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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0i8v | Problem:
Find the smallest $n$ such that $n!$ ends in 290 zeroes. | [
"Solution:\nEach 0 represents a factor of $10 = 2 \\cdot 5$. Thus, we wish to find the smallest factorial that contains at least 290 2's and 290 5's in its prime factorization. Let this number be $n!$, so the factorization of $n!$ contains 2 to the power $p$ and 5 to the power $q$, where\n$$\np = \\left\\lfloor \\f... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1170 | |
0hzo | Problem:
In a cube with side length $6$, what is the volume of the tetrahedron formed by any vertex and the three vertices connected to that vertex by edges of the cube? | [
"Solution:\n\nWe have a tetrahedron whose base is half a face of the cube and whose height is the side length of the cube, so its volume is\n$$\n\\frac{1}{3} \\cdot \\left(\\frac{1}{2} \\cdot 6^{2}\\right) \\cdot 6 = \\mathbf{36}\n$$"
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Volume",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 36 | |
0791 | Problem:
In triangle $ABC$ with $CA = CB$, point $E$ lies on the circumcircle of $ABC$ such that $\angle ECB = 90^\circ$. The line through $E$ parallel to $CB$ intersects $CA$ in $F$ and $AB$ in $G$. Prove that the centre of the circumcircle of triangle $EGB$ lies on the circumcircle of triangle $ECF$. | [
"Solution:\n\n\n\nWe have $FG = FA$ since $FG$ is parallel to $BC$. But also $\\triangle GAE$ is a right angle triangle. Thus, if $F'$ is the midpoint of $GE$, then $\\angle GAF = \\angle FGA = \\angle F'GA = \\angle GAF'$, which implies $F \\equiv F'$. Thus, $F$ is the midpoint of $GE$.\n\... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00x9 | Problem:
Show that in a non-obtuse triangle the perimeter of the triangle is always greater than two times the diameter of the circumcircle. | [
"Solution:\n\nLet $K$, $L$, $M$ be the midpoints of the sides $AB$, $BC$, $AC$ of a non-obtuse triangle $ABC$ (see Figure 2). Note that the centre $O$ of the circumcircle is inside the triangle $KLM$ (or at one of its vertices if $ABC$ is a right-angled triangle). Therefore $|AK| + |KL| + |LC| > |AO| + |OC|$ and he... | Baltic Way | Baltic Way 1992 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ffj | Problem:
Sea $A$ uno de los dos puntos de intersección distintos de dos círculos distintos $C_{1}$, $C_{2}$ de centros $O_{1}$, $O_{2}$, respectivamente.
Una de las tangentes comunes a los dos círculos toca a $C_{1}$ en $P_{1}$ y a $C_{2}$ en $P_{2}$, mientras que la otra toca a $C_{1}$ en $Q_{1}$ y a $C_{2}$ en $Q_{2... | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angl... | null | proof only | null | |
0csf | Sergey chose two different positive integers $a$ and $b$. Then he calculated all six pairwise products of the four numbers $a$, $a+2$, $b$, and $b+2$. Find the maximal number of perfect squares among the six calculated numbers. (S. Berlov) | [
"**Answer.** Two.\n\nNote that no two squares of natural numbers differ by $1$, since $x^2 - y^2 = (x - y)(x + y)$, where the second factor is greater than one. Therefore, the numbers $a(a+2) = (a+1)^2 - 1$ and $b(b+2) = (b+1)^2 - 1$ are not squares. Moreover, the numbers $ab$ and $a(b+2)$ cannot both be squares, o... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | 2 | |
0jit | Problem:
Let $ABC$ be a triangle with $AB = 13$, $BC = 14$, and $CA = 15$. Let $\Gamma$ be the circumcircle of $ABC$, let $O$ be its circumcenter, and let $M$ be the midpoint of minor arc $\widehat{BC}$. Circle $\omega_1$ is internally tangent to $\Gamma$ at $A$, and circle $\omega_2$, centered at $M$, is externally t... | [
"Solution:\n\nAnswer: $\\quad \\dfrac{1235}{108}$\n\nLet $N$ be the midpoint of $BC$. Notice that $BS - CS = \\dfrac{4}{15}$ means that $NS = \\dfrac{2}{15}$. Let lines $MN$ and $AS$ meet at $P$, and let $D$ be the foot of the altitude from $A$ to $BC$. Then $BD = 5$ and $AD = 12$, so $DN = 2$ and $DS = \\dfrac{32}... | United States | HMMT 2014 | [
"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 > Distance chasing",... | null | proof and answer | 1235/108 | |
03iq | Problem:
Let $p$ be a permutation of the set $S_{n} = \{1, 2, \ldots, n\}$. An element $j \in S_{n}$ is called a fixed point of $p$ if $p(j) = j$. Let $f_{n}$ be the number of permutations having no fixed points, and $g_{n}$ be the number with exactly one fixed point. Show that $\left|f_{n} - g_{n}\right| = 1$. | [] | Canada | Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0k67 | Problem:
Let
$$
\begin{aligned}
& A=(1+2 \sqrt{2}+3 \sqrt{3}+6 \sqrt{6})(2+6 \sqrt{2}+\sqrt{3}+3 \sqrt{6})(3+\sqrt{2}+6 \sqrt{3}+2 \sqrt{6})(6+3 \sqrt{2}+2 \sqrt{3}+\sqrt{6}) \\
& B=(1+3 \sqrt{2}+2 \sqrt{3}+6 \sqrt{6})(2+\sqrt{2}+6 \sqrt{3}+3 \sqrt{6})(3+6 \sqrt{2}+\sqrt{3}+2 \sqrt{6})(6+2 \sqrt{2}+3 \sqrt{3}+\sqrt{6})... | [
"Solution:\nNote that\n$$\n\\begin{aligned}\n& A=((1+2 \\sqrt{2})(1+3 \\sqrt{3}))((2+\\sqrt{3})(1+3 \\sqrt{2}))((3+\\sqrt{2})(1+2 \\sqrt{3}))((3+\\sqrt{3})(2+\\sqrt{2})) \\\\\n& B=((1+3 \\sqrt{2})(1+2 \\sqrt{3}))((2+\\sqrt{2})(1+3 \\sqrt{3}))((3+\\sqrt{3})(1+2 \\sqrt{2}))((2+\\sqrt{3})(3+\\sqrt{2}))\n\\end{aligned}... | United States | HMMT February 2019 | [
"Algebra > Intermediate Algebra > Other"
] | null | final answer only | 1 | |
0aej | Определи ја 2008-та цифра по децималната запирка во децималниот запис на бројот $\frac{1}{41}$. | [
"Децималниот запис на бројот $\\frac{1}{41}$ е $\\frac{1}{41} = 0,(02439)$. Значи групата од пет цифри 02439 по децималната запирка периодично се повторува бесконечно многу пати. Бројот 2008 можеме да го запишеме во облик $2008 = 5 \\cdot 401 + 3$. Според тоа, 2008-та цифра по децималната запирка е третата цифра од... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | Macedonian, English | proof and answer | 4 | |
0jmx | Problem:
Let $S$ be the set of positive integers $n$ such that the inequality
$$
\phi(n) \cdot \tau(n) \geq \sqrt{\frac{n^{3}}{3}}
$$
holds, where $\phi(n)$ is the number of positive integers $k \leq n$ that are relatively prime to $n$, and $\tau(n)$ is the number of positive divisors of $n$. Prove that $S$ is finite. | [
"Solution:\nLet $S$ be the set of all positive integers $n$ such that\n$$\n\\phi(n) \\cdot \\tau(n) \\geq \\sqrt{\\frac{n^{3}}{3}}\n$$\nDefine a function $\\Phi$ on all positive integers $n$ by\n$$\n\\Phi(n)=\\frac{\\phi(n)^{2} \\cdot \\tau(n)^{2}}{n^{3}}\n$$\nAn important observation is that $\\Phi$ has the proper... | United States | HMMT February 2015 | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
024u | Problem:
Em uma promoção, Joana comprou blusas por $\mathrm{R}\$ 15,00$ cada uma e calças por $\mathrm{R}\$ 17,00$ cada uma, gastando, ao todo, $\mathrm{R}\$ 143,00$. Quantas blusas e calças Joana comprou? | [
"Solution:\n\nSejam $b$ e $c$ o número de blusas e calças compradas, respectivamente. Logo, temos $15 b + 17 c = 143$, sendo $b$ e $c$ números inteiros positivos. Observe que $b < 10$ e $c < 9$, pois tanto $15 \\times 10$ quanto $17 \\times 9$ são maiores do que $143$. A partir deste ponto, apresentamos duas possib... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 5 blouses and 4 pants | |
03fo | Let $ABC$ be an acute triangle with midpoint $M$ of $AB$. The point $D$ lies on the segment $MB$ and $I_1, I_2$ denote the incenters of $\triangle ADC$ and $\triangle BDC$. Given that $\angle I_1MI_2 = 90^\circ$, show that $CA = CB$. | [] | Bulgaria | 4 Bulgarian National Olympiad - Regional Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0bp1 | Problem:
Fie $ABC$ un triunghi ascuţitunghic cu $AB \neq AC$, $M$ mijlocul laturii $[BC]$, $H$ ortocentrul triunghiului $ABC$, $O_{1}$ mijlocul lui $[AH]$, iar $O_{2}$ centrul cercului circumscris triunghiului $BCH$. Demonstraţi că $O_{1}AMO_{2}$ este paralelogram. | [
"Solution:\n\nCercul circumscris triunghiului este simetricul cercului circumscris triunghiului $ABC$ faţă de $BC$, prin urmare $O_{2}$ este simetricul lui $O$ faţă de $BC$. Se ştie că $AO_{1} = MO = MO_{2}$ şi, cum $AO_{1} \\parallel MO_{2}$, rezultă concluzia."
