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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08gh | Problem:
Su un foglio di carta sono disegnati due esagoni regolari. Il più piccolo ha area $18$, e una diagonale minore dell'esagono più grande coincide con una diagonale maggiore dell'esagono più piccolo. Quanto misura l'area dell'unione dei due esagoni? | [
"Solution:\n\nLa risposta è $29$. La figura disegnata sul foglio si può rappresentare come segue:\n\n\n\nIl punto $O$ in figura è dato dal centro dell'esagono maggiore $A B C D E F$. Poiché l'esagono $A B C D E F$ è regolare, il triangolo $O C D$ è equilatero. Inoltre, essendo $D A$ il diam... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 29 | |
0e91 | Let $a$ and $b$ be positive real numbers, such that their product is $1$ and the sum of their squares is $4$. Find the exact value of the expression $a^{-3} + b^{-3}$. | [
"From $a^2 + b^2 = 4$ and $ab = 1$ we get $(a+b)^2 = a^2 + b^2 + 2ab = 4 + 2 = 6$, or $a+b = \\sqrt{6}$ since $a$ and $b$ are positive. This implies\n$$\n\\frac{1}{a^3} + \\frac{1}{b^3} = \\frac{a^3 + b^3}{a^3 b^3} = \\frac{(a+b)^3 - 3ab(a+b)}{a^3 b^3} = \\frac{6\\sqrt{6} - 3 \\cdot 1 \\cdot \\sqrt{6}}{1} = 3\\sqrt... | Slovenia | National Math Olympiad 2013 - First Round | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 3*sqrt(6) | |
01jd | Let $\triangle ABC$ be an acute triangle with $|AB| < |AC|$ and incenter $I$. Let $D$ be the projection of $I$ onto $BC$. Let $H$ be the orthocenter of $\triangle ABC$. Prove that if $\angle IDH = \angle CBA - \angle ACB$ then $|AH| = 2 \cdot |ID|$. | [
"Let $H'$ be the reflection of $H$ in $BC$. It is well-known (and easy to prove) that $H'$ lies on the circumcircle of $\\triangle ABC$. Let $O$ be the circumcenter of $\\triangle ABC$. We have\n$$\n\\begin{align*}\n\\angle OH'A &= \\angle HAO = \\angle BAC - \\angle BAH - \\angle OAC \\\\\n&= \\angle BAC - 2(90^\\... | Baltic Way | Baltic Way 2023 Shortlist | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02hk | Problem:
Qual a menor das raízes da equação $2(x-3 \sqrt{5})(x-5 \sqrt{3})=0$ ? | [
"Solution:\nA equação já é dada na forma fatorada $a(x-m)(x-n)=0$, logo as raízes são $m=3 \\sqrt{5}$ e $n=5 \\sqrt{3}$. Devemos decidir qual delas é a maior.\n\nSabemos que $\\sqrt{5}<2,3$ e $1,7<\\sqrt{3}$, logo $3 \\sqrt{5}<3 \\times 2,3=6,9$ e $5 \\sqrt{3}>5 \\times 1,7=8,5$. Como $6,9$ é menor do que $8,5$, co... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | final answer only | 3√5 | |
079c | We have drawn a polygon with $2n$ sides without picking up the pencil from the paper. If we name the sides from $1$ to $2n$ with the order of drawing them, then the odd sides are all vertical and are drawn bottom-up. Prove that this polygon intersects itself. | [
"**First Solution:**\nIf a polygon does not intersect itself, then by moving clockwise (or counterclockwise) along its sides, the inside (or outside) of the polygon must always be at our right-side. But if the said polygon doesn't intersect itself, then the right-side of its rightmost side (because it is drawn upwa... | Iran | 27th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kn2 | Problem:
Let $p$, $q$, $r$ be primes such that $2p + 3q = 6r$. Find $p + q + r$. | [
"Solution:\nFirst, it is known that $3q = 6r - 2p = 2(3r - p)$, thus $q$ is even. The only even prime is $2$ so $q = 2$.\n\nFurther, $2p = 6r - 3q = 3(2r - q)$, which means that $p$ is a multiple of $3$ and thus $p = 3$.\n\nThis means that $2 \\cdot 3 + 3 \\cdot 2 = 6r \\Longrightarrow r = 2$.\n\nTherefore, $p + q ... | United States | HMMT November 2021 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 7 | |
02b7 | Problem:
No interior de um triângulo $ABC$, toma-se um ponto $E$ tal que $AE = BE$ e $AB = EC$. Se $A\hat{BE} = \alpha = E\hat{CA}$, $E\hat{AC} = 2\alpha$ e $E\hat{BC} = 5\alpha$, determine $\alpha$. | [
"Solution:\n\nSabendo que a soma dos ângulos internos de um triângulo é $180^{\\circ}$, obtemos\n\n$$\n\\begin{cases}\nA\\hat{E}B = 180^{\\circ} - (\\alpha + \\alpha) = 180^{\\circ} - 2\\alpha \\\\\nA\\hat{E}C = 180^{\\circ} - (\\alpha + 2\\alpha) = 180^{\\circ} - 3\\alpha\n\\end{cases}\n$$\n\nAssim, temos que\n\n$... | Brazil | null | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 15° | |
0jiz | Problem:
An up-right path from $(a, b) \in \mathbb{R}^{2}$ to $(c, d) \in \mathbb{R}^{2}$ is a finite sequence $\left(x_{1}, y_{1}\right), \ldots,\left(x_{k}, y_{k}\right)$ of points in $\mathbb{R}^{2}$ such that $(a, b)=\left(x_{1}, y_{1}\right),(c, d)=\left(x_{k}, y_{k}\right)$, and for each $1 \leq i<k$ we have tha... | [
"Solution:\n\nThe number of up-right paths from $(0,0)$ to $(4,4)$ is $\\binom{8}{4}$ because any such up-right path is identical to a sequence of 4 $U$'s and 4 $R$'s, where $U$ corresponds to a step upwards and $R$ corresponds to a step rightwards. Therefore, the total number of pairs of (possibly intersecting) up... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 1750 | |
0ik9 | Problem:
A finite set of circles in the plane is called nice if it satisfies the following three conditions:
(i) No two circles intersect in more than one point;
(ii) For every point $A$ of the plane there are at most two circles passing through $A$;
(iii) Each circle from the set is tangent to exactly 5 other circles ... | [
"Solution:\n(a) The answer is yes. The following picture shows that there is a nice set consisting of exactly 12 circles. It is also possible to construct a nice set with 22 circles (the picture can be found below). Since $2006 = 158 \\cdot 12 + 5 \\cdot 22$ we can make a nice set of 2006 by making a union of 158 d... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | a) Yes. b) No. | |
048g | Four integers $a, b, c, d$ satisfy $a + b = c$, $a + d = 2c$. Prove that there exists a rectangular triangle with integer sides whose area equals $abcd$. | [
"From the given equations we have $a = c - b$, $d = 2c - a = b + c$. So\n$$\nabcd = (c-b) \\cdot b \\cdot c \\cdot (b+c) = bc(c^2-b^2) = \\frac{1}{2}(2bc)(c^2-b^2).\n$$\nTherefore $abcd$ is the area of a right triangle with legs $2bc$ and $c^2 - b^2$.\nThe hypotenuse of that triangle equals\n$$\n\\sqrt{(2bc)^2 + (c... | Croatia | CroatianCompetitions2011 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
0k18 | Problem:
Lily has a $300 \times 300$ grid of squares. She now removes $100 \times 100$ squares from each of the four corners and colors each of the remaining 50000 squares black and white. Given that no $2 \times 2$ square is colored in a checkerboard pattern, find the maximum possible number of (unordered) pairs of s... | [
"Solution:\n\nFirst we show an upper bound. Define a grid point as a vertex of one of the squares in the figure. Construct a graph as follows. Place a vertex at each grid point and draw an edge between two adjacent points if that edge forms a black-white boundary. The condition of there being no $2 \\times 2$ check... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 49998 | |
0015 | Cada casilla de un tablero de $3 \times 3$ contiene un botón luminoso que puede estar apagado o prendido. Al apretar el botón del centro cambia el estado de sus ocho vecinos pero no de él mismo. Al apretar cualquier otro botón cambia su estado y el de cada botón vecino.
