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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0a1s | Problem:
Bepaal het grootste reële getal $M$ zodanig dat voor elke oneindige rij $x_{0}, x_{1}, x_{2}, \ldots$ van reële getallen die voldoet aan
a) $x_{0}=1$ en $x_{1}=3$,
b) $x_{0}+x_{1}+\cdots+x_{n-1} \geq 3 x_{n}-x_{n+1}$,
geldt dat
$$
\frac{x_{n+1}}{x_{n}}>M
$$
voor alle $n \geq 0$. | [
"Solution:\nAntwoord: de grootst mogelijke constante waarvoor dit geldt is $M=2$.\n\nDeze opgave is een typisch voorbeeld waarbij het baat om een sterkere inductiehypothese te nemen dan je strikt nodig hebt: we laten met inductie zien dat $x_{n+1}>2 x_{n}>x_{n}+x_{n-1}+\\ldots+x_{0}$.\n\nVoor $n=0$ staat hier dat $... | Netherlands | IMO-selectietoets | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2 | |
0g4v | Problem:
Soit $ABC$ un triangle tel que $2 \cdot \angle CBA = 3 \cdot \angle ACB$. Les points $D$ et $E$ sont sur le côté $AC$, tels que $BD$ et $BE$ divisent $\angle CBA$ en trois angles égaux et que $D$ soit entre $A$ et $E$. De plus, soit $F$ l'intersection de $AB$ et de la bissectrice de $\angle ACB$. Montrer que ... | [
"Solution:\n\nLes conditions du problème impliquent que $\\angle DCF = \\angle FCB = \\angle EBD = \\angle DBF$. Comme $\\angle DCF = \\angle DBF$, le quadrilatère $FDCB$ est cyclique. Ainsi, $\\angle FDB = \\angle FCB = \\angle EBD$, ce qui montre bien que $BE$ et $DF$ sont parallèles.",
"Solution:\n\nLes condit... | Switzerland | Deuxième tour 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0504 | Let $ABC$ be a triangle with integral side lengths. The angle bisector drawn from $B$ and the altitude drawn from $C$ meet at point $P$ inside the triangle. Prove that the ratio of areas of triangles $APB$ and $APC$ is a rational number. | [
"Let $H$ be the foot of the altitude drawn from $C$. First prove that $|AH|$ and $|BH|$ are rational numbers. For that, use the Pythagorean theorem for triangles $ACH$ and $BCH$ to obtain $|AH|^2 + |CH|^2 = |AC|^2$ and $|BH|^2 + |CH|^2 = |BC|^2$. Therefore $|AC|^2 - |BC|^2 = |AH|^2 - |BH|^2 = (|AH| - |BH|) \\cdot (... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0582 | Find all positive integers $k$ for which there is a right triangle with legs of integral lengths and hypotenuse of length $\sqrt{88\ldots822\ldots2}$, where the number under the root consists of exactly $k$ eights and exactly $k$ twos. | [
"Let the lengths of legs be $a$ and $b$. By the Pythagorean theorem,\n$$\na^2 + b^2 = 88\\ldots822\\ldots2.\n$$\nIf $k=1$ then one can choose $a=9$ and $b=1$ as $9^2 + 1^2 = 82$. If $k \\ge 2$ then $88\\ldots822\\ldots2 \\equiv 6 \\pmod{8}$ since $822 \\equiv 6 \\pmod{8}$ and $222 \\equiv 6 \\pmod{8}$. On the other... | Estonia | Estonian Math Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 1 | |
0fhp | Problem:
Justificar razonadamente que en cualquier triángulo el diámetro de la circunferencia inscrita no es mayor que el radio de la circunferencia circunscrita. | [
"Solution:\n\nPrimera solución\nEl área en función del radio de la circunferencia inscrita es $S = \\frac{(a + b + c) r}{2}$, de donde $r = \\frac{2S}{a + b + c}$.\nEl área en función del radio de la circunferencia circunscrita es $S = \\frac{abc}{4R}$, de donde $R = \\frac{abc}{4S}$.\nTenemos que probar que $\\fra... | Spain | OME 29 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
0cxg | Let $ABC$ be an acute triangle and let $MNPQ$ be a square inscribed in the triangle such that $M, N \in BC$, $P \in AC$, $Q \in AB$. Prove that
$$
\operatorname{area}[MNPQ] \leq \frac{1}{2} \operatorname{area}[ABC]
$$ | [
"\nDenote by $x$ the length of sides of square $MNPQ$, $a = BC$, $h_a = AA'$, where $AA' \\perp BC$. The triangle $AQP$ and $ABC$ are similar, hence we have $\\frac{x}{a} = \\frac{h_a - x}{h_a}$. We get\n$$\nx = \\frac{a h_a}{a + h_a} \\leq \\frac{a h_a}{2 \\sqrt{a h_a}} = \\frac{1}{2} \\sq... | Saudi Arabia | SAMC | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0fb5 | Problem:
Los enteros positivos $x, y, z$ cumplen
$$
x+2 y=z, \quad x^{2}-4 y^{2}+z^{2}=310
$$
Halla todos los posibles valores del producto $x y z$. | [
"Solution:\nPodemos despejar $2 y$ de la primera ecuación y sustituir en la segunda, con lo que ha de cumplirse\n$$\n310=x^{2}-(z-x)^{2}+z^{2}=2 z x, \\quad z x=155=5 \\cdot 31\n$$\nLuego al ser $5,31$ primos, se tiene que $z$ ha de tomar uno de los valores $155,31,5,1$, tomando $x$ respectivamente los valores $1,5... | Spain | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 11935, 2015 | |
0asw | Problem:
For each positive integer $n$, let $S_{n}$ be the sum of the infinite geometric series whose first term is $n$ and whose common ratio is $\frac{1}{n+1}$. Determine the least value of $n$ such that
$$
S_{1}+S_{2}+\cdots+S_{n}>5150
$$ | [
"Solution:\n\n101"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 101 | |
0jb3 | Acute triangle $ABC$ is inscribed in circle $\omega$. Let $H$ and $O$ denote its orthocenter and circumcenter, respectively. Let $M$ and $N$ be the midpoints of sides $AB$ and $AC$, respectively. Rays $MH$ and $NH$ meet $\omega$ at $P$ and $Q$, respectively. Lines $MN$ and $PQ$ meet at $R$. Prove that $OA \perp RA$.
!... | [
"Note that there is a dilation centered at $A$ with ratio $2$ sending triangle $AMN$ to $ABC$. Hence the circumcircles of triangles $ABC$ and $AMN$ are tangent at $A$. Denote their common tangent at $A$ by $\\ell$; we note that $\\ell$ is the radical axis of these two circles. We now have a key lemma.\n\n**Lemma 1.... | United States | Team Selection Test Selection Test | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
0blr | Find all two-digit positive integers $\overline{ab}$, with $a < b$, which are equal to the sum of the integers from $a$ to $b$ ($a$ and $b$ included). | [
"The solutions are $27$ and $15$.\n\nWe start by noticing that $\\overline{ab} = a + (a+1) + \\dots + b \\le 1 + 2 + \\dots + 9 = 45$, which yields $a \\le 4$.\n\nIf $a = 4$, then $\\overline{ab} \\ge 45 = 1 + 2 + \\dots + 9$, leading to $a = 1$ – impossible.\n\nIf $a = 3$, then $3+4+\\dots+(b-1)+b = \\overline{3b}... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 27 and 15 | |
0g3d | Problem:
Let $A B C$ be a triangle where $\angle B A C=90^{\circ}$, with circumcenter $O$ and incenter $I$. The angle bisector of $\angle B A C$ intersects the circumcircle of $A B C$ in $A$ and $P$. Let $Q$ be the projection of $P$ onto $A B$, and $R$ be the projection of $I$ onto $P Q$. Prove that $R O$ bisects $C I... | [
"Solution:\n\nBecause $O$ is the midpoint of $\\overline{B C}$, it suffices to show that $R O$ is parallel to $B I$, as the result follows by simply looking at a homothety centered at $C$.\n\nSo first of all by looking at the problem we instantly recognise the point $P$ as the circumcenter of $B I C$, as this is a ... | Switzerland | IMO Selection | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasi... | null | proof only | null | |
0fai | Problem:
Show that $\left(x + y + z\right)^2 / 3 \geq x\sqrt{yz} + y\sqrt{zx} + z\sqrt{xy}$ for all non-negative reals $x$, $y$, $z$. | [
"Solution:\n\nBy AM/GM, $xy + yz \\geq 2x\\sqrt{yz}$. Adding the similar results gives\n$$\n2(xy + yz + zx) \\geq 2\\left(x\\sqrt{yz} + y\\sqrt{zx} + z\\sqrt{xy}\\right).\n$$\n\nBy AM/GM, $x^2 + x^2 + y^2 + z^2 \\geq 4x\\sqrt{yz}$. Adding the similar results gives\n$$\nx^2 + y^2 + z^2 \\geq x\\sqrt{yz} + y\\sqrt{zx... | Soviet Union | 25th ASU | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
03o4 | Problem:
Centuries ago, the pirate Captain Blackboard buried a vast amount of treasure in a single cell of an $M \times N$ ($2 \leq M, N$) grid-structured island. You and your crew have reached the island and have brought special treasure detectors to find the cell with the treasure. For each detector, you can set it ... | [
"Solution:\n\nLet $m=\\left\\lceil\\frac{M}{2}\\right\\rceil$ and $n=\\left\\lceil\\frac{N}{2}\\right\\rceil$. We claim that the minimal $Q$ is $m+n$.\n\nFor the construction, start with $m$ detectors covering $[i, i+m-1] \\times [1, N]$ for $1 \\leq i \\leq m$. For every pair of rows, there is a detector that cove... | Canada | CMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ceil(M/2) + ceil(N/2) | |
0a4b | Problem:
Bepaal alle drietallen $(x, y, p)$ van positieve gehele getallen zo dat $p$ een priemgetal is, $x^2 = p - 1$ en $y^2 = 2p^2 - 1$. | [
"Solution:\n\nHet enige drietal dat voldoet is $(2, 7, 5)$.\nWe rekenen eerst uit dat\n$$\n(y + x)(y - x) = y^2 - x^2 = (2p^2 - 1) - (p - 1) = 2p^2 - p = p(2p - 1). \\quad (1)\n$$\nDat betekent in het bijzonder dat $p \\mid x + y$ of $p \\mid x - y$.\nStel dat $p \\mid y + x$. Dan geldt dat $y = kp - x$ voor een ze... | Netherlands | IMO-selectietoets III | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 7, 5) | |
05w8 | Problem:
Un ensemble $A$ d'entiers est dit admissible s'il vérifie la propriété suivante : pour tous $x, y \in A$ (non nécessairement distincts), et pour tout $k \in \mathbb{Z}$, on a $x^{2}+k x y+y^{2} \in A$.