] | Romania | Primul test de selecţie pentru OBMJ | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0jwf | Problem:
Lisa considers the number
$$
x=\frac{1}{1^{1}}+\frac{1}{2^{2}}+\cdots+\frac{1}{100^{100}} .
$$
Lisa wants to know what $x$ is when rounded to the nearest integer. Help her determine its value. | [
"Solution:\n\nThe answer is $1$. Indeed, note that\n$$\nx \\leq 1+\\frac{1}{2^{2}}+\\frac{1}{2^{3}}+\\cdots+\\frac{1}{2^{100}} .\n$$\nBy the formula for the sum of a geometric series, we see that\n$$\nx \\leq 1+\\frac{1}{2}-\\frac{1}{2^{101}} .\n$$\nThus $x<3 / 2$, and the closest integer to $x$ is $1$."
] | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 1 | |
05ie | Problem:
Les sept dixièmes de la surface de la Terre sont couverts par l'océan. Montrer qu'il existe un diamètre de la Terre dont les deux extrémités baignent dans l'océan. | [
"Solution:\n\nSoit $S$ la surface couverte par l'océan et $S'$ son symétrique par rapport au centre de la Terre. Si les surfaces $S$ et $S'$ étaient disjointes, elles occuperaient ensemble les $14/10$ de la surface de la Terre, ce qui est impossible, car $14/10 > 1$. Donc $S$ et $S'$ ont au moins un point d'interse... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0f5i | Problem:
The positive integer $k$ has $n$ digits. It is rounded to the nearest multiple of $10$, then to the nearest multiple of $100$ and so on ($n-1$ roundings in all). Numbers midway between are rounded up. For example, $1474$ is rounded to $1470$, then to $1500$, then to $2000$. Show that the final number is less ... | [] | Soviet Union | 17th ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
07aj | Nonnegative real numbers $p_1, p_2, \dots, p_n$ and $q_1, q_2, \dots, q_n$ are given such that
$$
p_1 + p_2 + \dots + p_n = q_1 + q_2 + \dots + q_n.
$$
Among all matrices with nonnegative real entries for which sum of entries of the $i$th row is $p_i$ and sum of entries of the $j$th column is $q_j$, find the maximum va... | [
"We use the notation $A$ for matrices and $A_{ij}$ for the entry of row $i$ and column $j$.\n\nFirst, observe that $A_{ii} \\le p_i$, since the sum of the $i$th row is $p_i$ and the entries are nonnegative. Similarly, $A_{ii} \\le q_i$. Thus, $A_{ii} \\le \\min(p_i, q_i)$ and it follows that $\\text{trace}(A) \\le ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | min(p_1, q_1) + min(p_2, q_2) + ... + min(p_n, q_n) | |
0a1y | Problem:
Gegeven een natuurlijk getal $n$ definiëren we $\tau(n)$ als het aantal natuurlijke getallen dat $n$ deelt, en definiëren we $\sigma(n)$ als de som van deze delers. Vind alle natuurlijke getallen $n$ waarvoor geldt dat
$$
\sigma(n)=\tau(n) \cdot\lceil\sqrt{n}\rceil .
$$
Voor een reëel getal $x$ bedoelen we m... | [
"Solution:\nOplossing I. Antwoord: dit geldt alleen voor 1, 3, 5, 6 .\nVoor deze vier gevallen geldt inderdaad respectievelijk dat $1=1 \\cdot 1, 4=2 \\cdot\\lceil\\sqrt{3}\\rceil, 6=2 \\cdot\\lceil\\sqrt{5}\\rceil$ en $12=4 \\cdot\\lceil\\sqrt{6}\\rceil$. Vanaf nu nemen we aan dat $n \\neq 1$.\n\nAls $n$ een kwadr... | Netherlands | IMO-selectietoets | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 1, 3, 5, 6 | |
0ca6 | Problem:
Fie $a \in \mathbb{N}$, $a > 2$. Să se arate că
a) Există un număr $n \in \mathbb{N}^* \setminus \{1\}$, care nu este prim, astfel încât $a^n \equiv 1 \pmod{n}$.
b) Dacă $p$ este cel mai mic număr din $\mathbb{N}^* \setminus \{1\}$ pentru care $a^p \equiv 1 \pmod{p}$, atunci $p$ este prim.
c) Nu există num... | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof only | null | |
0i28 | Problem:
There are some red and blue marbles in a box. We are told that there are twelve more red marbles than blue marbles, and we experimentally determine that when we pick a marble randomly we get a blue marble one quarter of the time. How many marbles are there in the box? | [
"Solution:\n\nCall the number of blue marbles $x$, so the number of red marbles is $x + 12$ and the total number of marbles is $2x + 12$. The probability of picking a blue marble is $\\frac{x}{2x + 12} = \\frac{1}{4}$. Solving, $x = 6$, so $2x + 12 = 24$."
] | United States | Harvard-MIT Math Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 24 | |
0kqy | Problem:
Given positive integers $a_{1}, a_{2}, \ldots, a_{2023}$ such that
$$
a_{k}=\sum_{i=1}^{2023}\left|a_{k}-a_{i}\right|
$$
for all $1 \leq k \leq 2023$, find the minimum possible value of $a_{1}+a_{2}+\cdots+a_{2023}$. | [
"Solution:\nWithout loss of generality, let $a_{1} \\leq a_{2} \\leq \\cdots \\leq a_{2023}$. Then, note that\n$$\n\\begin{aligned}\na_{k+1}-a_{k} & =\\sum_{i=1}^{2023}\\left|a_{k+1}-a_{i}\\right|-\\left|a_{k}-a_{i}\\right| \\\\\n& =k\\left(a_{k+1}-a_{k}\\right)-(2023-k)\\left(a_{k+1}-a_{k}\\right) \\\\\n& =(2 k-20... | United States | HMMT November 2022 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2046264 | |
0jss | Problem:
What is the smallest possible perimeter of a triangle whose side lengths are all squares of distinct positive integers? | [
"Solution:\n\nThere exist a triangle with side lengths $4^{2}$, $5^{2}$, $6^{2}$, which has perimeter $77$. If the sides have lengths $a^{2}$, $b^{2}$, $c^{2}$ with $0 < a < b < c$, then $a^{2} + b^{2} > c^{2}$ by the triangle inequality. Therefore $(b-1)^{2} + b^{2} \\geq a^{2} + b^{2} > c^{2} \\geq (b+1)^{2}$. So... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 77 | |
07u7 | The commuter village of Abelia has a population of $2022$ individuals. Households vary but the average household size is $3$ people. Houses are numbered consecutively.