Apretar un botón se cuenta como un paso.
Dos co... | [] | Argentina | XII Olimpíada Matemática Rioplatense | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Linear transformations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | español | proof and answer | 4 | |
01fn | Let $G$ be a graph of diameter $2$ on $2019$ vertices. $62$ cops and $1$ robber are placed on the vertices of graph $G$. Everybody sees all others, they move in turn. Each move each person can stay in his vertex or go to any of the adjacent vertices (each cop stay or go independently of the others and they coordinate t... | [
"The numbers $62$ and $2019$ have the following secret property:\n$$\n64 + 63 + \\dots + 3 = \\frac{64 \\cdot 65}{2} - 3 = 2077 > 2019.\n$$\nLet before the cops move the robber be in the vertex $A$.\nIf $\\deg A \\le 62$, the cops can catch the robber in one, two or three moves (due to diameter $2$ property): denot... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof only | null | |
02mr | Problem:
Certo número deixa resto $1$ quando dividido por $3$, deixa resto $2$ quando dividido por $4$, deixa resto $3$ quando dividido por $5$ e deixa resto $4$ quando dividido por $6$. Qual é o menor número inteiro positivo que satisfaz essas propriedades? | [] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | final answer only | 58 | |
0i5q | Problem:
If a positive integer multiple of $864$ is picked randomly, with each multiple having the same probability of being picked, what is the probability that it is divisible by $1944$? | [
"Solution:\nThe probability that a multiple of $864 = 2^{5} 3^{3}$ is divisible by $1944 = 2^{3} 3^{5}$ is the same as the probability that a multiple of $2^{2}$ is divisible by $3^{2}$, which since $4$ and $9$ are relatively prime is $\\frac{1}{9}$."
] | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 1/9 | |
07r3 | Let $n$ be a positive integer and $a_1, \dots, a_n$ positive real numbers. Let
$$
s_k = a_1^k + \dots + a_n^k
$$
for $k = 1, 2, 3, \dots$. Prove that
$$
\frac{s_5 s_1^3}{5} - \frac{s_4 s_2 s_1^2}{4} + \frac{s_2^4}{20} \ge 0.
$$ | [
"$$\n4s_5 s_1^3 + s_2^4 \\geq 5s_4 s_2 s_1^2.\n$$\nUsing the AM-GM inequality, $4s_5 s_1^3 + s_2^4 \\geq 5 [s_5 s_1^{12} s_2^4]^{1/5}$, so it suffices to prove that\n$$\ns_5^4 s_2^4 s_1^{12} \\geq s_4^5 s_2^5 s_1^{10},\n$$\nand thus, using the positivity of $s_k$, that $s_5 s_1^2 \\geq s_4 s_2$.\n\nObserve that $s_... | Ireland | Ireland_2017 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Muirhead / majorization"
] | English | proof only | null | |
0cn4 | $n$ lines are drawn in the plane, and all the intersection points of these lines are marked. It is appeared that one of $n$ lines contains exactly 1 marked point, one of $n$ lines contains exactly 3 marked points, and one of $n$ lines contains exactly 5 marked points. Find all the possible values of $n$. (I. Bogdanov) | [
"Answer: $8$ lines.\n\n\n\nFig. 14\n\n\n\nFig. 15\n\nLet $l_1$, $l_3$, $l_5$ be the lines on which $1$, $3$, and $5$ points are marked, respectively, and let $A$ be the only marked point on $l_1$. Then all other lines either pass through this point (call these lines... | Russia | Russian mathematical olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English; Russian | proof and answer | 8 | |
084g | Problem:
Quante sono le coppie ordinate $(x, y)$ di interi positivi $x$ e $y$ che soddisfano la relazione $x y + 5(x + y) = 2005$? | [
"Solution:\n\nLa risposta è 10. Sommando 25 ad entrambi i membri si ottiene la relazione equivalente $x y + 5(x + y) + 25 = 2030$ da cui, raccogliendo al primo membro, si ricava $(x + 5)(y + 5) = 2030 = 2 \\cdot 5 \\cdot 7 \\cdot 29$. Poiché $x$ è intero, allora anche $x + 5$ è intero, e lo stesso vale per $y$. Dun... | Italy | Progetto Olimpiadi di Matematica 2005 GARA di SECONDO LIVELLO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | final answer only | 10 | |
079j | Let $m$, $n$ be two relatively prime integers. Prove that the following equation has an infinite number of solutions:
$$
x^m t^n + y^m s^n = v^m r^n
$$ | [
"First note that if $(x, t, y, s, v, r)$ is a solution, then\n$$\n(x \\cdot a^{kn},\\ t \\cdot a^{km},\\ y \\cdot a^{kn},\\ s \\cdot a^{km},\\ v \\cdot a^{kn},\\ r \\cdot a^{km})\n$$\nis another solution. So it is enough to find a primary solution. We know that\n$$\n\\begin{aligned}\n2 + 3 = 5 &\\Rightarrow 2^{mn+1... | Iran | 27th Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0e9l | Problem:
Poišči vse polinome $p$ z realnimi koeficienti, za katere velja
$$
p(p(x)) = \left(x^{2} + x + 1\right) p(x)
$$
za vse $x \in \mathbb{R}$. | [
"Solution:\n\nNičelni polinom je očitno rešitev naloge. Naj bo $p$ neničeln polinom in pišimo $p(x) = a_{n} x^{n} + a_{n-1} x^{n-1} + \\cdots + a_{0}$, kjer je $a_{n} \\neq 0$. Vodilni člen na levi strani enakosti je tedaj enak $a_{n} (a_{n} x^{n})^{n} = a_{n}^{n+1} x^{n^{2}}$, vodilni člen na desni strani pa je en... | Slovenia | 58. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | p(x) = 0 or p(x) = x^2 + x | |
0bj7 | Let $ABC$ be a triangle and let the points $D \in (BC)$, $E \in (AC)$, $F \in (AB)$ be such that
$$
\frac{DB}{DC} = \frac{EC}{EA} = \frac{FA}{FB}.