Déterminer tous les couples d'entiers non nuls ( $m, n$ ) tels que le seul ensemble admissible contenant à ... | [
"Solution:\n\nRemarquons que si $d=\\operatorname{gcd}(m, n)>1$ alors l'ensemble $A$ des multiples de $d$ est un ensemble admissible contenant $n$ et $m$ mais n'étant pas égal à $\\mathbb{Z}$. En effet, si $x$ et $y$ sont multiples de $d$, alors pour tout entier $k$, $x^{2}+k x y+y^{2}$ est bien divisible par $d$ e... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | All pairs of nonzero integers that are coprime (their greatest common divisor is one). | |
0eyr | Problem:
The sequence $\{a_n\}$ is defined as follows: $a_1 = 1$, $a_{n+1} = a_n + 1 / a_n$ for $n \geq 1$. Prove that $a_{100} > 14$. | [
"Solution:\nFirst we must notice that for $1 \\leq a, b$ we have $a < b$, then $a + 1 / a < b + 1 / b$. This is basic to any estimation.\n\nThe obvious approach is to notice that if $a_i \\leq n$, then $a_{i+1} \\geq a_i + 1 / n$. Hence it takes at most $n$ steps to get from $n-1$ to $n$. Unfortunately, this does n... | Soviet Union | 2nd ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
065i | Let $\alpha$, $\beta$, $\gamma$ are positive integers such that the number
$$
A = \frac{\alpha\sqrt{2} + \beta\sqrt{3}}{\beta\sqrt{2} + \gamma\sqrt{3}}
$$
is rational. Prove that the number
$$
B = \frac{\alpha^2 + \beta^2 + \gamma^2}{\alpha + \beta + \gamma}
$$
is integer. | [
"First of all, it is easy to see that:\n$$\n\\alpha_1\\sqrt{2} + \\alpha_2\\sqrt{3} = \\alpha_3\\sqrt{2} + \\alpha_4\\sqrt{3}, \\text{ with } \\alpha_1, \\alpha_2, \\alpha_3, \\alpha_4 \\in \\mathbb{Q}^* \\Leftrightarrow \\alpha_1 = \\alpha_3 \\text{ and } \\alpha_2 = \\alpha_4.\n$$\nIn fact, we can write the first... | Greece | SELECTION EXAMINATION | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Linear Algebra > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0e20 | Problem:
Tone je rezervoar za gorivo napolnil do vrha in se $z$ avtom podal na dolgo pot. Podatki, ki jih je Tone razbral na elektronskem števcu avtomobila, so prikazani v tabeli.
| število km | 35824 | 36149 | 36449 |
| :--- | :--- | :--- | :---: |
| podatki o gorivu | poraba $17,2 \ell$ | poraba $39,3 \ell$ | ostal... | [
"Solution:\n\nUgotovimo, da poraba goriva predstavlja linearno funkcijo odvisno od prevoženih kilometrov $f(x) = k \\cdot x$. Izračunamo $k = \\frac{39,3 - 17,2}{36149 - 35824} = \\frac{22,1}{325} = 0,068$ litrov na kilometer oziroma $6,8$ litrov na $100$ kilometrov.\n\nZa nadaljnjih $300$ kilometrov potrebuje $f(3... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 65 liters | |
02ia | Problem:
$2-2\{2-2[2-2(4-2)]\}$ é igual a:
(A) 0
(B) 2
(C) -2
(D) 4
(E) -10 | [
"Solution:\n\n(E) As ordens de prioridade para resolver uma expressão são:\n$$\n\\begin{aligned}\n& \\underbrace{\\text{parênteses}}_{1^{\\circ}} \\rightarrow \\underbrace{\\text{colchete}}_{2^{\\circ}} \\rightarrow \\underbrace{\\text{chaves}}_{3^{\\circ}} \\text{ e } \\underbrace{\\text{multiplicações e divisões}... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | MCQ | E | |
059i | Two positive integers together contain each digit $0$, $1$, $\ldots$, $9$ exactly once. Find the largest possible common divisor that these two numbers can have. | [
"*Answer:* $48651$.\n\nSuppose that both numbers contain $5$ digits. A common divisor of two different numbers cannot exceed half of the larger one; thus the greatest common divisor of two such numbers must be less than $50000$. It means that if the greatest common divisor has $5$ digits then the first digit is at ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 48651 | |
00vj | Let $x_1, x_2, x_3$ and $x_4$ be positive real numbers. Prove the inequality:
$$
\frac{x_1 + 3x_2}{x_2 + x_3} + \frac{x_2 + 3x_3}{x_3 + x_4} + \frac{x_3 + 3x_4}{x_4 + x_1} + \frac{x_4 + 3x_1}{x_1 + x_2} \ge 8.
$$ | [
"Let us denote\n$$\nL(x_1, x_2, x_3, x_4) = \\frac{x_1 + 3x_2}{x_2 + x_3} + \\frac{x_2 + 3x_3}{x_3 + x_4} + \\frac{x_3 + 3x_4}{x_4 + x_1} + \\frac{x_4 + 3x_1}{x_1 + x_2}.\n$$\nNotice that this function is cyclic, i.e. $L(x_1, x_2, x_3, x_4) = L(x_2, x_3, x_4, x_1) = L(x_3, x_4, x_1, x_2) = L(x_4, x_1, x_2, x_3)$. H... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0dmo | Problem:
Нека је $k$ уписана кружница неједнакокраког $\triangle A B C$, чији је центар $S$. Кружница $k$ додирује странице $B C, C A, A B$ у тачкама $P, Q, R$, редом. Права $Q R$ сече праву $B C$ у тачки $M$. Нека кружница која садржи тачке $B$ и $C$ додирује $k$ у тачки $N$. Описана кружница $\triangle M N P$ сече п... | [
"Solution:\n\nПосматрајмо хомотетију са центром $N$ која слика круг $k$ у круг $B C N$; нека она слика тачку $P$ у $P_{1}$. Тангента на круг $B C N$ у $P_{1}$ је паралелна тангенти на $k$ у $P$, тј. правој $B C$, што значи да је $P_{1}$ средиште лука $B C$ круга $B C N$. Дакле, $N P$ је симетрала угла $C N B$, па ј... | Serbia | Српска математичка олимпијада | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: c... | null | proof only | null | |
0960 | Problem:
Să se arate, că dacă numerele reale $a, b, c \in [0,1]$, atunci
$$
\frac{3 a^{2}-2 a}{1+b^{3}+c^{3}}+\frac{3 b^{2}-2 b}{1+a^{3}+c^{3}}+\frac{3 c^{2}-2 c}{1+a^{3}+b^{3}} \leqslant 1
$$ | [
"Solution:\n\nPentru $a=b=c=0$ inegalitatea este adevărată: $0 \\leqslant 1$. În continuare, vom presupune că măcar unul dintre aceste numere este nenul.\nDacă $x \\in [0,1]$, atunci este adevărată inegalitatea\n$$\nx^{3} \\geqslant 3 x^{2}-2 x\n$$\nÎntr-adevăr,\n$$\n(*) \\Leftrightarrow x^{3}-3 x^{2}+2 x \\geqslan... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
04w3 | Ten boys and ten girls met at a party. Suppose that every boy likes a different (positive) number of girls and that every girl likes a different (positive) number of boys. Find the largest non-negative integer $n$ such that it is always possible to form $n$ disjoint couples of a boy and a girl that like each other.