Every inhabitant of Abelia takes a return train trip to a nearby city. There are four trains from Abelia to the city in the morning and four trains from... | [
"There are $4 \\times 4 = 16$ possible combinations of morning and evening trains. As $2022 > 2016 = 16 \\times 126$ there must be one combination of morning and evening trains with at least $127$ Abelians on both trains. Identify these Abelians, for example by painting them purple.\n\nSuppose for a contradiction t... | Ireland | IRL_ABooklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0lcs | Prove that for all positive integers $n$, the polynomial
$$
P(x) = (x^2 - 7x + 6)^{2n} + 13
$$
can not be written as a product of $(n+1)$ non-constant polynomials with integer coefficients. | [
"Clearly, polynomial $P(x)$ has degree $4n$ and has no real root. Thus, any factor of $P(x)$ has even degree. Suppose that $P(x)$ can be expressed as the product of $n + 1$ polynomials with degree greater than 0, so\n$$\nP(x) = P_1(x) \\cdot P_2(x) \\cdots P_{n+1}(x)\n$$\nthen $P_i(x)$ with $i = 1, 2, \\dots, n$ ha... | Vietnam | VMO | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
01p1 | Two parallelograms $ABCD$ and $NMKD$ are placed in the plane as it is shown in the figure; $T$ is the intersection point of the segments $BC$ and $MN$. Prove that the points $D$, $T$, and the intersection point of $AM$ and $BK$ are collinear if the points $A$, $T$, and $K$ are collinear.
 | [
"Construct the segment $AK$. By condition, $T$ lies on $AK$. Let $R$ be the intersection point of lines $DT$ and $KM$ (see the figures). Let $AN = x$, $ND = y$, $RM = z$, $AT : TK = 1 : k$. By Thales' theorem, $k = TK : AT = y : x$ (since $NT \\parallel DK$). Since $NMKD$ is a parallelogram, we have $MK = ND = y$, ... | Belarus | Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ij7 | Problem:
Let $A$, $B$, $C$, and $D$ be points on a circle such that $AB = 11$ and $CD = 19$. Point $P$ is on segment $AB$ with $AP = 6$, and $Q$ is on segment $CD$ with $CQ = 7$. The line through $P$ and $Q$ intersects the circle at $X$ and $Y$. If $PQ = 27$, find $XY$. | [
"Solution:\n\nSuppose $X$, $P$, $Q$, $Y$ lie in that order. Let $PX = x$ and $QY = y$. By power of a point from $P$, $x \\cdot (27 + y) = 30$, and by power of a point from $Q$, $y \\cdot (27 + x) = 84$. Subtracting the first from the second, $27 \\cdot (y - x) = 54$, so $y = x + 2$. Now, $x \\cdot (29 + x) = 30$, a... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 31 | |
0l9y | Given a regular 2007-polygon. Find the smallest positive integer $k$ satisfying the following property: In every set of $k$ vertices there are 4 vertices which form a quadrilateral with 3 edges of the given 2007-polygon. | [
"Denote the vertices of the regular 2007-polygon by $A_1, A_2, \\dots, A_{2007}$. Note that every quadrilateral has 3 edges of the given polygon if and only if its 4 vertices are consecutive vertices of the polygon.\n\nDenote by $A$ the set of the vertices except $A_{4k}$ ($k=1, 2, \\dots, 501$) and $A_{2007}$, in ... | Vietnam | Vijetnam 2007 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1506 | |
07jr | Determine all sequences $(a_n)$ of positive integers such that
$$
\frac{1}{2} < \frac{\gcd(a_r, a_s)}{\gcd(r, s)} < 2,
$$
for all positive integers $r, s$. | [
"For a moment, assume $a_2 \\neq a_4$. Through the following solution, we shall several times use the fact that $\\frac{n}{2} < a_n < 2n$. We divide the solution into three parts;\n\ni. *Large primes part*; We prove that for each prime $p \\ge 5$ the only prime less than or equal to $p$ that divides $a_p, a_{p^2}, ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > ... | null | proof and answer | a_n = n for all n | |
036d | Problem:
Let $ABCD$ be a parallelogram. The points $E$ and $F$ on the sides $AB$ and $BC$, respectively, are such that $DE$ is the bisector of $\angle ADF$ and $AE + CF = DF$. The line through $C$ and perpendicular to $DE$ meets the side $AD$ at $L$ and the diagonal $BD$ at $H$. Set $N = DE \cap AC$. Prove that:
a) $... | [] | Bulgaria | Bulgarian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
04gt | A chessboard is a board with $8$ rows and $8$ columns, whose squares are coloured alternately black and white, so that the square in the first row and first column is coloured black. An integer is inscribed in each square. It is known that the sum of all numbers on white squares equals $26$, and the sum of all numbers ... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 17 | |
0d6m | Let $k$ be a positive integer. Prove that there exist integers $x, y$, neither of which divisible by $7$ such that $x^{2} + 6 y^{2} = 7^{k}$. | [
"Take $x_{1} = y_{1} = -1$, and the following relations\n$$\nx_{k+1} = x_{k} - 6 y_{k}, \\quad y_{k+1} = x_{k} + y_{k}.\n$$\nfor $k = 1, 2, 3, \\ldots$\n\nOne can check that\n$$\nx_{k+1}^{2} + 6 y_{k+1}^{2} = 7 \\left( x_{k}^{2} + 6 y_{k}^{2} \\right)\n$$\nand\n$$\nx_{k} \\equiv y_{k} \\equiv (-1)^{k} \\pmod{7}.\n$... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof only | null | |
0lbe | **Bài 1.** (7 điểm). Cho đường tròn $(O)$ và hai điểm $B$, $C$ cố định trên $(O)$ sao cho $BC$ không là đường kính của $(O)$. Điểm $A$ thay đổi trên $(O)$ ($A \neq B, C$). Gọi $D$, $K$, $J$ tương ứng là trung điểm các cạnh $BC$, $CA$, $AB$; $E$, $M$, $N$ tương ứng là hình chiếu vuông góc của $A$, $B$, $C$ trên $BC$, $D... | [] | Vietnam | Kì thi chọn học sinh vào Đội tuyển Quốc gia Dự thi IMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | Vietnamese | proof only | null | |
0fg5 | Problem:
Sea $\mathcal{M}$ un conjunto de 1985 enteros positivos distintos, ninguno de los cuales tiene divisores primos mayores que 26. Demostrar que $\mathcal{M}$ contiene como mínimo un subconjunto de cuatro elementos distintos, cuyo producto es la cuarta potencia de un entero. | [] | Spain | International Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0fwg | Problem:
Sei $a_{1}, a_{2}, \ldots, a_{2007}$ eine Folge, die jede der Zahlen $1,2, \ldots, 2007$ genau einmal enthält. Es wird nun wiederholt folgender Operation ausgeführt: Ist das erste Folgeglied gleich $n$, dann wird die Reihenfolge der ersten $n$ Folgeglieder umgekehrt. Zeige, dass die Folge nach endlich vielen ... | [
"Solution:\n\nJede der Folgen ist eine Permutation von $1,2, \\ldots, 2007$. Da nur endlich viele solche Permutationen existieren und da jede Folge ihren Nachfolger bestimmt, müssen sich die Folgen ab einem bestimmten Zeitpunkt periodisch wiederholen. Nehme an, dass keine der Folgen mit einer $1$ beginnt und sei $N... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
08lp | Problem:
Find all the positive integers $x$ and $y$ that satisfy the equation
$$
x(x-y)=8y-7
$$ | [
"Solution:\nThe given equation can be written as:\n$$\n\\begin{aligned}\n& x(x-y)=8y-7 \\\\\n& x^{2}+7=y(x+8)\n\\end{aligned}\n$$\nLet $x+8=m$, $m \\in \\mathbb{N}$. Then we have: $x^{2}+7 \\equiv 0 \\pmod{m}$, and $x^{2}+8x \\equiv 0 \\pmod{m}$. So we obtain that $8x-7 \\equiv 0 \\pmod{m}$ \\quad (1).\nAlso we obt... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (63, 56) | |
0hi6 | In a triangle $ABC$ with an angle $\angle A = 60^\circ$, the Euler line intersects the circumcircle of the triangle at the points $U$ and $V$. $AD$ is the height $\Delta ABC$ and $\omega$ is the nine-point circle. The circumcircle $\Omega$ of triangle $\Delta UVD$ intersects the $\omega$ for the second time at the poin... | [
"Let $H$ and $O$ be the orthocentre and the centre of the circumcircle of $\\triangle ABC$ respectively, and $E$ be the centre of $\\omega$. Then, as you know $E$ be the centre of $HO$. Let $M$ be the midpoint of $BC$. Let the circle $\\Omega$ intersect $BC$ for the second time at point $X$. Since $\\angle CAB = 60... | Ukraine | Problems from Ukrainian Authors | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane G... | English | proof only | null | |
00q5 | Suppose that $k$ is a positive integer. A bijective map $f: \mathbb{Z} \to \mathbb{Z}$ is said to be $k$-jumpy if $|f(z) - z| \le k$ for all integers $z$.