$$
The halflines $(AD, BE$, and $(CF$ intersect the circumcircle of $ABC$ at points $M, N$ and $P$. Prove that the triangles $ABC$ and $MNP$ share the same centroid if and ... | [
"First, observe that $\\overrightarrow{AD} + \\overrightarrow{BE} + \\overrightarrow{CF} = 0$ ($*$). Denoting the areas of the triangles $ABC, BMC, CNA$ and $APB$ by $s, s_a, s_b$, and $s_c$, respectively, we have $\\frac{DM}{AD} = \\frac{s_a}{s}$, hence $\\frac{AM}{AD} = \\frac{s+s_a}{s}$, which implies $\\overrig... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
03la | Problem:
Consider a standard twelve-hour clock whose hour and minute hands move continuously. Let $m$ be an integer, with $1 \leq m \leq 720$. At precisely $m$ minutes after 12:00, the angle made by the hour hand and minute hand is exactly $1^{\circ}$. Determine all possible values of $m$. | [
"Solution:\n\nThe minute hand makes a full revolution of $360^{\\circ}$ every 60 minutes, so after $m$ minutes it has swept through $\\frac{360}{60} m = 6m$ degrees. The hour hand makes a full revolution every 12 hours (720 minutes), so after $m$ minutes it has swept through $\\frac{360}{720} m = m/2$ degrees. Sinc... | Canada | 2003 CMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 262, 458 | |
0g4q | Problem:
Pendant la coupe du monde, $n$ autocollants Panini sont à collectionner. Les amis de Marco veulent compléter leurs collections, mais personne n'a encore de collection complète! Une paire de deux amis est dite complète si leur collection commune contient au moins un de chaque autocollant. Marco connaît les con... | [
"Solution:\n\nPour montrer qu'au moins $n$ tables différentes sont nécessaires, supposons que nous avons $n$ personnes, et chacun d'entre eux manque exactement un sticker, avec chaque personne manquant un sticker différent. N'importe quelle paire d'amis est complète, donc les amis doivent être assis sur des tables ... | Switzerland | Deuxième tour 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n | |
0c0d | Let $ABC$ be an acute scalene triangle, and let $H$ and $O$ be its ortho-center and circumcenter, respectively. The line $OA$ crosses the altitudes from $B$ and $C$ of the triangle $ABC$ at $P$ and $Q$, respectively. Show that the center of the circle $HPQ$ lies on one of the medians of the triangle $ABC$. | [
"\n\n*Solution.* Without loss of generality, we may and will assume $AB < AC$. Begin by noticing that the triangles $ABC$ and $HPQ$ are similar. Indeed, $\\angle HQP = 90^\\circ - \\angle QAB = 90^\\circ - \\angle OAB = \\frac{1}{2}\\angle AOB = \\angle ACB$, and similarly $\\angle HPQ = \\... | Romania | 69th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0a2q | We call a positive integer a *reverse difference* if it can be written as a positive integer whose last digit is not a $0$, minus the number consisting of the same digits in reverse order. For example, $2178$ is a reverse difference because $4202 - 2024 = 2178$.
Which of the following numbers is *not* a reverse differ... | [] | Netherlands | Junior Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | D | |
04e4 | Prove that the sum of all three-digit numbers whose decimal representation consists of three different digits different from zero has at least three different prime divisors. | [
"Let us consider all three-digit numbers whose digits are all different and nonzero. Each such number can be written as $100a + 10b + c$, where $a$, $b$, $c$ are digits from $1$ to $9$, all different.\n\nThe total number of such numbers is $9 \\times 8 \\times 7 = 504$.\n\nLet $S$ be the sum of all such numbers.\n\... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0hgj | We chose several numbers among $1, 2, \ldots, 2022$. It turned out that sum of any two of the chosen numbers isn't divisible by the difference between any two of the chosen numbers. What largest possible number of numbers could be selected?
(Oleksii Masalitin) | [
"Suppose that more than $674$ numbers were chosen. Then there exists a triple $3n + 1, 3n + 2, 3n + 3$, among which at least two numbers were chosen, so the absolute difference between some two of the chosen numbers doesn't exceed $2$. Clearly, there exist some two chosen numbers with the same parity, so their sum ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 674 | |
0f54 | Problem:
Given a quadratic equation $ax^2 + bx + c$. If it has two real roots $A \leq B$, transform the equation to $x^2 + Ax + B$. Show that if we repeat this process we must eventually reach an equation with complex roots. What is the maximum possible number of transformations before we reach such an equation? | [] | Soviet Union | 17th ASU | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3 | |
0hm2 | Problem:
Find all prime numbers $p$ such that $p^{2} + 2007p - 1$ is prime as well. | [
"Solution:\n\nIf $p = 3$, then $p^{2} + 2007p - 1 = 6029$ which is a prime.\n\nFor $p \\neq 3$, we know that $p = 3k \\pm 1$ for some integer $k$. Hence\n$$\np^{2} + 2007p - 1 = (3k \\pm 1)^{2} + 2007(3k \\pm 1) - 1 = 9k^{2} \\pm 6k + 1 + 2007 \\cdot 3k \\pm 2007 - 1 = 9k^{2} \\pm 6k + 2007 \\cdot 3k \\pm 2007.\n$$... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | p = 3 | |
0ea7 | The bisector of the inner angle $\angle ACB$ in an acute triangle $ABC$ intersects the side $AB$ at $D$. The circumcircle of the triangle $ADC$ intersects the side $BC$ in two distinct points, $C$ and $E$. The line through $B$ parallel to the line $AE$ intersects the line $CD$ at $F$. Prove that $AFB$ is an isosceles t... | [
"Write $\\angle ACB = \\gamma$. Then $\\angle ACF = \\angle FCB = \\frac{\\gamma}{2}$, since $CF$ is the bisector of the angle $\\angle ACB$. The points $A$, $D$, $E$ and $C$ are concyclic, so $\\angle DAE = \\angle DCE = \\frac{\\gamma}{2}$. The line $BF$ is parallel to the line $AE$, so $\\angle ABF = \\angle DAE... | Slovenia | National Math Olympiad in Slovenia | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0imb | Problem:
Two reals $x$ and $y$ are such that $x - y = 4$ and $x^{3} - y^{3} = 28$. Compute $x y$. | [
"Solution:\n\nAnswer: $-3$. We have $28 = x^{3} - y^{3} = (x - y)\\left(x^{2} + x y + y^{2}\\right) = (x - y)\\left((x - y)^{2} + 3 x y\\right) = 4 \\cdot (16 + 3 x y)$, from which $x y = -3$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -3 | |
013y | Problem:
Let $p$ be a prime number and let $n$ be a positive integer. Let $q$ be a positive divisor of $(n+1)^{p}-n^{p}$. Show that $q-1$ is divisible by $p$. | [
"Solution:\n\nIt is sufficient to show the statement for $q$ prime. We need to prove that\n$$\n(n+1)^{p} \\equiv n^{p} \\quad(\\bmod q) \\Longrightarrow q \\equiv 1 \\quad(\\bmod p) .\n$$\nIt is obvious that $\\operatorname{gcd}(n, q)=\\operatorname{gcd}(n+1, q)=1$ (as $n$ and $n+1$ cannot be divisible by $q$ simul... | Baltic Way | Baltic Way 2005 | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0gih | $2024^{2024}$ divided by $102$ leaves a remainder of **3** **4** **5**.