(Jo... | [
"We shall prove that the answer is $n = 1$.\n\nTo begin with, note that the problem statement implies that the boys like $1, 2, \\ldots, 10$ girls in some order, so there exists a boy that likes all the girls. Analogously, there must exist a girl that likes all the boys, so putting the two together always yields an... | Czech Republic | Final Round of the 73rd Czech and Slovak Mathematical Olympiad (March 17–20, 2024) | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1 | |
02on | We call a number *pal* if it doesn't have a zero digit and the sum of the squares of the digits is a perfect square. For example, $2115522$ is pal (because $2^2 + 1^2 + 1^2 + 5^2 + 5^2 + 2^2 + 2^2 = 8^2$) but $304$ and $12$ are not pal.
a. What is the greatest two-digit pal number?
b. Does there exist a $2011$-digit ... | [
"a. First notice that $86$ is pal. Then it's not hard to check by hand that every number from $87$ to $99$ is not pal.\n\nb. The answer is *yes*. First consider the $2011$-digit number $11\\ldots1$. The sum of its digits is $2011$. The smallest perfect square greater than $2011$ is $45^2 = 2025$. Since $2025 - 2011... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) 86; b) Yes | |
0hh6 | Assume $P_1, P_2, \dots, P_n$ ($n > 10$) are pairwise different polynomials with coefficients $1$, $0$, or $-1$ and such that do not have integer roots. Additionally, $\forall i = \overline{1,n}$ $|P_i(5)| \le \frac{n^2}{2}$. Prove that $P_i + P_j = P_k + P_l$ for some $1 \le i,j,k,l \le n$ and $\{i, j\} \neq \{k, l\}$... | [
"Since the polynomials $P_1, \\dots, P_n$ do not have integer roots then $P_i(0) \\neq 0$ for any $i = \\overline{1,n}$ and $P_i(0) = \\pm 1$. Let us consider all pairs of polynomials $P_k$ and $P_j$, such that $k \\neq j$ and $P_k(0) = P_j(0)$. For each pair of polynomials we determine $Q_{k,j} = P_k - P_j$. If $a... | Ukraine | Problems from Ukrainian Authors | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
03ou | The straight line $\frac{x}{4} + \frac{y}{3} = 1$ intersects the ellipse $\frac{x^2}{16} + \frac{y^2}{9} = 1$ at two points $A$ and $B$. There is a point $P$ on this ellipse such that the area of $\triangle PAB$ is equal to $3$. There is/are ( ) such point/points $P$.
(A) 1 (B) 2 (C) 3 (D) 4 | [
"Suppose that there is a point $P(4\\cos \\alpha, 3\\sin \\alpha)$ on the ellipse. When $P$ and the origin $O$ are not on the same side of $AB$, the distance from $P$ to $AB$ is\n$$\n\\begin{aligned}\n& \\frac{3(4\\cos\\alpha) + 4(3\\sin\\alpha) - 12}{5} \\\\\n&= \\frac{12}{5}(\\cos\\alpha + \\sin\\alpha - 1) \\\\\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | MCQ | B | |
0775 | Problem:
Let $A_{1} B_{1} C_{1} D_{1} E_{1}$ be a regular pentagon. For $2 \leq n \leq 11$, let $A_{n} B_{n} C_{n} D_{n} E_{n}$ be the pentagon whose vertices are the midpoints of the sides of the pentagon $A_{n-1} B_{n-1} C_{n-1} D_{n-1} E_{n-1}$. All the 5 vertices of each of the 11 pentagons are arbitrarily coloured... | [
"Solution:\nWe first observe that all the eleven pentagons are regular. Moreover, there are 5 fixed directions and all the 55 sides are in one of these directions. If we consider any two sides which are parallel, they are the parallel sides of an isosceles trapezium, which is cyclic.\n\nIf we consider any pentagon,... | India | Indian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
05hb | Problem:
Caractériser les entiers $n \geq 2$ tels que pour tout entier $a$ on ait $a^{n+1} \equiv a \pmod{n}$. | [
"Solution:\n\nVoici les $n$ vérifiant cette propriété : $2$, $2 \\cdot 3$, $2 \\cdot 3 \\cdot 7$, $2 \\cdot 3 \\cdot 7 \\cdot 43$.\n\nPour prouver que c'est exhaustif, on procède de la façon suivante : on commence par remarquer que $n$ n'a pas de facteur carré. En effet, si $p^{2}$ divise $n$, alors $p^{n+1}-p$ est... | France | Envoi 1 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2, 6, 42, 1806 | |
0jlh | Problem:
If four fair six-sided dice are rolled, what is the probability that the lowest number appearing on any die is exactly $3$? | [
"Solution:\nThe probability that all the die rolls are at least $3$ is $\\left(\\frac{4}{6}\\right)^4$. The probability they are all at least $4$ is $\\left(\\frac{3}{6}\\right)^4$. The probability of being in the former category but not the latter is thus $\\left(\\frac{4}{6}\\right)^4 - \\left(\\frac{3}{6}\\right... | United States | HMMT 2014 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 175/1296 | |
07a2 | The escalator of “Champion Butcher” metro station has this property that if $m$ persons are on it, its speed is $m^{-\alpha}$ where $\alpha$ is a positive constant number. Suppose that $n$ persons want to go upstairs by the escalator. If the length of the escalator is $l$, what is the least time required for these pers... | [
"In every moment consider the number of persons that are on the escalator at that time. Now consider the intervals such that in every time of such intervals the number of persons on the escalator is equal to a fixed integer. Suppose that we have $k$ intervals $I_1, I_2, \\dots, I_k$ and for $1 \\le i \\le k$, $a_i$... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | If alpha ≥ 1, the least time is n l. If alpha < 1, the least time is n^alpha l. | |
0c5l | We consider $A$, the set of all positive integers with exactly $2019$ positive integer divisors, and, for each $n \in A$, we denote
$$
S_n = \frac{1}{d_1 + \sqrt{n}} + \frac{1}{d_2 + \sqrt{n}} + \dots + \frac{1}{d_{2019} + \sqrt{n}},
$$
where $d_1, d_2, \dots, d_{2019}$ are the positive divisors of $n$.
Determine the m... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 673/2^337 | |
06vr | Find all triples $(a, b, c)$ of positive integers such that $a^{3} + b^{3} + c^{3} = (a b c)^{2}$. | [
"Solution 1. We will start by proving that $c=1$. Note that\n$$\n3 a^{3} \\geqslant a^{3} + b^{3} + c^{3} > a^{3}\n$$\nSo $3 a^{3} \\geqslant (a b c)^{2} > a^{3}$ and hence $3 a \\geqslant b^{2} c^{2} > a$. Now $b^{3} + c^{3} = a^{2} (b^{2} c^{2} - a) \\geqslant a^{2}$, and so\n$$\n18 b^{3} \\geqslant 9 (b^{3} + c^... | IMO | IMO 2019 Shortlisted Problems | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All permutations of (1, 2, 3) | |
074d | Let $ABC$ be a triangle in which $\angle A = 60^\circ$. Let $T$ be the point where the in-circle and the nine-point circle of $ABC$ touch each other. If $r$ is the in-radius of $ABC$, prove that $AT = r$. | [
"\nWe first observe that $AI = r \\cos\\frac{A}{2} = 2r$. Let $T'$ be point on $AI$ such that $AT' = T'I = r$. Obviously $T'$ lies on the in-circle of $ABC$. We show that $T'$ also lies on the nine-point circle of $ABC$.\nLet $D, E, F$ be the midpoints of $BC, CA, AB$ respectively. Note tha... | India | Indija TS 2009 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
0ira | Problem:
Let $DEF$ be a triangle and $H$ the foot of the altitude from $D$ to $EF$. If $DE = 60$, $DF = 35$, and $DH = 21$, what is the difference between the minimum and the maximum possible values for the area of $DEF$? | [
"Solution:\n\nObserve that the two possible configurations come from $DEF$ being obtuse and $DEF$ being acute. In either case, we have that $HF = \\sqrt{35^2 - 21^2} = 28$ and $EH = \\sqrt{60^2 - 21^2} = 9\\sqrt{39}$. This means that $EF$ can be either $FH + EH$ or $|EH - FH|$. The difference in area is hence $\\fr... | United States | 1st Annual Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 588 | |
02wy | Problem:
O triângulo $ABC$ abaixo tem um ponto $D$ no seu lado $AB$, tal que $AB = CD$, $\angle ABC = 100^{\circ}$ e $\angle DCB = 40^{\circ}$.
a) Qual a medida do ângulo $B\hat{D}C$ ?