Is it that case that for every $k$, each $k$-jumpy map is a composition of 1-jumpy maps?
*It is well known that this is the case when the support of the map is finit... | [
"Yes, it is true. Suppose that $f$ is $k$-jumpy. A number of the form $z + \\frac{1}{2}$ where $z$ is an integer is called fence. Select a fence. Count the number of integers which $f$ causes to jump from left to right over the fence, minus the number of integers it causes to jump from right to left over the same f... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Yes | |
002u | Hallar todos los números de la forma $11\ldots1$ que tienen un múltiplo de la forma $10\ldots01$. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Español | proof and answer | 11 | |
0fq8 | Problem:
Determinar los números reales $x>1$ para los cuales existe un triángulo cuyos lados tienen longitudes
$$
x^{4}+x^{3}+2 x^{2}+x+1, \quad 2 x^{3}+x^{2}+2 x+1, \quad x^{4}-1
$$ | [
"Solution:\nJustificaremos que para todo número real $x>1$ existe un tal triángulo probando que el lado mayor es menor que la suma de los otros dos. En efecto, para cualquier real $x>1$, tenemos:\n\n(i) $x^{4}+x^{3}+2 x^{2}+x+1>0, \\quad 2 x^{3}+x^{2}+2 x+1>0, \\quad x^{4}-1>0$.\n\n(ii) $x^{4}+x^{3}+2 x^{2}+x+1>2 x... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | All real numbers greater than 1 | |
05yd | Problem:
Soient $ABC$ un triangle et $I$ le centre de son cercle inscrit. La médiatrice de $[BC]$ coupe $(AI)$ en $S$ et $(BI)$ en $T$. Montrer que $C$, $I$, $S$ et $T$ sont cocycliques. | [
"Solution:\n\nPour montrer que $C, I, S$ et $T$ sont cocycliques, nous allons montrer que $\\widehat{SIC} = \\widehat{STC}$.\nOn rappelle que les bissectrices intérieures d'un triangle $ABC$ sont concourantes et que leur point d'intersection $I$ est le centre du cercle inscrit dans le trian... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"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 chasing"
] | null | proof only | null | |
03xf | As seen in Fig. 1.1, the circumcenter of acute triangle $ABC$ is $O$, $K$ is a point (not the midpoint) on the side $BC$, $D$ is a point on the extended line of segment $AK$, lines $BD$ and $AC$ intersect at point $N$, and lines $CD$ and $AB$ intersect at point $M$. Prove if $OK \perp MN$, then $A$, $B$, $D$, $C$ are c... | [
"By reduction to absurdity, assume that $A, B, D, C$ are not concyclic. Let the circumcircle (with radius $r$) of $ABC$ intersect $AD$ at point $E$. Join $BE$ and extend it to intersect line $AN$ at point $Q$; join $CE$ and extend it to intersect $AM$ at $P$. Join $PQ$, as seen in Fig. 1.2.\n\n | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > M... | English | proof only | null | |
0i3c | Problem:
You take a wrong turn on the way to MIT and end up in Transylvania, where $99\%$ of the inhabitants are vampires and the rest are regular humans. For obvious reasons, you want to be able to figure out who's who. On average, nine-tenths of the vampires are correctly identified as vampires and nine-tenths of hum... | [
"Solution:\nConsider a sample of $1000$ inhabitants. On average, $990$ are vampires and $10$ are people. $99$ vampires are identified as human and $9$ humans are identified as human. So out of the $108$ who pass, only $\\frac{1}{12}$ are human."
] | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 1/12 | |
061b | Problem:
In einem Schritt kann man vom Punkt $A(i|j|k)$, mit $i, j, k \in \{0,1,2,3\}$, zu einem weiteren Punkt des Würfelgitters gelangen, indem man stets genau eine der Koordinaten um $1$ vergrößert.
Man ermittle die Anzahl aller kürzesten Wege, die vom Ursprung $O(0|0|0)$ in den Punkt $P(3|3|3)$ führen. | [] | Germany | Auswahlwettbewerb zur IMO 2001 | [
"Statistics > Probability > Counting Methods > Permutations",
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 1680 | |
0i0s | Problem:
Let $p(x)$ be a polynomial of degree exactly $3$, with real coefficients. For each real number $a$, let $q_{a}(x)$ be the unique polynomial of degree $2$ or less such that $p(x)-q_{a}(x)$ is divisible by $(x-a)^{3}$. Prove that for $a \neq b$, the graphs of $q_{a}(x)$ and $q_{b}(x)$ do not intersect. Can you ... | [
"Solution:\n\nWe prove the general case: if $p$ has real coefficients and odd degree $d$, let $q_{a}(x)$ be the polynomial of degree at most $d-1$ such that $p(x)-q_{a}(x)$ is divisible by $(x-a)^{d}$, i.e. $q_{a}(x)$ is the remainder upon dividing $p(x)$ by $(x-a)^{d}$. We will show, for $a \\neq b$, that $q_{a}(x... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
082e | Problem:
Determinare per quali valori di $n$ tutte le soluzioni dell'equazione $X^{3}-3 X+n=0$ sono numeri interi. | [
"Solution:\n\nSe $a, b, c$ sono le tre soluzioni (non necessariamente distinte) dell'equazione $X^{3}-3 X+n=0$, allora\n$$\nX^{3}-3 X+n=(X-a)(X-b)(X-c)=X^{3}-(a+b+c) X^{2}+(a b+a c+b c) X-a b c .\n$$\nNe segue che\n$$\n\\left\\{\\begin{array}{l}\na+b+c=0 \\\\\na b+a c+b c=-3 \\\\\na b c=-n .\n\\end{array}\\right.\n... | Italy | Cesenatico Gara Individuale | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | n = -2 or n = 2 | |
0l96 | Let be given an acute triangle $ABC$ inscribed in a circle with center $O$ and two points $M$, $N$ on the line $AC$ such that $\overline{MN} = \overline{AC}$. Let $D$ be the orthogonal projection of $M$ on the line $BC$, $E$ be that of $N$ on the line $AB$.
1/ Prove that the orthocenter $H$ of triangle $ABC$ lies on t... | [
"1/ Let $K$ be the point of intersection of $MD$ and $NE$. It is clear that the circle with diameter $BK$ circumscribes about $\\triangle BED$.\n\nWe have: $AH \\parallel MK$ and $CH \\parallel NK$.\n\nIt implies: $\\angle HAC = \\angle KMN$ and $\\angle ACH = \\angle MNK$.\n\nFrom these equalities and from $AC = M... | Vietnam | 2003 Vietnamese Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0cbz | Let $(G, \cdot)$ be a group with the unit element $e$, and $H$ and $K$ two proper subgroups of $G$, such that $H \cap K = \{e\}$ and the set $(G \setminus (H \cup K)) \cup \{e\}$ is closed with respect to the operation in $G$. Show that $x^2 = e$, for any $x \in G$. | [
"Consider $L = (G \\setminus (H \\cup K)) \\cup \\{e\\}$. Because $x \\in H \\cup K \\iff x^{-1} \\in H \\cup K$, it follows that $x \\in L \\iff x^{-1} \\in L$, so that $L$ is a proper subgroup of $G$.\nAlso, $L \\cap H = L \\cap K = H \\cap K = \\{e\\}$, $G = H \\cup K \\cup L$ and it follows that for any permuta... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0e6f | We have $10$ intervals of numbers $I_1 = [a_1, b_1], I_2 = [a_2, b_2], \dots, I_{10} = [a_{10}, b_{10}]$ such that $I_1 \cup I_2 \cup \dots \cup I_{10} = [0, 1]$. Prove that there is a $k \le 10$ such that we can choose $k$ intervals $I_{n_1}, I_{n_2}, \dots, I_{n_k}$ among $I_1, I_2, \dots, I_{10}$ for which no two di... | [
"We shall construct two sets of intervals $A$ and $B$ such that the intervals in $A$ will be pairwise disjoint, the intervals in $B$ will be pairwise disjoint, and the union of all the intervals in $A$ and $B$ will be equal to $[0, 1]$. This construction will prove the claim of the problem because then either the s... | Slovenia | Selection Examinations for the IMO 2012 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0asu | Problem:
The positive integers are grouped as follows:
$$
A_{1} = \{1\}, \quad A_{2} = \{2, 3, 4\}, \quad A_{3} = \{5, 6, 7, 8, 9\}, \text{ and so on. }
$$
In which group does $2009$ belong to? | [
"Solution:\n\n$A_{45}$"
] | Philippines | Philippines Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | final answer only | A45 | |
03ir | Problem:
If $a$, $b$ and $c$ are the roots of the equation $x^{3} - x^{2} - x - 1 = 0$,
i. show that $a$, $b$ and $c$ are distinct:
ii. show that
$$
\frac{a^{1982} - b^{1982}}{a - b} + \frac{b^{1982} - c^{1982}}{b - c} + \frac{c^{1982} - a^{1982}}{c - a}
$$
is an integer. | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0aeu | Даден е произволен триаголник **АВС**. На страните **АВ**, **ВС** и **СА** се избрани произволни точки **C**₁, **A**₁ и **B**₁. Нека со **P**₁, **P**₂ и **P**₃ се означени плоштините на триаголниците **АС**₁**B**₁, **ВС**₁**A**₁ и **СА**₁**B**₁ соодветно, а со **P** е означена плоштината на триаголникот **АВС**. Докажи... | [
"За плоштините на триаголниците $\\overline{AB_1C_1}$ и $\\overline{ABC}$ важи:\n$$\nP_1 = \\frac{1}{2} \\overline{AB_1} \\overline{AC_1} \\sin \\angle A \\text{ и } P = \\frac{1}{2} \\overline{AB} \\overline{AC} \\sin \\angle A.\n$$\nСпоред тоа $\\frac{P_1}{P} = \\frac{\\overline{AB_1}}{\\overline{AC}} \\cdot \\fr... | North Macedonia | Републички натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Macedonian, English | proof only | null | |
049r | Determine the greatest possible ratio of the volume of a sphere and the volume of a right circular cone inscribed in it. | [] | Croatia | Hrvatska 2011 | [
"Geometry > Solid Geometry > Volume"
] | English | proof and answer | 27/8 | |
0b0s | Problem:
In the figure below, an equilateral triangle of height $1$ is inscribed in a semicircle of radius $1$. A circle is then inscribed in the triangle. Find the fraction of the semicircle that is shaded.
