\n\n**Case 1.** $f(0) \\neq 0$ implies $f(x) = x + c$ for all $x \\in \\mathbb{R}$ with $c$ is constant, which is not a solution.\n\n**Case 2.** $f(0) = 0$ implies $f(-1) = 0$.\nBy pluggi... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 0 for all real x; f(x) = x(x + 1) for all real x | |
0a3r | For a five-digit number $n = abcde$, we define the twisted sum of $n$ as $bcdea + cdeab + deabc + eabcd$. For example, the twisted sum of $20253$ is $02532 + 25320 + 53202 + 32025 = 113079$.
Let $m$ and $n$ be two five-digit numbers with the same twisted sum. Prove that $m = n$. | [
"We define $S(n)$ as the sum of the digits of $n$ and $T(n)$ as the twisted sum of $n$. If we look at the five terms of $T(n) + n$, we see that each digit of $n$ appears exactly once in each of the spots (ten thousand, thousand, hundred, ten, unit). This means that\n$$\nT(n) + n = (10000 + 1000 + 100 + 10 + 1)(a + ... | Netherlands | BxMO/EGMO Team Selection Test | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0bs6 | Prove that $2015 \in M$ and $2016 \notin M$, where
$$
M = \{ x_1 + 2x_2 + 3x_3 + \dots + 2015x_{2015} \mid x_1, x_2, \dots, x_{2015} \in \{-2, 3\} \}.
$$ | [
"An integer $n$ belongs to $M$ if and only if there exists disjoint subsets $A$, $B$ of the set $S = \\{1, 2, \\dots, 2015\\}$, with $A \\cup B = S$, so that $-2a + 3b = n$, where $a$ is the sum of the elements of $A$ and $b$ is the sum of the elements of $B$ (the sum of the elements of the empty set being $0$).\n\... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
03iz | Problem:
Prove that the sum of the squares of $1984$ consecutive positive integers cannot be the square of an integer. | [
"Solution:\nLet the $1984$ consecutive positive integers be $n, n+1, n+2, \\ldots, n+1983$.\n\nThe sum of their squares is:\n$$\nS = n^2 + (n+1)^2 + (n+2)^2 + \\cdots + (n+1983)^2\n$$\n\nThis can be written as:\n$$\nS = \\sum_{k=0}^{1983} (n+k)^2 = \\sum_{k=0}^{1983} (n^2 + 2nk + k^2) = 1984 n^2 + 2n \\sum_{k=0}^{1... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kh3 | Problem:
For each positive integer $1 \leq m \leq 10$, Krit chooses an integer $0 \leq a_{m} < m$ uniformly at random. Let $p$ be the probability that there exists an integer $n$ for which $n \equiv a_{m} \pmod{m}$ for all $m$. If $p$ can be written as $\frac{a}{b}$ for relatively prime positive integers $a$ and $b$, ... | [
"Solution:\n\nTuples of valid $a_{m}$ correspond with residues mod $\\operatorname{lcm}(1,2, \\ldots, 10)$, so the answer is\n$$\n\\frac{\\operatorname{lcm}(1,2, \\ldots, 10)}{10!} = \\frac{2^{3} \\cdot 3^{2} \\cdot 5 \\cdot 7}{2^{8} \\cdot 3^{4} \\cdot 5^{2} \\cdot 7} = \\frac{1}{1440}\n$$"
] | United States | HMMT Spring 2021 Guts Round | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 1540 | |
0fwz | Problem:
Sei $ABCD$ ein Quadrat mit Seitenlänge $1$ in der Ebene. Bestimme den geometrischen Ort aller Punkte $P$ mit der Eigenschaft
$$
AP \cdot CP + BP \cdot DP = 1
$$ | [
"Solution:\n\nDer gesuchte Ort ist die Vereinigung der beiden Diagonalen des Quadrates. Wir führen Koordinaten ein und setzen $A=(0,0)$, $B=(1,0)$, $C=(1,1)$, $D=(0,1)$ und $P=(a, b)$. Es gilt dann nach $\\mathrm{CS}$\n$$\n\\begin{aligned}\nAP \\cdot CP + BP \\cdot DP &= \\sqrt{a^2 + b^2} \\sqrt{(1-a)^2 + (1-b)^2} ... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations ... | null | proof and answer | The union of the two diagonals of the square. | |
0c1r | Let $(a_n)_{n \ge 1}$ be a sequence of integers larger than $1$ and
$$
b_n = 1 - \frac{1}{a_1} + \frac{1}{a_1 a_2} - \dots + (-1)^n \frac{1}{a_1 a_2 \dots a_n}, \quad n = 1, 2, 3, \dots
$$
Prove that:
a) the sequence $(b_n)_{n \ge 1}$ is convergent;
b) if $(a_n)_{n \ge 1}$ is unbounded, then the limit of the sequence $... | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kz0 | Let $x_n = \sin^2(n°)$. What is the mean of $x_1, x_2, x_3, \dots, x_{90}$?
(A) $\frac{11}{45}$ (B) $\frac{22}{45}$ (C) $\frac{89}{180}$ (D) $\frac{1}{2}$ (E) $\frac{91}{180}$ | [
"**Answer (E):** The required mean is\n$$\n\\frac{1}{90} \\sum_{n=1}^{90} \\sin^2(n°).\n$$\n\nGroup the summands into 44 pairs plus two additional terms as follows:\n$$\n\\begin{align*}\n\\sin^2(1^\\circ) + \\sin^2(89^\\circ) &= \\sin^2(1^\\circ) + \\cos^2(1^\\circ) = 1 \\\\\n\\sin^2(2^\\circ) + \\sin^2(88^\\circ) ... | United States | 2024 AMC 12 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | E |
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