$2024^{2024}$ 除以 $102$ 的餘數是③④⑤。 | [
"答案：052\n\n$102 = 2 \\times 3 \\times 17$ and $2024 \\equiv 1 \\pmod{17}$ so the answer can be rather easy to calculate, even without the knowledge Fermat's little theorem. However, if the student apply Euler's totient theorem blindly, he will get something like $2024^{2024} \\equiv 86^8 \\pmod{102}$, which is stil... | Taiwan | APMO Taiwan Preliminary Round 1 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Chinese; English | final answer only | 052 | |
066a | Determine the non-positive integers $m$, $n$ satisfying the equation:
$$
\frac{n(n+2)}{4} = m^4 + m^2 - m + 1
$$ | [
"For $n = 0$, the equation becomes: $m^4 + m^2 - m + 1 = 0$, which is impossible, because $m^4 + m^2 - m + 1 = m^4 + m(m-1) + 1 > 0$, for all $m \\in \\mathbb{Z}_+$. \n\nFor $m = 0$, the equation becomes: $n(n+2) = 4$, impossible in the set of non-negative integers. \n\nFor $m, n \\neq 0$, the equation becomes:\n$$... | Greece | Selection Examination A | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (m, n) = (1, 2) | |
09ui | The sequence of Fibonacci numbers $F_0, F_1, F_2, \ldots$ is defined by $F_0 = F_1 = 1$ and $F_{n+2} = F_n + F_{n+1}$ for all $n \ge 0$. For example, we have
$$
F_2 = F_0 + F_1 = 2, \quad F_3 = F_1 + F_2 = 3, \quad F_4 = F_2 + F_3 = 5, \quad F_5 = 8.
$$
The sequence $a_0, a_1, a_2, \ldots$ is defined by
$$
a_n = \frac{... | [
"Note that for all $n \\ge 0$ the number $a_n$ can be rewritten as follows:\n$$\n\\begin{aligned}\na_n &= \\frac{1}{F_n F_{n+2}} = \\frac{F_{n+1}}{F_n F_{n+2}} \\cdot \\frac{1}{F_{n+1}} = \\frac{F_{n+2} - F_n}{F_n F_{n+2}} \\cdot \\frac{1}{F_{n+1}} \\\\\n&= \\left( \\frac{1}{F_n} - \\frac{1}{F_{n+2}} \\right) \\cdo... | Netherlands | Final Round, September 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
08jn | Problem:
Se dă un romb $ABCD$ cu latura $1$. Pe laturile $BC$ şi $CD$ există punctele $M$, respectiv $N$, astfel încât $MC + CN + NM = 2$ şi $\angle MAN = \frac{1}{2} \angle BAD$.
Să se afle unghiurile rombului. | [] | JBMO | Olimpiada Naţională de Matematică | [
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof and answer | The rhombus is a square: all angles are ninety degrees. | |
0ilq | Problem:
A subset $S$ of the nonnegative integers is called supported if it contains $0$, and $k+8, k+9 \in S$ for all $k \in S$. How many supported sets are there? | [
"Solution:\nAnswer: $1430$. Note that every supported set $S$ contains $0,8,9,16,17,18,24$-$27,32$-$36,40$-$45,48$-$54$, and all $n \\geq 55$. Now define $\\bar{S} := \\mathbb{Z}^{+} \\setminus S$, which is a subset of $\\{1$-$7,10$-$15,19$-$23,28$-31,37,38,39,46,47,55\\}$ satisfying the opposite property that $k \... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 1430 | |
0ij2 | Problem:
A pebble is shaped as the intersection of a cube of side length $1$ with the solid sphere tangent to all of the cube's edges. What is the surface area of this pebble? | [
"Solution:\nImagine drawing the sphere and the cube. Take a cross section, with a plane parallel to two of the cube's faces, passing through the sphere's center. In this cross section, the sphere looks like a circle, and the cube looks like a square (of side length $1$) inscribed in that circle. We can now calculat... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Solid Geometry > Surface Area",
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | ((6*sqrt(2)-5)/2) * pi | |
01gv | Let $ABCDE$ be a convex pentagon inscribed in a circle $\omega$ such that $CD \parallel BE$. The line tangent to $\omega$ at $B$ intersects the line $AC$ at a point $F$ such that $A$ lies between the points $C$ and $F$. The lines $BD$ and $AE$ intersect at $G$. Prove that the line $FG$ is tangent to the circumcircle of... | [
"Since $CD \\parallel BE$, we have $BC = DE$. Note that\n$$\n\\angle BFC = \\angle BAC - \\angle ABF = \\angle DBE - \\angle AEB = \\angle BGA,\n$$\nhence $ABGF$ is cyclic. It follows that $\\angle AGF = \\angle ABF = \\angle ADG$, hence the circumcircle of $ADG$ is tangent to $FG$."