 | [] | Philippines | Philippines 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 > Circles > Tangents"
] | null | final answer only | 2/(π√3) - 2/9 | |
0i9a | Problem:
In triangle $A B C$ with area $51$, points $D$ and $E$ trisect $A B$ and points $F$ and $G$ trisect $B C$. Find the largest possible area of quadrilateral $D E F G$. | [
"Solution:\n\nAssume $E$ is between $D$ and $B$, and $F$ is between $G$ and $B$ (the alternative is to switch two points, say $D$ and $E$, which clearly gives a non-convex quadrilateral with smaller area).\n\nIf two triangles have their bases on the same line and the same opposite vertex, then it follows from the $... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 17 | |
000m | Cintia tiene una larga tira de papel donde están escritos todos los números naturales de 20 dígitos, ordenados de menor a mayor (desde $00\ldots01$ hasta $999\ldots99$) sin espacios entre números consecutivos. Cintia elige un número entero positivo $k$ y se lo dice a Elicita. A continuación Elicita elige $k$ dígitos co... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Other"
] | español | proof and answer | 40 | |
09f0 | Let $M$ be the midpoint of the side $BC$ of acute triangle $ABC$ and $H$ be the orthocenter of $ABC$. Prove that if $D$ is base of perpendicular dropped from the vertex $A$ to the line $HM$, then the intersection point of bisectors of the angles $DBH$, $DCH$ lies on the line $HM$. | [
"Let $BB'$, $CC'$ be altitudes. Then points $D$, $B'$, $C'$ lie on the circle with diameter $AH$. Since $\\angle C'DM = \\angle C'DH = \\angle C'AH = \\angle C'CM$, points $D$, $C'$, $M$, $C$ are cyclic. Thus we have $\\triangle C'HM \\sim \\triangle DHC \\Rightarrow \\frac{C'M}{CD} = \\frac{HM}{HC} \\Rightarrow \\... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0etn | $ABCD$ is a cyclic quadrilateral with $\angle BAD = 90^\circ$ and $\angle ABC > 90^\circ$. $AB$ is extended to a point $E$ such that $\angle AEC = 90^\circ$. If $AB = 7$, $BE = 9$, and $EC = 12$, calculate $AD$. | [
"Extend $AE$ and $DC$ until they meet in $F$, say. (These two lines will definitely meet: since $ABCD$ is a cyclic quadrilateral, $\\angle ABC > 90^\\circ$ implies that $\\angle ADC < 90^\\circ$.) See Figure 1.\n\n\n\nFigure 1\n\nWe have $\\angle FBC = \\angle FDA = \\angle FCE$ (note that ... | South Africa | The South African Mathematical Olympiad Third Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 24 | |
00op | Let $n$ be a positive integer.
Prove that $a(n) = n^5 + 5^n$ is divisible by $11$ if and only if $b(n) = n^5 \cdot 5^n + 1$ is divisible by $11$. | [
"If $n$ is a multiple of $11$, both sides of the equivalence are wrong, so the equivalence is true.\n\nIf $n$ is not a multiple of $11$, Fermat's little theorem implies that $n^{10} - 1$ is a multiple of $11$. The equivalence now follows from\n$$\nn^5 a(n) = n^{10} + n^5 \\cdot 5^n \\equiv b(n) \\pmod{11}.\n$$"
] | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | English | proof only | null | |
06zz | Problem:
$ABCD$ is a cyclic quadrilateral. A circle whose center is on the side $AB$ touches the other three sides. Show that $AB = AD + BC$. What is the maximum possible area of $ABCD$ in terms of $|AB|$ and $|CD|$? | [
"Solution:\n\n\n\nLet the circle have center $O$ on $AB$ and radius $r$. Let $\\angle OAD = \\theta$, $\\angle OBC = \\varphi$. Since $ABCD$ is cyclic, $\\angle ADC = 180^\\circ - \\varphi$, so $\\angle ODA = 90^\\circ - \\varphi / 2$.\n\nIf $AD$ touches the circle at $X$, then\n$$\nAD = AX... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > An... | null | proof and answer | AB = AD + BC; maximum area = (AB + CD)/4 * sqrt(2*AB*CD - CD^2). | |
09bl | $2x(xy - 2y - 3) = (x + y)(3x + y)$ тэгшитгэлийн бүх натурал шийдийг ол. | [
"Өгөгдсөн тэгшитгэлээ $y$-ийн хувьд квадрат тэгшитгэл хэлбэртэй бичвэл\n$$\ny^2 + (8x - 2x^2)y + (3x^2 + 6x) = 0 \\text{ болно.}\n$$\nЭнэ тэгшитгэлийн дискриминант\n$$\nD = (8x - 2x^2)^2 - 4(3x^2 + 6x) = 4x(x - 1)^2(x - 6)\n$$\nба $D$ бүтэн квадрат байх ёстой гэдгээс\n$$\nx(x-6) = z^2,\\ x \\in \\mathbb{Z}\n$$\nбол... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | Mongolian | proof and answer | (6,12) | |
04ae | Circles $k_1$ and $k_2$, with radii $r, R$ ($r < R$) touch each other internally in $A$. Let $p$ be a line parallel to their common tangent and let $B$ be an intersection of the line $p$ with the circle $k_1$, and $C$ an intersection of $p$ with $k_2$, so that the points $B$ and $C$ lie on the same side of the line thr... | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof and answer | sqrt(rR) | |
0ih6 | Problem:
If $f$ is a continuous real function such that $f(x-1)+f(x+1) \geq x+f(x)$ for all $x$, what is the minimum possible value of $\int_{1}^{2005} f(x) d x$? | [
"Solution:\n\nLet $g(x) = f(x) - x$. Then\n$$\ng(x-1) + x - 1 + g(x+1) + x + 1 \\geq x + g(x) + x\n$$\nor $g(x-1) + g(x+1) \\geq g(x)$. But now,\n$$\ng(x+3) \\geq g(x+2) - g(x+1) \\geq -g(x)\n$$\nTherefore\n$$\n\\begin{aligned}\n\\int_{a}^{a+6} g(x) d x & = \\int_{a}^{a+3} g(x) d x + \\int_{a+3}^{a+6} g(x) d x \\\\... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 2010012 | |
07kk | Let $p_1, p_2, p_3$ and $p_4$ be four different prime numbers satisfying the equations
$$
2p_1 + 3p_2 + 5p_3 + 7p_4 = 162,
$$
$$
11p_1 + 7p_2 + 5p_3 + 4p_4 = 162.