b) Qual a medida do ângulo $A\hat{C}D$ ? | [
"Solution:\n\na) Analisando a soma dos ângulos internos do triângulo $\\triangle BDC$, temos $100^{\\circ} + 40^{\\circ} + \\angle BDC = 180^{\\circ}$, ou seja, $\\angle BDC = 40^{\\circ}$.\n\nb) Em virtude do item anterior, $\\triangle BDC$ é isósceles de base $CD$ e então $BC = BD$. Construa o triângulo auxiliar ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) 40°; b) 10° | |
02k2 | Problem:
A tabela mostra as temperaturas máximas e mínimas durante 5 dias seguidos em certa cidade. Em qual dia ocorreu o maior variação de temperatura?
| Dia | Temperatura máxima em $^{\circ} \mathrm{C}$ | Temperatura mínima em $^{\circ} \mathrm{C}$ |
|:----------:|:------------------------------------------... | [
"Solution:\n\nA variação de temperatura é a diferença entre a máxima e a mínima. Temos:\n\n| Dia | Temperatura máxima em $^{\\circ} \\mathrm{C}$ | Temperatura mínima em $^{\\circ} \\mathrm{C}$ | Variação |\n|:----------:|:------------------------------------------:|:-----------------------------------------... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 6a-feira | |
02zz | Problem:
Seja $ABCD$ um retângulo tal que $AB = \sqrt{2} BC$. Seja $E$ um ponto sobre o semicírculo com diâmetro $AB$, como indicado na figura a seguir. Sejam $K$ e $L$ as interseções de $AB$ com $ED$ e $EC$, respectivamente. Se $AK = 2\ \mathrm{cm}$ e $BL = 9\ \mathrm{cm}$, calcule, em $\mathrm{cm}$, o comprimento do... | [
"Solution:\n\nSejam $x$ e $y$ os comprimentos das projeções ortogonais dos segmentos $EK$ e $EL$ sobre o segmento $AB$ e $P$ a projeção ortogonal de $E$ sobre $AB$. Além disso, seja $h = EP$. Por semelhança de triângulos, temos\n$$\n\\frac{2}{x} = \\frac{BC}{h} \\text{ e } \\frac{9}{y} = \\frac{BC}{h}\n$$\nPortanto... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 6 | |
01xq | Find all functions $f(x): \mathbb{R} \to \mathbb{R}$ satisfying the equality
$$
x = -\frac{1}{2}f(|x|) + |f(x)|
$$
for all real numbers $x$. | [
"$f(x) = 2x$ for all $x \\ge 0$ and $f(x) = 0$ for all $x < 0$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | f(x) = 2x for x ≥ 0, and f(x) = 0 for x < 0 | |
056d | Find all triplets of consecutive integers, such that one of these numbers is the sum of the two others. | [
"Let the consecutive numbers be $x$, $x+1$, and $x+2$. There are three cases based on which of the numbers is the sum of the other two.\n\nIf $x+2 = x + (x+1)$, then $x = 1$, which gives $(1, 2, 3)$.\n\nIf $x+1 = x + (x+2)$, then $x = -1$, which gives $(-1, 0, 1)$.\n\nIf $x = (x+1) + (x+2)$, then $x = -3$, which gi... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | (-3, -2, -1), (-1, 0, 1), (1, 2, 3) | |
0kox | Problem:
Let $z_{1}, z_{2}, z_{3}, z_{4}$ be the solutions to the equation $x^{4}+3 x^{3}+3 x^{2}+3 x+1=0$. Then $|z_{1}|+|z_{2}|+|z_{3}|+|z_{4}|$ can be written as $\frac{a+b \sqrt{c}}{d}$, where $c$ is a square-free positive integer, and $a, b, d$ are positive integers with $\operatorname{gcd}(a, b, d)=1$. Compute $... | [
"Solution:\n\nNote that $x=0$ is clearly not a solution, so we can divide the equation by $x^{2}$ to get $\\left(x^{2}+2+\\frac{1}{x^{2}}\\right)+3\\left(x+\\frac{1}{x}\\right)+1=0$. Letting $y=x+\\frac{1}{x}$, we get that $y^{2}+3y+1=0$, so $y=x+\\frac{1}{x}=\\frac{-3 \\pm \\sqrt{5}}{2}$. Since $\\frac{-3+\\sqrt{5... | United States | HMMT February | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | final answer only | 7152 | |
0hpn | Problem:
Art and Ben play a game while sharing an $m \times n$ chocolate bar. They take turns breaking the bar into two rectangular pieces along one of the lines and eating the smaller piece. (If the two pieces are equal, they can choose which piece to eat.) Whoever is left with the last $1 \times 1$ square of chocola... | [
"Solution:\n\nThe answer is all pairs $(m, n)$ such that the ratio\n$$\nR=\\frac{m+1}{n+1}\n$$\nis an integer power of two, that is, the pairs $\\left(a-1,2^{k} a-1\\right)$ and $\\left(2^{k} a-1, a-1\\right)$ for $a \\geq 2$ and $k \\geq 0$. We will show that\n\na. If $R$ is a power of two, then it will not remain... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | All pairs with (m+1)/(n+1) equal to an integer power of two; equivalently, (a−1, 2^k a−1) or (2^k a−1, a−1) for integers a ≥ 2 and k ≥ 0. | |
0kpv | Problem:
Let $\pi$ be a uniformly random permutation of the set $\{1,2, \ldots, 100\}$. The probability that $\pi^{20}(20)=20$ and $\pi^{21}(21)=21$ can be expressed as $\frac{a}{b}$, where $a$ and $b$ are relatively prime positive integers. Compute $100a+b$. (Here, $\pi^{k}$ means $\pi$ iterated $k$ times.) | [
"Solution:\nWe look at the cycles formed by $\\pi$. Let $\\operatorname{ord}_{\\pi}(n)$ denote the smallest $m$ such that $\\pi^{m}(n)=n$. In particular, the condition implies that $\\operatorname{ord}_{\\pi}(20) \\mid 20$ and $\\operatorname{ord}_{\\pi}(21) \\mid 21$.\n\nClaim 1. $20$ and $21$ cannot be in the sam... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 1025 | |
0d3v | In a tournament each player played exactly one game against each of the other players. In each game the winner was awarded 1 point, the loser got 0 points, and each of the two players earned $\frac{1}{2}$ point if the game was a tie. After the completion of the tournament, it was found that exactly half of the points e... | [
"Suppose that there were $n$ players and assume $s_{1} \\geq s_{2} \\geq \\cdots \\geq s_{n}$ be the scores of each player. Because each player played exactly one game against each of the other players and each game worths 1 point, the total sum of scores is\n$$\ns_{1}+s_{2}+\\cdots+s_{n}=\\binom{n}{2}=\\frac{n(n-1... | Saudi Arabia | SAMC | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English, Arabic | proof and answer | 25 | |
00xh | Problem:
Let $a=\sqrt[1992]{1992}$. Which number is greater:
$$
a^{a^{a^{a}}}
$$
or $1992$? | [
"Solution:\nThe first of these numbers is less than\n\n$$\na^{a^{a^{\\cdots}}}^{1992} = a^{a^{a^{\\cdots}}}^{1991} = \\ldots = 1992.\n$$"
] | Baltic Way | Baltic Way 1992 | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1992 | |
006c | Se tiene un tablero rectangular de $9 \times 2007$ dividido en cuadritos de $1 \times 1$. Inicialmente todos los cuadritos son blancos. En cada paso se colorean de negro 4 cuadritos blancos que estén en la intersección de dos filas y dos columnas del tablero. Cuando ya no queden 4 cuadritos blancos en la intersección d... | [] | Argentina | Argentina 2008 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 2007 for 9×2007; 2007 for 99×2007 | |
01pn | Find all pairs $(p; q)$ of prime numbers $p$ and $q$ satisfying the equality
$$
p^{5} + p^{3} + 2 = q^{2} - q.
$$ | [
"Answer: $(p; q) = (2; 7)$ and $(p; q) = (3; 17)$.\nIf $p = 2$ or $p = 3$, then the obtained quadratic equations with respect to $q$ have two prime solutions: $q = 7$ and $q = 17$.\n\nNow let $p > 3$. We rewrite the initial equation in the form $p^3(p^2 + 1) = (q + 1)(q - 2)$. Note that the greatest common divisor ... | Belarus | BelarusMO 2013_s | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | [[2, 7], [3, 17]] | |
0a5i | Problem:
Let $a$, $b$, $c$, $d$ be integers such that $a > b > c > d \geq -2021$ and
$$
\frac{a + b}{b + c} = \frac{c + d}{d + a}
$$
(and $b + c \neq 0 \neq d + a$). What is the maximum possible value of $ac$? | [
"Solution:\nWe claim that the maximum value of $ac$ is $2 \\times 505^{2} = 510050$, and this is uniquely achieved by $(a, b, c, d) = (1010, 506, 505, -2021)$.\n\nTo prove this we start by rearranging the expression $\\frac{a + b}{b + c} = \\frac{c + d}{d + a}$ to get $(c + b)(c + d) = (a + b)(a + d)$. Now expand t... | New Zealand | NZMO Round One | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 510050 | |
07rl | Let $A$, $B$, $C$, $D$, $E$ be five points in the plane with distances given (in some units) as
$$
|AB| = 12,\ |BC| = 50,\ |CD| = 38,\ |AD| = 100,\ |BE| = 30,\ |CE| = 40.