] | Baltic Way | Baltic Way 2020 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00la | Let $ABC$ be a triangle. Its incircle meets the sides $BC$, $CA$ and $AB$ in the points $D$, $E$ and $F$, respectively. Let $P$ denote the intersection point of $ED$ and the line perpendicular to $EF$ and passing through $F$, and similarly let $Q$ denote the intersection point of $EF$ and the line perpendicular to $ED$... | [
"Let $H$ be the common point of $PF$ and $QD$, as can be seen in Figure 3. Since $\\angle EDH$ and $\\angle HFE$ are both right angles, $HE$ is a diameter of the incircle of $ABC$. Now let $X$ denote the common point of $EH$ and $PQ$. We see that $H$ is the orthocenter of the triangle $EPQ$, and $X$, $D$ and $F$ ar... | Austria | National Competition | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
000l | Daniel elige un entero positivo $n$ y se lo dice a Ana. Con esta información, Ana elige un entero positivo $k$ y se lo dice a Daniel. Daniel traza $n$ circunferencias en un papel y elige $k$ puntos distintos con la condición de que cada uno de ellos pertenezca a alguna de las circunferencias que trazó. Luego borra las ... | [] | Argentina | XI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | español | proof and answer | 2n + 1 | |
04qc | Let $M$ be the midpoint of the base $\overline{BC}$ in the isosceles triangle $ABC$. Let $P$ be the point different from $A$ such that $PA \parallel BC$. Points $X$ and $Y$ lie on the rays $PB$ and $PC$, respectively, so that $B$ lies between $P$ and $X$, $C$ lies between $P$ and $Y$, and $\angle PXM = \angle PYM$ hold... | [
"Let $N$ be the other intersection of the circles circumscribed to the triangles $BMX$ and $CMY$.\n\n\n\nSince $\\angle BNM = \\angle BXM = \\angle CYM = \\angle CNM$, the line $MN$ is the angle bisector of $\\angle BNC$. The segment $NM$ is a median in the triangle $BNC$, hence $|BN| = |CN... | Croatia | Croatian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00g0 | Find all positive integers $a$ and $b$ such that
$$
\frac{a^{2}+b}{b^{2}-a} \text{ and } \frac{b^{2}+a}{a^{2}-b}
$$
are both integers. | [
"By the symmetry of the problem, we may suppose that $a \\leq b$. Notice that $b^{2}-a \\geq 0$, so that if $\\frac{a^{2}+b}{b^{2}-a}$ is a positive integer, then $a^{2}+b \\geq b^{2}-a$. Rearranging this inequality and factorizing, we find that $(a+b)(a-b+1) \\geq 0$. Since $a, b>0$, we must have $a \\geq b-1$.\n\... | Asia Pacific Mathematics Olympiad (APMO) | XIV APMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | (2,2), (3,3), (1,2), (2,3), (2,1), (3,2) | |
046f | Find the largest positive integer $m$ such that there is a way to color some of the cells red in a $70 \times 70$ grid, satisfying the following two conditions:
(1) There do not exist two red cells such that the number of red cells in the row they belong to and the number of red cells in the column they belong to are t... | [
"Let the row with the maximum number of red squares (let's assume it's the first row) have $a$ red squares, and the column with the maximum number of red squares (let's assume it's the first column) have $b$ red squares. If $a > b$, consider the $a$ columns in which the $a$ red squares of the first row are located.... | China | 2023 Chinese IMO National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 32 | |
02e4 | Find the number of ways that a positive integer $n$ can be represented as a sum of one or more consecutive positive integers. | [
"The sum of $k$ consecutive integers is $(m+1) + (m+2) + \\dots + (m+k) = \\frac{k(2m+k+1)}{2}$. So we require $k(2m+k+1) = 2n$. Note that $k$ and $2m+k+1$ have opposite parity and that $k < 2m+k+1$.\n\nNow suppose $2n = ab$ with $a$ odd. Then $b$ must be even. So $a$ and $b$ cannot be equal. Take $k$ to be the sma... | Brazil | VIII OBM | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | the number of odd divisors of 2n | |
0739 | Show that in a non-equilateral triangle, the following are equivalent:
(a) the angles of the triangle are in arithmetic progression;
(b) the common tangent to the nine-point circle and the in-circle is parallel to the Euler line. | [
"Let $A$, $B$, $C$ be the vertices, $I$ the in-center, $H$ the ortho-centre, $O$ the circum-center and $N$ the nine-point center of the triangle.\n\n**Claim 1:** (b) $\\Leftrightarrow IH = IO$ (for a non-equilateral triangle).\n\nThe line $NI$ joins the centre of nine-point circle and the centre of in-circle, so $N... | India | Indija TS 2007 | [
"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 > Quadrilaterals > Cyclic quad... | null | proof only | null | |
0ihz | Problem:
Let $f(x) = x^{2} - 2x$. How many distinct real numbers $c$ satisfy $f(f(f(f(c)))) = 3$? | [
"Solution:\nWe see the size of the set $f^{-1}\\left(f^{-1}\\left(f^{-1}\\left(f^{-1}(3)\\right)\\right)\\right)$. Note that $f(x) = (x-1)^{2} - 1 = 3$ has two solutions: $x = 3$ and $x = -1$, and that the fixed points $f(x) = x$ are $x = 3$ and $x = 0$. Therefore, the number of real solutions is equal to the numbe... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 9 | |
06tu | Let $n \geqslant 2$ be an integer. In the plane, there are $n$ segments given in such a way that any two segments have an intersection point in the interior, and no three segments intersect at a single point. Jeff places a snail at one of the endpoints of each of the segments and claps his hands $n-1$ times. Each time ... | [
"(a) For odd $n$, we travel along the circumference of the disk and mark each of the points $A_{i}$ or $B_{i}$ 'in' and 'out' alternately. Since every pair of lines intersect in the disk, there are exactly $n-1$ points between $A_{i}$ and $B_{i}$ for any fixed $1 \\leqslant i \\leqslant n$. As $n$ is odd, this mean... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | English | proof only | null | |
0g0j | Problem:
An einer EGMO-Prüfung gibt es drei Aufgaben, wobei bei jeder Aufgabe eine ganzzahlige Punktzahl zwischen $0$ und $7$ erreicht werden kann. Zeige, dass es unter $49$ Schülerinnen immer zwei gibt, sodass die eine in jeder der drei Aufgaben mindestens so gut war wie die andere. | [
"Solution:\n\nWir nehmen an, dass wir möglichst viele Schülerinnen haben, sodass wir keine zwei finden, die die Voraussetzung erfüllen. Wir zeigen, dass es dann maximal $48$ Schülerinnen geben kann.\n\nWir teilen die Schülerinnen in bis zu $8$ Gruppen ein gemäss der Punktzahl der ersten Aufgabe. Sei $a_{i}$ die Grö... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07hu | Given a cyclic quadrilateral $ABCD$ with circumcenter $O$. Let $E$ be the intersection of $AD$ and $BC$, and $F$ be the intersection of $AC$ and $BD$. Circle $\omega$ is tangent to $BD$ and $AC$. $QP$ is a diameter of $\omega$ such that $F$ is the orthocenter of triangle $QEP$. Prove that $EO$ passes through the center... | [
"Let $H$ be the projection of $P$ to the line $EQ$ and $Z$ be the projection of $E$ to the line $PQ$. Let $AD$ and $BC$ touch $\\omega$ at $X$ and $Y$, respectively. Let $O'$ be the center of $\\omega$ and $\\omega_1$ be the circle with diameter $O'F$.\n\n\n\nSince $\\angle O'XF = \\angle O... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | null | proof only | null | |
01y8 | Find the minimal positive integer $k$ for which there exist pairwise distinct positive integers $a$, $b$ and $c$ such that all three numbers: $4a^2 + kb + c$, $4b^2 + kc + a$ and $4c^2 + ka + b$ are perfect squares. | [
"Since the expressions $4a^2 + kb + c$, $4b^2 + kc + a$ and $4c^2 + ka + b$ are invariant under cyclic permutations of $a$, $b$ and $c$, we can assume that $c = \\max(a, b, c)$. Then for $k \\le 3$\n$$\n(2c)^2 < 4c^2 + ka + b \\le 4c^2 + 3c + c < 4c^2 + 4c + 1 = (2c + 1)^2.\n$$\nHence $4c^2 + ka + b$ is not a perfe... | Belarus | BY 2020-2021 tst for Navid | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4 | |
0iqh | Problem:
For how many ordered triples $(a, b, c)$ of positive integers are the equations $a b c + 9 = a b + b c + c a$ and $a + b + c = 10$ satisfied? | [
"Solution:\n\nSubtracting the first equation from the second, we obtain\n$$\n1 - a - b - c + a b + b c + c a - a b c = (1 - a)(1 - b)(1 - c) = 0.\n$$\nSince $a$, $b$, and $c$ are positive integers, at least one must equal $1$. Note that $a = b = c = 1$ is not a valid triple, so it suffices to consider the cases whe... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 21 | |
0jcu | Let $n$ be a positive integer. Suppose we are given $2^n + 1$ distinct sets, each containing finitely many objects. Place each set into one of two categories, the red sets and the blue sets, so that there is at least one set in each category. We define the *symmetric difference* of two sets as the set of objects belong... | [
"Call an element *a diverse* if there is a red set containing $a$, a red set not containing $a$, a blue set containing $a$, and a blue set not containing $a$. We consider two cases based on whether a diverse element exists. We use $A\\Delta B$ to denote the symmetric difference of sets $A$ and $B$.\n\n*Case 1:* Sup... | United States | TSTST | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ho4 | Problem:
How many ways are there to list the numbers $1$ to $10$ in some order such that every number is either greater or smaller than all the numbers before it? | [
"Solution:\n\nThe number of lists is $2^{9} = 512$. Imagine building a list from the end. First, note that the last number must be $1$ or $10$, since otherwise it would be less than some numbers before it and greater than others. Thus, there are two choices for the last number.\n\nSimilarly, after choosing the last... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 512 | |
0emo | A triangular grid is obtained by tiling an equilateral triangle of side length $n$ into $n^2$ equilateral triangles of side length $1$. Determine the number of parallelograms bounded by the line segments of the grid. | [
"The parallelograms can be naturally partitioned into three sets, based on the orientation of their sides in the grid. By symmetry there will be the same number of parallelograms of each orientation, so we can restrict ourselves to the 'diamond' case (as in the diagram).\n\n\n\nExtend the t... | South Africa | South-Afrika 2011-2013 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 3*binom(n+2,4) | |
04p1 | Find all pairs $(a, n)$ of positive integers such that
$$
3a^2 + 2^n = a^4.