$$
Find all possible values of the product $p_1p_2p_3p_4$. | [
"There are many ways to go about this. One can get upper bounds on each of $p_1, p_2, p_3, p_4$ and work from there. Alternatively, you can use simple number theory ideas.\nAs $p_1, p_2, p_3, p_4$ are all different there can be at most one of them even. If they were all odd, the LHS of the first equation would be o... | Ireland | Irish Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 570 | |
0fxi | Problem:
Finde ein Paar $(m, n)$ ungerader natürlicher Zahlen mit $m, n>2009$ und
$$
m\mid n^{2}+8, \quad n\mid m^{2}+8
$$ | [
"Solution:\nDa $m$ und $n$ ungerade sind, folgt sofort $(m, n)=\\left(m, m^{2}+8\\right)=\\left(n, n^{2}+8\\right)=1$. Somit erfüllt ein Paar $(m, n)$ genau dann die Bedingung der Aufgabe, wenn $m n$ ein Teiler ist von $\\left(m^{2}+8\\right)\\left(n^{2}+8\\right)$. Da $m$ und $n$ ungerade sind, ist dies wiederum g... | Switzerland | IMO Selektion | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | final answer only | (19251, 3303) | |
01hy | Let $I$ be the incenter of a triangle $ABC$. Let the incircle of $ABC$ be tangent to $CA$ and $AB$ at $E$ and $F$, respectively. Lines $BI$ and $CI$ intersect line $EF$ at $Y$ and $Z$, respectively. Denote by $M$, $N$ midpoints of segments $BC$ and $YZ$, respectively. Prove that $MN$ is parallel to $AI$.
![](attached_... | [
"Refer to figure 22. Let us start with proving a known simple lemma stating that $\\angle BYC = 90^\\circ$. To that end, as $\\angle IEC = 90^\\circ$, it is enough to prove that $IEYC$ is cyclic. Indeed:\n$$\n\\begin{align*} \n\\angle CEY &= \\angle AEF \\\\ \n&= 90^\\circ - \\frac{1}{2} \\angle BAC \\\\ \n&= 90^\\... | Baltic Way | Baltic Way 2021 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations... | null | proof only | null | |
0841 | Problem:
Determinare se il seguente enunciato è vero o falso:
"Per ogni successione $x_{1}, x_{2}, x_{3}, \ldots$ di numeri reali maggiori o uguali a zero esistono due successioni $a_{1}, a_{2}, a_{3}, \ldots$ e $b_{1}, b_{2}, b_{3}, \ldots$ di numeri reali maggiori o uguali a zero tali che
- $x_{n}=a_{n}+b_{n}$ per o... | [
"Solution:\n\nL'enunciato è vero. Per dimostrarlo, indichiamo innanzitutto una strategia possibile.\nFissata una successione di numeri interi $0<n_{1}<n_{2}<\\ldots$, dividiamo gli indici in intervalli disgiunti: il primo intervallo comprende gli indici $1, \\ldots, n_{1}$, il secondo gli indici $n_{1}+1, \\ldots, ... | Italy | XX Olimpiade Italiana di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | true | |
04n1 | Let $n \ge 4$ be a positive integer. Prove that among any $n$ numbers from the set
$$
\{1, 2, \dots, 2n - 1\}
$$
there are some of them with the sum divisible by $2n$. (Ratko Višak) | [] | Croatia | Croatia_2018 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
006z | Rocío debe escribir en una línea 100 números enteros distintos elegidos desde $1$ hasta $199$ de manera que cada número, a partir del segundo y hasta el antecesor, sea mayor que por lo menos uno de sus dos vecinos. A continuación calcula la suma de los números de las posiciones pares que denotamos $P$ y la suma de los ... | [] | Argentina | Argentina 2009 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Spanish | proof and answer | 149 | |
08t7 | Determine all positive integers $n$ for which $8^n + n$ is divisible by $2^n + n$. | [
"To begin with, we transform $8^n + n$ into the following expression:\n$$\n8^n + n = (2n)^3 + n = (2^n + n)((2^n)^2 - n \\cdot 2^n + n^2) - (n^3 - n).\n$$\nSince $(2^n)^2 - n \\cdot 2^n + n^2$ is an integer, from the expression above it follows that the statement $8^n + n$ is divisible by $2^n + n$ is equivalent to... | Japan | Japan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | 1, 2, 4, 6 | |
08wq | A regular octagon $ABCDEFGH$ of side length $1$ is given. Let $I$ be the point of intersection of the lines $AD$ and $BF$. Find the area of the quadrilateral $AIGH$.
 | [
"$$\n\\boxed{\\frac{1 + \\sqrt{2}}{2}}\n$$\nSince the straight lines $CH$ and $AD$ are symmetrically positioned with respect to the straight line $BF$, the lines $CH$, $AD$, $BF$ intersect at a single point. Therefore, the $3$ points $C$, $I$, $H$ lie on the same straight line. Since the diagonals $DH$ and $CG$ are... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | (1+\sqrt{2})/2 | |
0l4s | Problem:
Compute the sum of the positive divisors (including 1) of $9!$ that have units digit $1$. | [
"Solution:\nThe prime factorization of $9!$ is $2^{7} \\cdot 3^{4} \\cdot 5 \\cdot 7$. Every divisor of $9!$ has prime factorization $2^{a} \\cdot 3^{b} \\cdot 5^{c} \\cdot 7^{d}$, where $0 \\leq a \\leq 7$, $0 \\leq b \\leq 4$, $0 \\leq c \\leq 1$, and $0 \\leq d \\leq 1$. If the divisor has units digit $1$, it ca... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)"
] | null | final answer only | 103 | |
02gb | Let $X$ be the set of all sequences $a = (a_1, a_2, \dots, a_{2000})$ such that $a_i \in \{0, 1, 2\}$ for $1 \le i \le 1000$ and $a_i \in \{0, 1\}$ for $1001 \le i \le 2000$. Given $a, b$ in $X$, we define the distance $d(a, b)$ between $a$ and $b$ as the number of values for $i$, $1 \le i \le 2000$, for which $a_i \ne... | [
"Let $a = f(0, 0, 0, \\dots, 0)$, $b = f(1, 0, 0, \\dots, 0)$ and $c = f(2, 0, 0, \\dots, 0)$, say $a = (a_1, a_2, \\dots, a_{2000})$, $b = (b_1, b_2, \\dots, b_{2000})$ and $c = (c_1, c_2, \\dots, c_{2000})$. It's easy to conclude that there exists a unique $1 \\le i_0 \\le 2000$ such that $a, b$ and $c$ differ at... | Brazil | XXII OBM | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | English | proof and answer | 1000!^2 * 12^{1000} | |
0cp1 | Ignoramus arranged 11 positive integers in a circle. For each two adjacent numbers, he calculated their (nonnegative) difference. Among the 11 results obtained, there are 4 ones, 4 twos, and 3 threes. Prove that Ignoramus made a mistake in his calculations.