$$
Find the distance $|ED|$. | [
"Since $|AD| = 100 = 12 + 50 + 38 = |AB| + |BC| + |CD|$, the points $A$, $B$, $C$, $D$ lie on a line in this order. Since\n$$\n|BE|^2 + |EC|^2 = 30^2 + 40^2 = 900 + 1600 = 2500 = 50^2 = |BC|^2,\n$$\nthe triangle $EBC$ has a right angle at $E$.\n\n\n\nLet $Q$ be the projection of $E$ on the ... | Ireland | Irish | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 74 | |
0les | Let $a$ be a non-negative real number and a sequence $(u_n)$ defined as
$$
u_1 = 6, u_{n+1} = \frac{2n + a}{n} + \sqrt{\frac{n+a}{n}}u_n + 4$$
for all positive integers $n$.
a) For $a = 0$, prove that $(u_n)$ has a finite limit and find its value.
b) For $a \ge 0$, prove that $(u_n)$ has a finite limit. | [
"a) For $a = 0$, we have $(u_n)$ defined by\n$$\nu_1 = 6, u_{n+1} = 2 + \\sqrt{u_n + 4}, \\forall n \\in \\mathbb{N}^*.$$ \nIt is clear that $u_n \\ge 2$ for all positive integers $n$. On the other hand, we get $u_2 < u_1$. By induction, we can point out that $(u_n)$ is decreasing. Hence, $(u_n)$ has a finite limit... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 5 | |
0hxu | Problem:
How many values of $x$, $-19 < x < 98$, satisfy
$$
\cos^2 x + 2 \sin^2 x = 1?
$$ | [
"Solution:\n\nAnswer: 38. For any $x$, $\\sin^2 x + \\cos^2 x = 1$. Subtracting this from the given equation gives $\\sin^2 x = 0$, or $\\sin x = 0$. Thus $x$ must be a multiple of $\\pi$, so $-19 < k\\pi < 98$ for some integer $k$, or approximately $-6.1 < k < 31.2$. There are 38 values of $k$ that satisfy this, s... | United States | Harvard-MIT Mathematics Tournament | [
"Precalculus > Trigonometric functions"
] | null | final answer only | 38 | |
05ts | Problem:
Soit $n$ un entier. On dispose de $n$ couleurs, et chaque point d'un cercle est colorié de l'une de ces couleurs. Montrer qu'il existe deux droites parallèles qui intersectent le cercle en 4 points distincts de même couleur. | [
"Solution:\n\nL'énoncé nous demande de démontrer une existence, on peut essayer de montrer cette existence à l'aide du principe des tiroirs. Gardant cette information en tête, on essaye de traduire le résultat demandé.\n\nConsidérons deux droites parallèles coupant le cercle en 4 points distincts. Ces 4 points déli... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geo... | null | proof only | null | |
0esc | Water freezes at $0^\circ$ Celsius and boils at $100^\circ$ Celsius. On the SAMO scale, water freezes at $20^\circ$ and boils at $170^\circ$. The two scales are linearly related. What is the temperature on the SAMO scale if it is $80^\circ$ on the Celsius scale? | [
"A temperature of $80^\\circ$ C is $80\\%$ of the distance from freezing point to boiling point, so on the SAMO scale the reading is\n$$\n20 + \\frac{80}{100}(170 - 20) = 20 + 120 = 140^\\circ.\n$$"
] | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | final answer only | 140 | |
023k | Problem:
Considere a tabela de números a seguir. A primeira linha possui os números de 1 até $n$. A segunda possui os números de 1 até $n$ com cada um multiplicado por 2. As linhas seguem este padrão até a última linha que apresenta $n$ vezes cada número de 1 até $n$.
Vamos usá-la para calcul... | [
"Solution:\n\n(a) Os números na linha $t$ são os números de 1 até $n$ multiplicados por $t$. A soma deles é\n$$\n\\begin{aligned}\nt+2 t+3 t+\\ldots+n t & =t(1+2+3+\\ldots+n) \\\\\n& =t \\frac{n(n+1)}{2}\n\\end{aligned}\n$$\nA soma dos números na tabela pode ser calculada usando a soma de todas as linhas. Como $\\f... | Brazil | NÍVEL 3 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 25502500 | |
0dou | Let $a(1), a(2), \dots$ be the infinite sequence defined by:
$a(1) = a(2) = 1$ and
$$
a(n) = a(a(n-1)) + a(n - a(n-1)) \text{ for } n \ge 3.
$$
Prove that $a(2n) \le 2a(n)$ for all $n \ge 1$. | [
"We first prove that:\n$$\n(1) \\qquad a(n+1) - a(n) = 0 \\text{ or } 1\n$$\nfor all positive integers $n$. It is true if $n = 1, 2$, since $a(3) = 2$. Assume that $n \\ge 3$ and (1) is true for all $k \\le n$. In particular, $1 \\le a(k) \\le k$ for all $k \\le n$\n\n$$\na(n+1) - a(n) = a(a(n)) - a(a(n-1)) + a(n+1... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0c57 | If $a_1, a_2, \dots, a_{10}$ are non-negative rational numbers with sum $6$, denote $S_k = a_k + a_{k+1} + a_{k+2} + a_{k+3}$, $1 \le k \le 7$, $M$ the largest element of the set $S = \{S_1, S_2, \dots, S_7\}$ and $m$ the smallest element of $S$. Find the largest possible $m$ and the smallest possible $M$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 70th NMO | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof and answer | Largest possible m = 3; Smallest possible M = 2 | |
01vo | In the parallelogram $ABCD$ ($BC \parallel AD$) the side $AB$ is a half length of the side $BC$. The bisector of the angle $ABC$ intersects the side $AD$ at $K$ and the diagonal $AC$ at $L$. The bisector of the angle $ADC$ intersects the extension of the side $AB$ beyond $B$ at point $M$. The line $ML$ intersects the s... | [
"Answer: $2 : 5$.\nLet the segment $MD$ intersect the side $AC$ at $N$ and the diagonal $AC$ at $P$.\nFrom $BC \\parallel AD$ we get $\\angle CBK = \\angle BKA$.\nSince $BK$ is the bisector, $\\angle ABK = \\angle CBK$\nhence $\\angle ABK = \\angle BKA$ and the triangle $ABK$ is isosceles with $AB = AK$. From $2AB ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 2 : 5 | |
0470 | The sequence $\{a_n\}$ is defined as follows: $a_1 = 0$, and for integer $n \ge 2$,
$$
a_n = \frac{1}{n} + \frac{1}{\lceil \frac{n}{2} \rceil} \sum_{k=1}^{\lceil \frac{n}{2} \rceil} a_k,
$$
where $\lceil \frac{n}{2} \rceil$ denotes the smallest integer not less than $\frac{n}{2}$. Find the maximum term of the sequence ... | [
"*Proof*. From the given definition, we have $a_2 = \\frac{a_1}{2} + \\frac{1}{2} = \\frac{a_1+a_2}{2} + \\frac{1}{3} = \\frac{7}{12}$.\nWe now prove by induction that $a_n \\le \\frac{7}{12}$, with equality if and only if $n = 3$. The cases $n = 1, 2, 3$ have been verified. Assume the statement holds for all $1, 2... | China | 2024 CGMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 7/12 | |
04s0 | We are given two circles $k_1(S_1, r_1)$ and $k_2(S_2, r_2)$ in the plane, with $|S_1S_2| > r_1 + r_2$. Find the locus of points $X$ which do not lie on the line $S_1S_2$ and possess the following property: The segments $S_1X$ and $S_2X$ intersect successively the circles $k_1$ and $k_2$ in such points whose distances ... | [
"In the first part of our solution, we will assume that $X$ is any point with the required property. It is clear that $X$ lies in the exteriors of the circles $k_1$ and $k_2$ and that the points $S_1$, $S_2$ and $X$ are vertices of a triangle whose sides $S_1X$, $S_2X$ are intersected successively by the circles $k... | Czech Republic | 63rd Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | If the radii are unequal, the locus is the circle of Apollonius for the two centers with distance ratio equal to the ratio of the radii, excluding the two homothety centers on the line of centers. If the radii are equal, the locus is the perpendicular bisector of the segment joining the centers, excluding its midpoint. | |
0j5f | Problem:
Let $ABCD$ be a quadrilateral inscribed in the unit circle such that $\angle BAD$ is $30$ degrees. Let $m$ denote the minimum value of $CP + PQ + QC$, where $P$ and $Q$ may be any points lying along rays $AB$ and $AD$, respectively. Determine the maximum value of $m$. | [
"\nFor a fixed quadrilateral $ABCD$ as described, we first show that $m$, the minimum possible length of $CP + PQ + QC$, equals the length of $AC$. Reflect $B$, $C$, and $P$ across line $AD$ to points $E$, $F$, and $R$, respectively, and then reflect $D$ and $F$ across $AE$ to points $G$ an... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geo... | null | proof and answer | 2 | |
0ik0 | Problem:
On your answer sheet, clearly mark at least seven points, as long as
(i) No three are collinear.