$$ | [] | Croatia | Croatian Mathematical Society Competitions | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (a, n) = (2, 2) | |
05kc | Problem:
Pour tout entier $k \geqslant 2$, on note $\mathrm{P}(k)$ le plus grand diviseur premier de $k$.
Prouver qu'il existe une infinité d'entiers $n$ tels que
$$
\mathrm{P}(n)<\mathrm{P}(n+1)<\mathrm{P}(n+2)
$$ | [
"Solution:\n\nSoit $p$ un nombre premier impair. Pour tout entier $k \\geqslant 1$, on note $n_{k}=p^{2^{k}}-1$. On a évidemment $\\mathrm{P}\\left(n_{k}+1\\right)=p$. D'autre part, pour tout entier $i$ tel que $0<i<k$, on a\n$$\np^{2^{k}}=\\left(p^{2^{i}}\\right)^{2^{k-i}}=(-1)^{2^{k-i}}=1 \\quad \\bmod p^{2^{i}}+... | France | Olympiades Françaises de Mathématiques | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0d5t | Let $n \geq 2$ be an integer and $p_{1} < p_{2} < \cdots < p_{n}$ prime numbers. Prove that there exists an integer $k$ relatively prime with $p_{1} p_{2} \cdots p_{n}$ and such that $\operatorname{gcd}\left(k + p_{1} p_{2} \cdots p_{i},\ p_{1} p_{2} \cdots p_{n}\right) = 1$ for all $i = 1, 2, \ldots, n-1$. | [
"The problem is equivalent to proving that $k$ is relatively prime with $p_{1} p_{2} \\cdots p_{n}$ and such that for all $i = 2, \\ldots, n$, none of the integers $k,\\ k + p_{1}, \\ldots, k + p_{1} \\cdots p_{i-1}$ is divisible by $p_{i}$.\n\nLet $i \\in \\{2, \\ldots, n\\}$. Notice that $p_{i} \\geq i + 1$. So t... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English, Arabic | proof only | null | |
08vy | Let $ABCDE$ be a convex pentagon, for which the quadrilateral $ABCD$ is a square. If $\angle AEC + \angle BED = 180^\circ$, show that the pentagon $ABCDE$ is circumscribed by a circle. Here, by a convex pentagon we mean a pentagon for which each of the inner angles is less than $180^\circ$. | [
"Take a point $E'$ on the opposite side of the line $BC$ from $A$ in such a way that the triangle $BCE'$ becomes congruent with the triangle $ADE$. Since $ABCD$ is a square, we have $AC = BD$ and $\\angle DAC = \\angle CBD$. We also have $AE = BE'$ and $\\angle EAD = \\angle E'BC$. Therefore, we get $\\angle EAC = ... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02o8 | Find all positive integers $a, b$ such that
$$
3^a = 2b^2 + 1.
$$ | [
"The only solutions are $(1, 1)$, $(2, 2)$ and $(5, 11)$.\n\nIf $a$ is even and greater than 2, the equation can be rewritten as $(3^{a/2} - 1) \\cdot (3^{a/2} + 1) = 2b^2$. But $\\gcd(3^{a/2} - 1, 3^{a/2} + 1) = \\gcd(3^{a/2} - 1, 2) = 2$, so $3^{a/2} + 1 = 4u^2$ and $3^{a/2} - 1 = 2v^2$ or $3^{a/2} + 1 = 2u^2$ an... | Brazil | Brazilian Math Olympiad | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof and answer | [[1, 1], [2, 2], [5, 11]] | |
08hb | Problem:
Solve the system
$$
\begin{cases}
x + y + z + t = 6, \\
\sqrt{1 - x^{2}} + \sqrt{4 - y^{2}} + \sqrt{9 - z^{2}} + \sqrt{16 - t^{2}} = 8.
\end{cases}
$$ | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | x = 3/5, y = 6/5, z = 9/5, t = 12/5 | |
015u | In a party of eight persons, each pair of persons either know each other or do not know each other. Each person knows exactly two of the other people.