Незнайка выписал по кругу 11 натуральных чисел. Для каждого д... | [
"Запишем каждую из наших разностей со знаком плюс, если в соответствующей паре чисел большее стоит перед меньшим по часовой стрелке, и со знаком минус в противном случае. У нас получились 11 разностей между числом и следующим за ним по часовой стрелке; значит, сумма всех этих чисел равна нулю, то есть чётному числу... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English; Russian | proof only | null | |
0aov | Problem:
Let $a$, $b$, and $c$ be real constants such that $x^{2}+x+2$ is a factor of $a x^{3}+b x^{2}+c x+5$, and $2 x-1$ is a factor of $a x^{3}+b x^{2}+c x-\frac{25}{16}$. Find $a+b+c$. | [
"Solution:\n$\\frac{45}{11}$\n\nUsing long division, when $a x^{3}+b x^{2}+c x+5$ is divided by $x^{2}+x+2$, the quotient is $a x+(b-a)$ and the remainder is $(c-a-b) x+5+2 a-2 b$. Since $x^{2}+x+2$ is a factor of $a x^{3}+b x^{2}+c x+5$, we must have $c-a-b=0$ and $5+2 a-2 b=0$. On the other hand, since $2 x-1$ is... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 45/11 | |
05z7 | Problem:
Soient $a$ et $b$ deux réels. On définit les suites $\left(a_{n}\right)$ et $\left(b_{n}\right)$ par $a_{0}=a,\ b_{0}=b$ et pour tout $n$ entier naturel, $a_{n+1}=a_{n}+b_{n}$ et $b_{n+1}=a_{n} b_{n}$. Déterminer toutes les paires $(a, b)$ telles que $a_{2022}=a_{0}$ et $b_{2022}=b_{0}$. | [
"Solution:\n\nSupposons que $(a, b)$ soit une solution de l'énoncé. On calcule\n$$\n\\sum_{i=1}^{2022} a_{i}=\\sum_{i=1}^{2022}\\left(a_{i-1}+b_{i-1}\\right)=\\sum_{i=0}^{2021} a_{i}+\\sum_{i=0}^{2021} b_{i}\n$$\nMais on a\n$$\n\\sum_{i=0}^{2021} a_{i}=a_{0}+\\sum_{i=1}^{2021} a_{i}=a_{2022}+\\sum_{i=1}^{2021} a_{i... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOi 2 : Algèbre | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | All real a with b = 0 | |
038y | Solve in integers the system
$$
\begin{cases} 3a^4 + 2b^3 = c^2 \\ 3a^6 + b^5 = d^2. \end{cases}
$$ | [
"We shall prove that $a = b = c = d = 0$. It is easy to see that if one of the numbers $a, b, c, d$ equals $0$, then the others equal $0$, too. Indeed, if $b = 0$, $a \\neq 0$, then $\\sqrt{3} = \\pm\\frac{c}{a^2}$ is a rational number, a contradiction. The case $a = 0$, $b \\neq 0$ is impossible by the same reason... | Bulgaria | Spring Mathematical Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a = b = c = d = 0 | |
0kdz | Problem:
Compute the value of $\sqrt{105^{3}-104^{3}}$, given that it is a positive integer. | [
"Solution:\n\nFirst compute $105^{3}-104^{3}=105^{2}+105 \\cdot 104+104^{2}=3 \\cdot 105 \\cdot 104+1=32761$. Note that $180^{2}=32400$, so $181^{2}=180^{2}+2 \\cdot 180+1=32761$ as desired.",
"Solution:\n\nWe have $105^{3}-104^{3}=105^{2}+105 \\cdot 104+104^{2}$. Thus\n$$\n104 \\sqrt{3}<\\sqrt{105^{3}-104^{3}}<1... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 181 | |
0hrs | Problem:
In triangle $ABC$, $\angle A = 50^\circ$, $\angle B = 60^\circ$, and $\angle C = 70^\circ$. A ray of light bounces from point $D$ on $BC$ to $E$ on $CA$ to $F$ on $AB$ and then back to $D$. Find the angles of $\triangle DEF$.
(Light always takes the shortest path between points, meaning it bounces off an edg... | [
"Solution:\n\nLet $\\angle BDF = \\angle CDE = x$, $\\angle CED = \\angle AEF = y$, and $\\angle AFE = \\angle BFD = z$.\n\n\n\nSince $\\triangle AEF$, $\\triangle BDF$, and $\\triangle CDE$ all have angle sum $180^\\circ$, we get\n$$\n\\begin{aligned}\n& x + y + 70^\\circ = 180^\\circ, \\\... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | ∠EDF = 80°, ∠DEF = 60°, ∠DFE = 40° | |
0csp | For a positive integer $N > 1$, let $m$ be its largest divisor smaller than $N$. Find all $N$ such that $N + m$ is a power of $10$. | [
"**Ответ.** $75$.\n\nПусть $m$ — наибольший делитель числа $N$, меньший, чем $N$. Тогда $N = m p$, где $p$ — наименьший простой делитель числа $N$. Имеем $N + m = 10^k$, то есть $m(p + 1) = 10^k$. Число в правой части не делится на $3$, поэтому $p > 2$. Отсюда следует, что $N$ — нечётное число, а тогда и $m$ нечётн... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 75 | |
0jvs | Problem:
Let $P_{1} P_{2} \ldots P_{8}$ be a convex octagon. An integer $i$ is chosen uniformly at random from $1$ to $7$, inclusive. For each vertex of the octagon, the line between that vertex and the vertex $i$ vertices to the right is painted red. What is the expected number of times two red lines intersect at a p... | [
"Solution:\n\nAnswer: $\\frac{54}{7}$\n\nIf $i=1$ or $i=7$, there are $0$ intersections. If $i=2$ or $i=6$ there are $8$. If $i=3$ or $i=5$ there are $16$ intersections. When $i=4$ there are $6$ intersections (since the only lines drawn are the four long diagonals).\n\nThus the final answer is $\\frac{8+16+6+16+8}{... | United States | HMMT November 2016 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 54/7 | |
07hc | Let $x_1$ be a fixed positive integer. Prove that there is no positive integer $m > 2500$ such that the sequence $x_{n+1} = x_n^{S(n)} + 1$ is eventually periodic modulo $m$. Where $S(n)$ is the sum of digits of $n$ in base 10.
(We say a sequence is eventually periodic mod $m$ if there are positive integers $N, T$ suc... | [
"We shall prove a more general statement. That is, we prove that $x_n$ is a-periodic modulo $m$, for each $m$. In doing so, we need the following lemmas.\n\n**Lemma 1.** Let $m \\ge 3$ be an integer which is not a power of 2. Let $x_1$ be a fixed positive integer and $y_n$ be a sequence of positive integers such th... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Algebra > Algebraic Expressions > Sequences ... | English | proof only | null | |
08so | There are $2008$ pieces each of red cards and white cards. These $4016$ cards are mixed and put on a pile. Then, each of the $2008$ participating players is dealt $2$ cards from the pile, and then all of the players are seated in a circular arrangement facing interior of the circle. At every turn of the game each playe... | [
"Pick some player and call him $A$. Let $A_1$ be $A$ and for $i = 2, 3, \\dots$ designate by $A_i$ the player sitting at the $i-1$-th position from the player $A$ counted clockwise. For a non-negative integer $k$ and a positive integer $i$ denote by $F(k, i)$ the number obtained by subtracting the total number of r... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 2007 | |
07d9 | Two circles $\omega_1(O)$ and $\omega_2$ intersect each other at $A, B$. And $O$ lies on $\omega_2$. Let $S$ be the foot of perpendicular line to $AB$ from $O$. Line $OS$ intersects $\omega_2$ for the second time at $P$. The bisector of $\widehat{ASP}$ intersects $\omega_1$ at $L$ ($A$ and $L$ are on the same side of t... | [
"Let $Q$ be the intersection point of $PK$, $AB$ and let $L'$ be the incenter of triangle $QSP$. We'll prove that $L \\equiv L'$.\n\nWe know that $\\widehat{QSP} = \\widehat{OKP} = 90^\\circ$ therefore $QKSO$ is a cyclic quadrilateral. And since $PS = PK$, it's also concluded that $PQ = PO$. Now we have\n$$\n\\wide... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
03mj | Let $G$ be the centroid of a right-angled triangle $ABC$ with $\angle BCA = 90^\circ$. Let $P$ be the point on ray $AG$ such that $\angle CPA = \angle CAB$, and let $Q$ be the point on ray $BG$ such that $\angle CQB = \angle ABC$. Prove that the circumcircles of triangles $AQG$ and $BPG$ meet at a point on side $AB$. | [
"Since $\\angle C = 90^\\circ$, the point $C$ lies on the semicircle with diameter $AB$ which implies that, if $M$ is the midpoint of side $AB$, then $MA = MC = MB$. This implies that triangle $AMC$ is isosceles and hence that $\\angle ACM = \\angle A$. By definition, $G$ lies on segment $M$ and it follows that $\\... | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneo... | null | proof only | null | |
04jm | A finite number of real numbers are ordered around a circle, and each number is coloured red, white or blue. Each red number is twice smaller than the sum of its two adjacent numbers, each white number is equal to the sum of its two adjacent numbers, and each blue number is twice larger than the sum of its two adjacent... | [
"Let the numbers be $a_1, a_2, \\ldots, a_n$ arranged around a circle. Each $a_i$ is colored red, white, or blue.\n\nLet $R$, $W$, $B$ be the sets of indices of red, white, and blue numbers, respectively.\n\n- For red $a_i$: $a_i = \\frac{1}{2}(a_{i-1} + a_{i+1})$ (since it is twice smaller than the sum of its neig... | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | -3/2 | |