(ii) No seven form a convex heptagon.
Please do not cross out any points; erase if you can do so neatly. If the graders deem that your paper is too messy, or if they determine that you violated one of those conditi... | [
"Solution:\nThis is the heptagon case of what is known as the \"Happy Ending\" or \"Erdős-Szekeres\" problem, which in general asks, For any integer $n \\geq 3$, what is the smallest $N(n)$, such that any $N(n)$ points in the plane in general position determine a convex $n$-gon? It is known that such an $N(n)$ alwa... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
05zo | Problem:
On définit la suite $\left(a_{n}\right)$ par $a_{0}=a_{1}=a_{2}=1$ et si $n \geqslant 3$ :
$$
a_{n}=\left\lfloor\frac{n}{a_{n-1} a_{n-2} a_{n-3}}\right\rfloor
$$
Déterminer $\mathrm{a}_{2022}$. | [
"Solution:\nEn calculant les premiers termes de la suite, on effectue la conjecture suivante, que l'on montre par récurrence sur $k$ :\n$$\n\\text{Si } k \\geqslant 1 \\text{, alors } a_{4 k}=1, a_{4 k+1}=1, a_{4 k+2}=\\left\\lfloor\\frac{4 k+2}{3}\\right\\rfloor \\text{ et } a_{4 k+3}=3.\n$$\nInitialisation : On a... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - Envoi 5 : Pot Pourri | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 674 | |
04zq | Inside a regular $2n$-gon, an arbitrary point is chosen and connected to every vertex of the $2n$-gon. The triangles obtained are colored alternately black and white so that triangles with a common side are of different color. Prove that the sum of the areas of all white triangles equals the sum of the areas of all bla... | [
"If $n = 2$, i.e., the $2n$-gon is a square, then the claim holds because the base sides of white triangles are the opposite sides of the square and the altitudes lie on the same line, so the total area of white triangles is a half of the area of the square.\n\nAssume in the following that $n > 2$. Consider the reg... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
03hp | Problem:
A function $f(x)$ is periodic if there is a positive number $p$ such that $f(x+p)=f(x)$ for all $x$. For example, $\sin x$ is periodic with period $2 \pi$. Is the function $\sin \left(x^{2}\right)$ periodic? Prove your assertion. | [
"Solution:\nSuppose $\\sin(x^2)$ is periodic with period $p > 0$. Then for all $x$, we have\n$$\n\\sin((x+p)^2) = \\sin(x^2).\n$$\nThis means\n$$\n(x+p)^2 = x^2 + 2xp + p^2,\n$$\nso\n$$\n\\sin(x^2 + 2xp + p^2) = \\sin(x^2).\n$$\nTherefore, for all $x$,\n$$\n\\sin(x^2 + 2xp + p^2) - \\sin(x^2) = 0.\n$$\nBut $\\sin(a... | Canada | Canadian Mathematical Olympiad | [
"Precalculus > Functions",
"Precalculus > Trigonometric functions"
] | null | proof and answer | It is not periodic. | |
0ap9 | Problem:
If $a$ and $b$ are integers such that $a \log_{250} 2 + b \log_{250} 5 = 3$, what is the value of $a + 2b$? | [
"Solution:\nApplying laws of logarithms to the given equation, we get\n$$\n\\log_{250}\\left(2^{a} 5^{b}\\right) = 3 \\text{ or } \\quad 2^{a} 5^{b} = 250^{3} = 2^{3} 5^{9}\n$$\nSince $a$ and $b$ are integers and $\\gcd(2,5) = 1$, we get $a = 3$ and $b = 9$, so that $a + 2b = 21$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 21 | |
0ck6 | Let $ABC$ be an acute-angled triangle inscribed in a circle with center $O$ and radius $R$, and let $H$ be the orthocenter of triangle $ABC$. Let $A_1$ be a point on the side $BC$ such that $HA_1 + A_1O = R$. Similarly, define points $B_1$ on $CA$ and $C_1$ on $AB$. If $\overrightarrow{AA_1} + \overrightarrow{BB_1} + \... | [
"Let $H_A$ be the reflection of $H$ across line $BC$. Then $H_A$ lies on the circumcircle of triangle $ABC$. Since $A_1H = A_1H_A$, we get $A_1O + A_1H_A = R = OH_A$. Hence, $A_1$ lies on segment $OH_A \\cap BC$, and is uniquely defined by the given condition. We now prove that $\\angle BHA_1 = \\angle B$ and $\\an... | Romania | 75th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > A... | English | proof only | null | |
0ho6 | Problem:
Let $C$ be a circle with center at the origin $O$ of a system of rectangular coordinates, and let $M O N$ be the quarter circle of $C$ in the first quadrant. Let $P Q$ be an arc of $C$ of fixed length that lies in the arc $M N$. Let $K$ and $L$ be the feet of the perpendiculars from $P$ and $Q$ to $O N$, and ... | [
"Solution:\n\nDraw $O P$, $O U$, and $O Q$, and note that $A+B$ is the area of rectangle $P V W U$, plus the area of rectangle $U K L Q$, plus twice the area of the triangle $P U Q$. But the area of triangle $P O U$ is half the area of rectangle $P V W U$, and the area of triangle $U O Q$ is half the area of rectan... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jam | Problem:
Find the smallest positive integer $n$ such that $\underbrace{2^{2^{2 \cdots}}}_{n} > 3^{3^{3^{3}}}$. (The notation $\underbrace{2^{2^{2 \cdots}}}_{n}$ is used to denote a power tower with $n$ $2$'s. For example, $\underbrace{2^{2^{2^{2}}}}_{n=4}$ would equal $2^{2^{2^{2}}}$.) | [
"Solution:\nAnswer: 6\n\nClearly, $n \\geq 5$. When we take $n=5$, we have\n$$\n2^{2^{2^{2^{2}}}} = 2^{2^{16}} < 3^{3^{27}} = 3^{3^{3^{3}}}.\n$$\nOn the other hand, when $n=6$, we have\n$$\n2^{2^{2^{2^{2^{2}}}}} = 2^{2^{65536}} = 4^{2^{65535}} > 4^{4^{27}} > 3^{3^{2^{27}}} = 3^{3^{3^{3}}}.\n$$\nOur answer is thus $... | United States | HMMT November 2012 | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 6 | |
0dgq | Point $P$ lies inside parallelogram $ABCD$ and satisfies $PC = BC$. Prove that the line joining midpoints of segments $AP$ and $CD$ is perpendicular to $BP$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0iv9 | Problem:
In a connected graph, it is possible to reach any vertex from any other vertex by following the edges. A tree is a connected graph with $n$ vertices and $n-1$ edges for some positive integer $n$. Suppose $n \geq 2$. What is the chromatic number of a tree having $n$ vertices? Prove your answer. | [
"Solution:\n\nThe chromatic number of any tree is $2$. We show this by induction on the size of the tree.\n\nA tree with $2$ nodes can clearly be $2$-colored.\n\nNow, suppose a tree of size $n-1$ can be colored in $2$ colors. Given a tree of size $n$, choose any leaf (node with only one edge coming out of it), say ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2 | |
0hga | Is it possible to write five integers on a board so that for any two numbers, there exists a pair of numbers among the remaining three whose sum is equal to the sum of the original two numbers? | [
"It's enough to choose the following numbers: $-2$, $-1$, $0$, $1$, $2$. Then we can write down the sets of integers, but we can also apply the following reasoning: for any two numbers, say, $a$, $b$, selected by Petrik, from one side exists pair of numbers $(-a, -b)$, whose sum is the opposite to the initial, and,... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | Yes; for example: −2, −1, 0, 1, 2. | |
0j5b | Problem:
For positive integers $n$, let $L(n)$ be the largest factor of $n$ other than $n$ itself. Determine the number of ordered pairs of composite positive integers $(m, n)$ for which $L(m) L(n) = 80$. | [
"Solution:\n\nLet $x$ be an integer, and let $p_{x}$ be the smallest prime factor of $x$. Then, if $L(a) = x$, we note that we must have $a = p x$ for some prime $p \\leq p_{x}$. (Otherwise, if $p > p_{x}$, then $\\frac{p x}{p_{x}} > x$. If $p$ is composite, then $k x > x$ for some factor $k$ of $x$.)\n\nSo we have... | United States | Harvard-MIT November Tournament | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 12 | |
09yz | We draw a rectangle in a grid. The four midpoints of the four sides of the rectangle turn out to be the vertices $(-3, 0)$, $(2, 0)$, $(5, 4)$, and $(0, 4)$.