Is the following situation possible: (i) no three persons know each other but (ii) there are no four persons such that no two of these know each other? | [
"\n\nAs each person in the party is acquainted to three others, the number of acquaintance relations must be $\\frac{1}{2} \\cdot 8 \\cdot 3 = 12$. Assume $A$ is one of the persons in the party. Denote by $B$, $C$ and $D$ the three persons $A$ knows. To meet condition (i), none of $B$, $C$ ... | Baltic Way | Baltic Way SHL | [
"Discrete Mathematics > Other",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Yes, it is possible. | |
0b4u | Problem:
A set of positive integers is said to be pilak if it can be partitioned into 2 disjoint subsets $F$ and $T$, each with at least 2 elements, such that the elements of $F$ are consecutive Fibonacci numbers, and the elements of $T$ are consecutive triangular numbers. Find all positive integers $n$ such that the s... | [
"Solution:\nThe only positive integer $n$ that satisfies this property is $30$.\nIn this case, $D=\\{1,2,3,5,6,10,15\\}$, and we can partition $D$ into $F=\\{1,2,3,5\\}$ and $T=\\{6,10,15\\}$. We will show that there are no other $n$.\n\nClaim 1: $1 \\in F$.\nProof: Obviously, $1 \\in D$. Assume that $1 \\in T$. Si... | Philippines | 25th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 30 | |
0ihe | Problem:
A triangular piece of paper of area $1$ is folded along a line parallel to one of the sides and pressed flat. What is the minimum possible area of the resulting figure? | [
"Solution:\n\nLet the triangle be denoted $ABC$, and suppose we fold parallel to $BC$. Let the distance from $A$ to $BC$ be $h$, and suppose we fold along a line at a distance of $c h$ from $A$. We will assume that neither angle $B$ nor $C$ is obtuse, for the area of overlap will only be smaller if either is obtuse... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2/3 | |
07fw | Let $n$ be a positive integer. There are $2n$ red lines and $n$ blue lines in general position given on the plane. Prove that there are at least $\frac{(n-1)(n-2)}{2}$ regions with a monochromatic perimeter. (An infinite region is also counted as a region with rays and segments forming its perimeter). | [
"First note that there are totally $\\binom{3n+1}{2} + 1$ regions formed by these lines. So we have to prove that there are at most $\\binom{3n+1}{2} + 1 - \\frac{(n-1)(n-2)}{2} = 4n^2 + 3n$ bi-colored regions.\n\nConsider an arbitrary point $X$ on the plane as \"the infinity point\", such that $X$ is neither insid... | Iran | 37th Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F"
] | English | proof only | null | |
07i5 | *Arman* starts with a number and calculates the sum of the cubes of its digits. He then repeats this procedure with the resulting number, and continues this procedure. *Arman* considers a number to be 'good' if, after a certain number of steps, it reaches $1$. Prove that there exists an arithmetic progression of length... | [
"We define $T(n)$ to be the sum of the cubes of digits of $n$.\n\nIn this solution first we give an example of an arithmetic progression of length $1402$ such that all the numbers have the same non-zero digits with the same amount of recurrence. Then we will have an arithmetic progression of $a_1, \\dots, a_{1402}$... | Iran | 40th Iranian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Primitive roots mod p / p^n",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0ddz | Let $x$, $y$ and $z$ be odd positive integers such that $\gcd(x, y, z) = 1$ and the sum $x^2 + y^2 + z^2$ is divisible by $x + y + z$. Prove that $x + y + z - 2$ is not divisible by $3$. | [
"Suppose there exists a prime divisor $p \\equiv 2 \\pmod{3}$ of $x + y + z$. Since $z \\equiv -(x + y) \\pmod{p}$ we have $2(x^2 + y^2 + xy) \\equiv 0 \\pmod{p}$. Multiplying by $x - y$ we get $x^3 \\equiv y^3 \\pmod{3}$ but this yields $x \\equiv y \\pmod{p}$, because $\\gcd(3, p-1) = 1$. Similarly $x \\equiv z \... | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
05r0 | Problem:
Dans un pays se trouvent 64 villes. Il existe des routes qui relient certaines paires de villes, mais Bob ne sait pas lesquelles (les routes ne se croisent pas mais peuvent éventuellement passer les unes au-dessus des autres ; de plus, les routes peuvent être parcourues dans les deux sens). Alice, elle, conna... | [
"Solution:\n\nOn remarque que $2016 = 64 \\times 63 / 2 = \\binom{64}{2}$, donc il y a exactement 2016 paires de villes. Le réseau routier est un graphe non orienté ayant 64 sommets. Chaque paire de villes correspond à une question, et on veut savoir si en moins de 2016 questions Bob peut déterminer si le graphe es... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07hb | Find all functions $f : \mathbb{Q}[X] \to \mathbb{R}$ such that for all polynomials $P(x)$ and $Q(x)$ in $\mathbb{Q}[X]$;
i. $f(P(Q(x))) = f(Q(P(x)))$;
ii. $f(P(x)Q(x)) = f(P(x)) + f(Q(x))$, where $P(x)Q(x) \neq 0$. | [
"Let $P(x)$ be a non-zero polynomial and $d = \\deg P$. We shall prove that $f(P(x)) = Cd$, for some constant $C$, furthermore $f(0) = 0$. It is easy to check that such a function properly works.\nLetting $(P(x), Q(x)) = (P(x), 1)$, $(b, P(x))$ in the second equation as well as $(P(x), Q(x)) = (a, 1)$ in the first ... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | All solutions are f(0) = 0 and f(P) = C·deg(P) for nonzero polynomials P, where C is a real constant. | |
03rd | Suppose that $a, b, c$ are positive real numbers, prove that
$$
1 < \frac{a}{\sqrt{a^2 + b^2}} + \frac{b}{\sqrt{b^2 + c^2}} + \frac{c}{\sqrt{c^2 + a^2}} \le \frac{3\sqrt{2}}{2}.
$$ | [
"Set $x = \\frac{b^2}{a^2}$, $y = \\frac{c^2}{b^2}$, $z = \\frac{a^2}{c^2}$, then $x, y, z \\in \\mathbb{R}^+$ and $xyz = 1$. It suffices to prove that\n$$\n1 < \\frac{1}{\\sqrt{1+x}} + \\frac{1}{\\sqrt{1+y}} + \\frac{1}{\\sqrt{1+z}} \\le \\frac{3\\sqrt{2}}{2}.\n$$\n\nWithout loss of generality, we assume that $x \... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
02py | Problem:
Considere uma urna que contém uma bola preta, quatro bolas brancas e algumas bolas azuis. Uma bola é retirada ao acaso dessa urna, sua cor é observada e a bola é devolvida à urna. Em seguida, retira-se novamente, ao acaso, outra bola dessa urna. Para quais quantidades de bolas azuis, a probabilidade das duas ... | [
"Solution:\n\nChamemos de $n$ o número de bolas azuis da caixa. Quando retiramos as duas bolas, elas podem ser:\n- Duas bolas pretas. A probabilidade é\n$$\n\\frac{1}{n+5} \\times \\frac{1}{n+5} = \\left(\\frac{1}{n+5}\\right)^2\n$$\n- Duas bolas brancas. A probabilidade é\n$$\n\\frac{4}{n+5} \\times \\frac{4}{n+5}... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 1 or 9 | |
0i6x | Problem:
A rubber band is 4 inches long. An ant begins at the left end. Every minute, the ant walks one inch rightwards along the rubber band, but then the band is stretched (uniformly) by one inch. For what value of $n$ will the ant reach the right end during the $n$th minute? | [
"Solution:\nThe ant traverses $1/4$ of the band's length in the first minute, $1/5$ of the length in the second minute (the stretching does not affect its position as a fraction of the band's length), $1/6$ of the length in the third minute, and so on. Since\n$$\n1/4 + 1/5 + \\cdots + 1/9 < 0.25 + 0.20 + 0.167 + 0.... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 7 | |
033c | Problem:
For any positive integer $n$ the sum $1+\frac{1}{2}+\cdots+\frac{1}{n}$ is written in the form $\frac{p_{n}}{q_{n}}$, where $p_{n}$ and $q_{n}$ are coprime numbers.
a) Prove that $3$ does not divide $p_{67}$.
b) Find all $n$, for which $3$ divides $p_{n}$. | [
"Solution:\n\na) Set $S_{n}=1+\\frac{1}{2}+\\cdots+\\frac{1}{n}$. We have that $S_{2}=\\frac{3}{2}$, $S_{7}=\\frac{3121}{140}$,\n$$\n\\begin{aligned}\nS_{22}-S_{7}= & \\frac{1}{8}+\\frac{1}{22}+\\frac{1}{10}+\\frac{1}{20}+\\frac{1}{11}+\\frac{1}{19}+\\frac{1}{14}+\\frac{1}{16}+\\frac{1}{13}+\\frac{1}{17}+ \\\\\n& \... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | Part a: 3 does not divide p_67. Part b: n = 2, 7, 22. | |
0avg | Problem:
A function $f(x)$ satisfies
$$
(2-x) f(x)-2 f(3-x)=-x^{3}+5 x-18
$$
for all real numbers $x$. Solve for $f(0)$. | [
"Solution:\nSet $x=0$ and $x=3$ to obtain the system\n$$\n\\begin{cases}\n2 f(0)-2 f(3) = -18 \\\\\n-f(3)-2 f(0) = -30\n\\end{cases}\n$$\nWe can solve for $f(0)$ from the system and obtain $f(0)=7$."