0fue | Problem:
Sei $ABCD$ ein Sehnenviereck, sodass gilt $|AB| + |CD| = |BC|$. Zeige, dass der Schnittpunkt der Winkelhalbierenden von $\angle DAB$ und $\angle CDA$ auf der Seite $BC$ zu liegen kommt. | [
"Solution:\n\nSei $P$ der Punkt auf $\\overline{BC}$ mit $|BP| = |BA|$. Nach Voraussetzung gilt dann auch $|CP| = |CD|$. Wir definieren $\\angle APB = \\alpha$ und $\\angle CPD = \\beta$. Mit Winkeljagd finden wir $\\angle DCP = 180^\\circ - 2\\beta$ und darum $\\angle BAD = 2\\beta$. Analog zeigen wir $\\angle ADC... | Switzerland | SMO Finalrunde | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
07i4 | A family of some *three-element* subsets of a $1000$-element set is considered. We know that the union of every $5$ of the subsets has at least $12$ elements. Find the largest possible value for the number of these subsets. | [
"Consider the selected subsets as vertices of a graph and connect two subsets if they have a common element. We claim that each connected component has at most $4$ elements. Otherwise, the union of $5$ subsets of this connected component has at most $11 = 3 + 2 + 2 + 2 + 2$ elements, which contradicts the assumptio... | Iran | 40th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 444 | |
051b | Is it possible that the perimeter of a triangle whose side lengths are integers, is divisible by the double of the longest side length? | [
"Let the side lengths of the triangle be integers $a$, $b$, $c$. Without loss of generality we may assume that $c \\ge a$ and $c \\ge b$. Suppose that the perimeter of the triangle $a+b+c$ is divisible by double of the longest side length $2c$. Since $0 < a+b+c \\le 3c < 2 \\cdot 2c$, the perimeter $a+b+c$ can be d... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0e2q | The diagonal $AC$ of a convex quadrilateral $ABCD$ is the bisector of the angle $\angle DCB$. Let $E$ be the intersection of the side $AB$ and the circumcircle of the triangle $ACD$. Let $F$ be the intersection of the side $AD$ and the circumcircle of the triangle $ABC$. Prove that the segments $AC, DE$ and $BF$ inters... | [
"Let $G$ be the intersection of the segments $AC$ and $BF$. The points $A, B, C$ and $F$ lie on the same circle, so $\\angle AFB = \\angle ACB$. The diagonal $AC$ bisects the angle $\\angle DCB$, so $\\angle ACB = \\angle DCA$. This implies $\\angle AFG = \\angle DCA$ and $\\angle GFD + \\angle DCG = \\pi$. We conc... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0e8g | Problem:
Turnir v namiznem tenisu poteka po naslednjem pravilu. V vsakem krogu v primeru lihega števila tekmovalcev najprej izžrebajo enega, ki se avtomatično uvrsti v naslednji krog. Ostale tekmovalce $z$ žrebom razporedijo v pare. Tekmovalca vsakega para se pomerita med seboj, zmagovalec iz vsakega para pa se uvrsti... | [
"Solution:\n\nČe je na turnirju 2013 tekmovalcev, potem gre v drugi krog $1 + \\frac{2012}{2} = 1007$ tekmovalcev. V tretji krog se uvrsti $1 + \\frac{1006}{2} = 504$ tekmovalcev, v četrti krog pa $\\frac{504}{2} = 252$ tekmovalcev. V peti krog nato napreduje $\\frac{252}{2} = 126$ tekmovalcev, v šestega $\\frac{12... | Slovenia | 57. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | f(2013) = 11; smallest n with f(n) = 11 is 1025 | |
0i8e | Problem:
What is the smallest number of regular hexagons of side length $1$ needed to completely cover a disc of radius $1$? | [
"Solution:\nFirst, we show that two hexagons do not suffice. Specifically, we claim that a hexagon covers less than half of the disc's boundary. First, a hexagon of side length $1$ may be inscribed in a circle, and this covers just $6$ points. Translating the hexagon vertically upward (regardless of its orientation... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 3 | |
0ix4 | Problem:
A set $S$ of positive integers is called magic if for any two distinct members of $S$, $i$ and $j$,
$$
\frac{i+j}{\operatorname{GCD}(i, j)}
$$
is also a member of $S$. The $\operatorname{GCD}$, or greatest common divisor, of two positive integers is the largest integer that divides evenly into both of them; fo... | [
"Solution:\nSuppose there are two members $i$ and $j$ whose $\\operatorname{GCD}$ is $1$. Then $i+j$ is also in $S$. But $\\operatorname{GCD}(j, i+j)$ is also $1$, so $i+2j$ is also in $S$. Continuing, $i+kj$ is in $S$ for all integers $k$, so the set is infinite.\n\nThe only remaining possibility for a finite magi... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All finite magic sets are either singletons {m} for any positive integer m, or two-element sets {i, i^2 − i} for integers i ≥ 3. | |
04ds | Determine the smallest positive integer $n$ such that a half of $n$ is a perfect square, a third of $n$ is a perfect cube, and a fifth of $n$ a fifth power of an integer. | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 30233088000000 | |
0k4w | Problem:
Tessa has a unit cube, on which each vertex is labeled by a distinct integer between $1$ and $8$ inclusive. She also has a deck of $8$ cards, $4$ of which are black and $4$ of which are white. At each step she draws a card from the deck, and
- if the card is black, she simultaneously replaces the number on ea... | [
"Solution:\n\nAnswer: $42648$\nThe order of the deck does not matter as black cards and white cards commute, therefore we can assume that the cards are alternating black and white, and only worry about the arrangement of the numbers. After each pair of black and white cards, each number is replaced by the sum of tw... | United States | HMMT November 2018 | [
"Algebra > Linear Algebra > Linear transformations",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 42648 | |
0d8w | For integer $n > 1$, consider $n$ cube polynomials $P_{1}(x), \ldots, P_{n}(x)$ such that each polynomial has 3 distinct real roots. Denote $S$ as the set of roots of the following equation
$$
P_{1}(x) P_{2}(x) P_{3}(x) \ldots P_{n}(x) = 0.
$$
It is also known that for each $1 \leq i < j \leq n$, $P_{i}(x) P_{j}(x) = 0... | [
"1) For $i \\in \\{1,2,3, \\ldots, n\\}$, denote $A_{i}$ as the set of roots of $P_{i}(x)$ then since the given condition, we can see that $|A_{i}| = 3$ and for all $i, j \\in \\{1,2,3, \\ldots, n\\}$ then $|A_{i} \\cup A_{j}| = 5 \\Leftrightarrow |A_{i} \\cap A_{j}| = 1$.\n\nSuppose that $|S| = m$ and $S = \\{a_{1... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Polynomials"
] | English | proof and answer | Under the uniqueness condition, n = 7. If n > 7, then |S| = 2n + 1. | |
0hk1 | Problem:
In the sequence
$$
1, 4, 7, 10, 13, 16, 19, \ldots
$$
each term is 3 less than the next term. Find the 1000th term of the sequence. | [
"Solution:\n\nTo get from the first term to the 1000th term, we must add three 999 times. That amounts to adding $3 \\cdot 999$ which is $2997$. So the 1000th term is $1 + 2997 = 2998$."
] | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Sequences and Series"
] | null | final answer only | 2998 | |
0a0c | A teacher makes a big fraction with the numbers $1$ up to and including $12$. She writes some of the numbers in the numerator and puts $\times$-signs between them. She puts the remaining numbers in the denominator, again with $\times$-signs between the numbers. In both the numerator and denominator, there is at least o... | [
"$231$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 231 | |
02ac | Problem:
Um baralho possui 32 cartas divididas em 4 tipos, cada um com 8 cartas. De quantas formas podemos escolher 6 cartas de modo que todos os quatro tipos de cartas estejam entre elas? | [
"Solution:\n\nVamos dividir as escolhas apropriadas de cartas em dois grupos:\n\na) Grupo $S_{1}$: Dois tipos são representados por duas cartas e os outros dois tipos restantes por apenas uma carta cada.\n\nb) Grupo $S_{2}$: Um tipo é representado por três cartas e os outros três tipos restantes por apenas uma cart... | Brazil | NÍVEL 3 | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 415744 | |
07nn | Find all pairs of positive integers $(a, b)$ for which
$$
\binom{a+1}{2} \binom{b+1}{2} = 630.
$$ | [
"Let $f(a) = \\left(\\frac{a+1}{2}\\right)$, then $f(1) = 1$ and $f(35) = 630$. Because $f(a) > 630$ for $a > 35$, any solution $(a, b)$ satisfies $1 \\le a, b \\le 35$ and $f(a)$ and $f(b)$ are divisors of $630 = 2 \\cdot 3^2 \\cdot 5 \\cdot 7$. If $a$ is odd, $a$ is a factor of $f(a)$. If $a$ is even, $a+1$ is od... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | [(3, 14), (14, 3), (2, 20), (20, 2), (1, 35), (35, 1)] |
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