What is the area of the rectangle? | [] | Netherlands | Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 40 | |
001b | Inicialmente en el pizarrón están escritos en una línea y en algún orden todos los números enteros del $1$ al $2002$ inclusive, sin repeticiones.
En cada paso se borran el primero y el segundo número de la línea y se escribe al principio de la línea el valor absoluto de la resta de los dos números que se acaba de borra... | [] | Argentina | XIX Olimpíada Matemática Argentina | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | español | proof and answer | all odd integers between 1 and 2001 inclusive | |
01wy | The diagonals of a quadrilateral $ABCD$ intersect at the point $N$. It is known that the incircles of the triangles $ABN$ and $CBN$ are tangent to each other. Moreover, the incircles of the triangles $ADN$ and $CDN$ are tangent to each other as well.
Find all possible values of $AN : NC$. | [
"$AN : NC = 1$."
] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof and answer | AN : NC = 1 | |
0aig | Find all $n \in \mathbb{N}$ divisible by $11$, such that all numbers that can be obtained from $n$ by an arbitrary rearrangement of its digits are again divisible by $11$.
Најди ги сите $n \in \mathbb{N}$ деливи со $11$, такви што сите броеви кои се добиваат со произволна прераспределба на цифрите на бројот $n$ повтор... | [
"From the condition $11|n$, the number $n$ must have at least two digits. Let $n = \\overline{a_k a_{k-1} \\dots a_0}$ where $a_i$, $0 \\le i \\le k$ are digits and $a_k \\ne 0$. From the former discussion we have $k \\ge 1$.\n\nWe will show that all digits in the number $n$ are equal. Namely, from the condition of... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | All numbers whose digits are all the same and whose number of digits is even; equivalently, numbers of the form a times a string of ones of even length with a from one to nine. | |
0bd8 | Evaluate $\lim_{n \to \infty} \int_0^1 e^{x^n} dx$. | [
"Since $e^{x^n} \\ge 1$ for all $x \\in [0, 1]$ we get $\\int_0^1 e^{x^n} dx \\ge 1$. Notice that $e^t \\le 1 + 3t$ for all $t \\in [0, 1]$ to obtain\n$$\n\\int_{0}^{1} e^{x^{n}} dx \\le \\int_{0}^{1} (1 + 3x^{n}) dx = 1 + \\frac{3}{n + 1}\n$$\nConsequently, the limit is equal to 1."
] | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Precalculus > Limits"
] | null | proof and answer | 1 | |
05rq | Problem:
Un coloriage de $\mathbb{Q}$ consiste à colorier tout nombre rationnel soit en rouge, soit en bleu. On dit qu'un coloriage de $\mathbb{Q}$ est harmonieux si, pour tous les rationnels $x$ et $y$ d'une même couleur, le rationnel $x+y$ est encore de la même couleur.
Trouver tous les coloriages de $\mathbb{Q}$ q... | [
"Solution:\n\nOn va montrer que les coloriages recherchés sont les suivants :\n\n$\\triangleright$ on colorie tous les rationnels avec la même couleur;\n\n$\\triangleright$ on colorie tous les rationnels positifs ou nuls avec une même couleur, et tous les rationnels strictement négatifs avec l'autre couleur;\n\n$\\... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | Exactly these harmonious colorings: (1) all rationals have the same color; (2) nonnegative rationals have one color and negative rationals have the other; (3) strictly positive rationals have one color and nonpositive rationals have the other. | |
08j5 | Problem:
Prove that amongst any 29 natural numbers there are 15 such that sum of them is divisible by 15.
Problem:
Demonstrați că într-o mulțime de 29 numere naturale există 15 a căror sumă este divizibilă cu 15. | [
"Solution:\nAmongst any 5 natural numbers there are 3 such that sum of them is divisible by 3. Amongst any 29 natural numbers we can choose 9 groups with 3 numbers such that sum of numbers in every group is divisible by 3. In that way we get 9 natural numbers such that all of them are divisible by 3. It is easy to ... | JBMO | 7th JBMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0aep | За множеството $S = \{-2,-1,0,1,2\}$ од правоаголен координатен систем се избрани 17 точки од множеството $S \times S$. Докажи дека постојат три точки $A,B,C$ од избраните, такви што $B$ е средина на отсечката $AC$. | [
"а) Координатниот почеток е во избраните 17 точки. Од преостанатите 24 точки ќе формираме 12 пара точки. Точките од еден пар се централно симетрични во однос на координатниот почеток. Бидејќи бројот на парови е 12 а бројот на избрани точки е поголем од 13, во еден пар двете точки ќе бидат од избраните точки. Коорди... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | Macedonian, English | proof only | null | |
0go9 | In an exam every question is solved by exactly four students, every pair of questions is solved by exactly one student, and none of the students solved all of the questions. Find the maximum possible number of questions in this exam. | [
"Suppose $S_1$ solved $Q_1, \\dots, Q_k$, but not $Q_{k+1}$ where $k > 4$. Since $Q_{k+1}$ is solved by exactly 4 students, and $Q_i$ and $Q_{k+1}$ are solved by exactly one student for each $1 \\le i \\le k$, there must be another student who solved two of the questions $Q_1, \\dots, Q_k$ besides $S_1$, a contradi... | Turkey | 15th Junior Turkish Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 13 | |
0fho | Problem:
Escrito el triángulo aritmético
donde cada número es igual a la suma de los dos que tiene encima (como es evidente, cada fila consta de un número menos que la anterior, y por lo tanto la última fila estará formada por un único número), razonar que el último número es múltiplo de $1... | [
"Solution:\n\nPrimera solución\nVamos a sumar los elementos de cada fila y llamaremos $S_{i}$ a la suma de los elementos de la fila $i$.\n$$\n\\begin{aligned}\nS_{1} & = \\frac{0+1993}{2} \\cdot 1994 = 997 \\cdot 1993 \\\\\nS_{2} & = 2 S_{1} - (0+1993) = 1993 \\\\\nS_{3} & = 2 S_{2} - (0+1993) = 1993 \\\\\n& \\ldot... | Spain | OME 29 | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
030b | Problem:
Determine o valor da soma
$$
S=\operatorname{sen}^{2} 1^{\circ}+\operatorname{sen}^{2} 2^{\circ}+\operatorname{sen}^{2} 3^{\circ}+\ldots+\operatorname{sen}^{2} 89^{\circ}
$$ | [
"Solution:\n\nComo $\\sen x = \\cos (90^{\\circ} - x)$ e $\\sen^{2} x + \\cos^{2} x = 1$, segue que\n$$\n\\sen^{2} 1^{\\circ} + \\sen^{2} 89^{\\circ} = \\sen^{2} 2^{\\circ} + \\sen^{2} 88^{\\circ} = \\ldots = \\sen^{2} 44^{\\circ} + \\sen^{2} 46^{\\circ} = 1\n$$\nPortanto, pareando os termos da soma em pares do tip... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 89/2 | |
00fo | Show that for any positive integers $a$ and $b$, $(36a + b)(a + 36b)$ cannot be a power of $2$. | [
"Suppose that $(36a + b)(a + 36b)$ is a power of $2$ for some positive integers $a$ and $b$. Write $36a + b = 2^{m} = r$ and $a + 36b = 2^{n} = s$. Then\n$$\n36r - s = 35 \\times 37a, \\quad \\text{and} \\quad 36s - r = 35 \\times 37b.\n$$\n\nHence\n$$\n\\frac{1}{36} < \\frac{r}{s} = 2^{m-n} < 36, \\quad \\text{or}... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
09uy | On a white strip that is $100$ mm long and $10$ mm wide, ten black squares are drawn that, from left to right, have sides of length $1$, $2$, $\ldots$, $10$ mm. The centre of each black square is in the middle of the strip and $5$, $15$, $\ldots$, $95$ mm from the start (left edge) of the strip.
A transparent square i... | [] | Netherlands | Junior Mathematical Olympiad, September 2019 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 61.125 mm | |
00xs | Problem:
Let's consider three pairwise non-parallel straight lines in the plane. Three points are moving along these lines with different non-zero velocities, one on each line (we consider the movement as having taken place for infinite time and continuing infinitely in the future). Is it possible to determine these st... | [
"\nFigure 5\n\nSolution:\nYes, it is. First, place the three points at the vertices of an equilateral triangle at the \"zero\" moment and let them move with equal velocities along the straight lines determined by the sides of the triangle as shown in Figure 5. Then, at any moment in the pas... | Baltic Way | Baltic Way 1993 | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | Yes | |
0f80 | Problem:
A book contains 30 stories. Each story has a different number of pages under 31. The first story starts on page 1 and each story starts on a new page. What is the largest possible number of stories that can begin on odd page numbers? | [
"Solution:\nAnswer: 23.\n\nCall stories with an odd number of pages odd stories and stories with an even number of pages even stories. There are 15 odd stories and 15 even stories. The odd stories change the parity of the starting page (in the sense that the following story starts on a page of opposite parity), whe... | Soviet Union | 22nd ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 23 | |
03kj | Problem:
Suppose that $u$ is a real parameter with $0 < u < 1$. Define
$$
f(x) = \begin{cases}
0 & \text{ if } 0 \leq x \leq u \\
1 - (\sqrt{u x} + \sqrt{(1-u)(1-x)})^{2} & \text{ if } u < x \leq 1
\end{cases}
$$
and define the sequence $\{u_n\}$ recursively as follows:
$$
u_1 = f(1), \text{ and } u_n = f(u_{n-1}) \te... | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0hrp | Problem:
Let $X$, $Y$, and $Z$ be points on one side of a line $AB$ such that
$$
\triangle XAB \sim \triangle BYA \sim \triangle ABZ
$$
Prove that $\triangle XYZ$ is similar to all these triangles. | [
"Solution:\nWe will prove that $\\triangle XYZ \\sim \\triangle XAB$. Because $\\angle AXB = \\angle YXZ$ if and only if $\\angle AXY = \\angle BXZ$ and\n$$\n\\frac{ZX}{BX} = \\frac{YX}{AX} \\quad \\text{if and only if} \\quad \\frac{ZX}{YX} = \\frac{BX}{AX}\n$$\nthis is the same as proving $\\triangle XAY \\sim \\... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity"
] | null | proof only | null | |
0a1a | We have a triangle $\triangle ABC$ and two points $D$ and $E$ on the line segment $BC$, such that $\angle BAD = \angle DAE = \angle EAC$. Furthermore, $|AB| = 6$, $|BD| = 4$, $|DE| = 2$, and $|EA| = 3$. See the picture on the right: be aware, it is not drawn to scale!
What is the length of the... | [
"E) 3"
] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | MCQ | E | |
06ca | $ABC$ is a triangle; $ABX$, $BCY$ and $CAZ$ are similar isosceles triangles outside $ABC$ such that $XA = XB$, $YB = YC$ and $ZC = ZA$.
Prove that $AY$, $BZ$ and $CX$ are concurrent. | [
"This is a special case of Jacobi's theorem. Alternatively, by Ceva's theorem, it suffices to prove\n$$\n\\frac{\\sin \\angle ACX}{\\sin \\angle BCX} \\times \\frac{\\sin \\angle BAY}{\\sin \\angle CAY} \\times \\frac{\\sin \\angle CBZ}{\\sin \\angle ABZ} = 1.\n$$\nApplying the sine law to $\\triangle ACX$ and $\\t... | Hong Kong | HKG TST | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0dlt | Given a square board of $8 \times 8$ and one colors exactly $7$ squares of the board. Prove that for every way of coloring, there always exists a sub-square size $2 \times 2$ of the board on which exactly one square is colored. | [
"We number rows numbered from $1$ to $8$ (from top to bottom), and columns numbered from $1$ to $8$ (from left to right). Suppose by contradiction that each $2 \\times 2$ square board does not contain exactly one colored square.\n\nObviously there must also be an empty column, let's say it is the $i$ column with $1... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
03bs | Given is a $\triangle ABC$ with incircle $k$ touching the sides $BC$ and $CA$ in points $P$ and $Q$ respectively. Let denote with $J$ the center of the excircle at side $AB$ in $\angle ABC$ and with $T$ – the second intersection point of the circumcircles of $\angle JBP$ and $\angle JAQ$. Prove that the circumcircle of... | [
"Let $R$ be the tangent point of $k$ to $AB$. We will use the standard notations about the angles of $\\triangle ABC$. We have\n$$\n\\angle PTJ = 180^{\\circ} - \\angle PBJ = 90^{\\circ} - \\frac{\\beta}{2} = \\angle PRB\n$$\nand\n$$\n\\angle QTJ = 180^{\\circ} - \\angle QAJ = 90^{\\circ} - \\frac{\\alpha}{2} = \\a... | Bulgaria | 55th IMO Team Selection Test | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ghk | 給定一個遞增函數 $f: N \to R_{>0}$，其中 $N$ 代表全體正整數所成的集合，$R_{>0}$ 為所有正實數所成的集合。我們給出下列定義：
- 若正整數 $m, n$ 滿足 $f(mn) \neq f(m)f(n)$，我們稱數對 $(m, n)$ 為『不服從的』。
- 正整數 $m$ 被稱為『極端不服從的』。若且唯若對於任意的非負整數 $N$，總存在無限多個正整數 $n$ 使得所有 $(m, n), (m, n + 1), \dots, (m, n + N)$ 都是『不服從的』。
證明：若存在一個『不服從的』數對，則必存在一個『極端不服從的』正整數。
Let $f: \mathbb{N} \to \mathbb{R... | [
"We will show the contrapositive of the statement. In particular, we will show that if there are no ultra-disobedient positive integers, then there exists a nonnegative integer $\\alpha$ with $f(n) = n^\\alpha$ for any $n \\in \\mathbb{N}$.\n\nWe first show that $f(1) = 1$. Since $1$ is not ultra-disobedient, there... | Taiwan | 2023 數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | Chinese (Traditional) | proof only | null | |
0eqe | The circle centre $O$ has radius $4$ cm. If the area of sector $OPQ$ is $\frac{1}{2}\pi$, and if the probability that a point chosen randomly in the circle lies in the shaded sector is $\frac{1}{n}$, then the value of $n$ is
 | [
"The area of the whole circle is $\\pi \\cdot 4^2 = 16\\pi$, but the area of the sector is $\\frac{1}{2}\\pi$, so the probability is $\\frac{\\frac{1}{2}\\pi}{16\\pi} = \\frac{1}{32}$"
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Circles"
] | English | final answer only | 32 | |
0c6d | At the math class 25 students of the fifth grade receive a paper stick with an integer on it. Each student divides his number to $24$ and communicates the remainder to the math teacher. The sum of all remainders is $288$. Dany, one of the students, remarks that all the remainders of the students are different two by tw... | [] | Romania | 70th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | a) 300; b) 7776 | |
0j8s | Problem:
Find the smallest $k$ such that for any arrangement of $3000$ checkers in a $2011 \times 2011$ checkerboard, with at most one checker in each square, there exist $k$ rows and $k$ columns for which every checker is contained in at least one of these rows or columns. | [
"Solution:\nAnswer: $1006$\nIf there is a chip in every square along a main diagonal, then we need at least $1006$ rows and columns to contain all these chips. We are left to show that $1006$ is sufficient.\n\nTake the $1006$ rows with greatest number of chips. Assume without loss of generality they are the first $... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1006 | |
0g7x | 有若干個正整數排成一列。某人每次選擇兩個相鄰的數字 $x$ 與 $y$, 其中 $x > y$, 並且 $x$ 在 $y$ 的左邊; 接著他把這個數對 $(x, y)$ 用 $(y + 1, x)$ 或 $(x - 1, x)$ 取代。然後他重複作這樣的選擇與取代的操作。試證他只能作有限次這樣的操作。 | [
"解一 注意到題設的操作不會改變原數列的最大值 $M$。設某一次操作完成後所得到的數列為 $a_1, a_2, \\dots, a_n$。考慮下列的和\n$$\nS = a_1 + 2a_2 + \\dots + n a_n.\n$$\n我們宣稱：在每次操作後，$S$ 都會增加某正整數的量。設某操作將數對 $(a_i, a_{i+1})$ 以 $(c, a_i)$ 取代，其中 $a_i > a_{i+1}$，且 $c$ 為 $a_{i+1} + 1$ 或 $a_i - 1$ 其中之一。於是 $S$ 的新值與舊值的差為\n$$\nd = (i c + (i + 1)a_i) - (i a_i + (i + 1)a_{i+1})... | Taiwan | 二〇一三數學奧林匹亞競賽第二階段選訓營 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0ixq | Problem:
Let $ABC$ be a triangle in the coordinate plane with vertices on lattice points and with $AB=1$. Suppose the perimeter of $ABC$ is less than $17$. Find the largest possible value of $1/r$, where $r$ is the inradius of $ABC$. | [
"Solution:\n\nLet $a$ denote the area of the triangle, $r$ the inradius, and $p$ the perimeter. Then $a = r p / 2$, so $r = 2a / p > 2a / 17$. Notice that $a = h / 2$ where $h$ is the height of the triangle from $C$ to $AB$, and $h$ is an integer since the vertices are lattice points. Thus we first guess that the i... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geome... | null | proof and answer | 1 + 5√2 + √65 | |
0b54 | Problem:
Şirul de numere reale $a_{0}, a_{1}, \ldots, a_{n}, \ldots$ satisface relaţia
$$
a_{m+n} + a_{m-n} - m + n - 1 = \frac{1}{2}\left(a_{2m} + a_{2n}\right)
$$
pentru orice numere naturale $m$ şi $n$, $m \geq n$.
Dacă $a_{1} = 3$, aflaţi $a_{2004}$. | [] | Romania | BMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 4018021 |
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