] | Philippines | 18th PMO National Stage Oral Phase | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 7 | |
0kff | Problem:
After viewing the John Harvard statue, a group of tourists decides to estimate the distances of nearby locations on a map by drawing a circle, centered at the statue, of radius $\sqrt{n}$ inches for each integer $2020 \leq n \leq 10000$, so that they draw 7981 circles altogether. Given that, on the map, the J... | [
"Solution:\n\nConsider a coordinate system on any line $\\ell$ where 0 is placed at the foot from $(0,0)$ to $\\ell$. Then, by the Pythagorean theorem, a point $(x, y)$ on $\\ell$ is assigned a coordinate $u$ for which $x^{2}+y^{2}=u^{2}+a$ for some fixed $a$ (dependent only on $\\ell$). Consider this assignment of... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 49 | |
0gkg | Let $\Omega$ be a circumcircle centered at $O$ of $\triangle ABC$ with $\angle B > 90^\circ$. Let $B_1$ be the intersection point of the line $AB$ and the tangent line to the circle $\Omega$ at the point $C$. Let $O_1$ be the circumcenter of $\triangle AB_1C$. Choose a point $B_2$ on the line segment $BB_1$ ($B_2 \neq ... | [
"**Solution.** Here we use directed angles measured in the counterclockwise direction.\nSince $AC_1$ is the common chord of the circles $O$ and $O_2$, we see that $OO_2 \\perp AC_1$. Therefore, $OO_2 \\perp AO_1$ is equivalent to that $O_1$ lies on $AC_1$.\nAssume that $O_1$ lies on $AC_1$. (Claim that $O_2$ lies o... | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0hq1 | Problem:
Find the largest integer that divides $m^{5}-5 m^{3}+4 m$ for all $m \geq 5$. | [
"Solution: $.5!=120$."
] | United States | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 120 | |
05v1 | Problem:
Soit $(a_{n})$ une suite strictement croissante d'entiers positifs telle que $a_{1}=1$ et pour tout $n \geqslant 1$, $a_{n+1} \leqslant 2n$. Montrer que pour tout entier $n \geqslant 1$, il existe deux indices $p$ et $q$ tels que $a_{p}-a_{q}=n$. | [
"Solution:\n\nSoit $n \\geqslant 1$ un entier. On va utiliser le principe des tiroirs. Construisons nos tiroirs de sorte que si deux nombres sont dans un même tiroir, ils vérifient la propriété de l'énoncé. Pour cela, on considère les tiroirs suivants : on réunit les entiers de $1$ à $2n$ en les couples de la forme... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0huz | Problem:
Some people like to write with larger pencils than others. Ed, for instance, likes to write with the longest pencils he can find. However, the halls of MIT are of limited height $L$ and width $L$. What is the longest pencil Ed can bring through the halls so that he can negotiate a square turn? | [
"Solution:\n\n$3L$."
] | United States | null | [
"Geometry > Solid Geometry > Other 3D problems"
] | null | final answer only | 3L | |
0du6 | Problem:
Naj bodo $a, b, c, d, e$ in $f$ taka pozitivna realna števila, da je zaporedje $a, b, c, d$ aritmetično, zaporedje $a, e, f, d$ pa geometrično. Dokaži, da velja
$$
b c \geq e f
$$ | [
"Solution:\n\nKer je $a, b, c, d$ aritmetično zaporedje, velja $b = a + \\frac{1}{3}(d - a)$ in $c = a + \\frac{2}{3}(d - a)$. Podobno je $e = a \\left(\\frac{d}{a}\\right)^{1/3}$ in $f = a \\left(\\frac{d}{a}\\right)^{2/3}$. Sledi\n$$\nbc - ef = \\left(\\frac{2}{3}a + \\frac{1}{3}d\\right)\\left(\\frac{1}{3}a + \\... | Slovenia | 45. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0cn3 | For every real numbers $1 < a \le b \le c$, prove the inequality
$$
\log_a b + \log_b c + \log_c a \le \log_b a + \log_c b + \log_a c. \qquad (D. Tereshin)
$$ | [
"Введем переменные $x = \\log_a b$, $y = \\log_b c$. В новых переменных неравенство принимает вид\n$$\nx + y + \\frac{1}{xy} \\le \\frac{1}{x} + \\frac{1}{y} + xy,\n$$\nчто после приведения к общему знаменателю переходит в\n$$\n\\frac{(x-1)(y-1)(xy-1)}{xy} \\ge 0.\n$$\nПоследнее неравенство верно, т.к. $x \\ge 1$, ... | Russia | Russian mathematical olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Other"
] | English; Russian | proof only | null | |
05bi | Acute triangle $ABC$ with $AB < AC$ has circumcircle $\omega$. Let $D$, $E$, $F$ be the midpoints of sides $BC$, $CA$, $AB$ correspondingly. Ray $AD$ intersects $\omega$ at point $K$ ($K \neq A$) and the circumcircle of triangle $AEF$ at point $M$ ($M \neq A$). On circle $\omega$ a point $T$ such that $AT \parallel BC$... | [
"Let $O$ be the center of $\\omega$ and $Q$ be the intersection point of lines $DO$ and $AT$. Since $ABCT$ is a trapezoid and a cyclic quadrilateral, it is an isosceles trapezoid and the points $A$ and $T$ are symmetric with respect\n\nFig. 47\n\nFig. 48\n\nto line ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
09qm | Problem:
Vind alle drietallen reële getallen $(x, y, z)$ die voldoen aan
$$
x^{2}+y^{2}+z^{2}+1=x y+y z+z x+|x-2 y+z|
$$ | [
"Solution:\nOplossing I. We herschrijven de gegeven vergelijking als\n$$\n\\frac{1}{2} x^{2}-x y+\\frac{1}{2} y^{2}+\\frac{1}{2} y^{2}-y z+\\frac{1}{2} z^{2}+\\frac{1}{2} z^{2}-z x+\\frac{1}{2} x^{2}+1=|x-2 y+z|,\n$$\noftewel als\n$$\n\\frac{1}{2}(x-y)^{2}+\\frac{1}{2}(y-z)^{2}+\\frac{1}{2}(z-x)^{2}+1=|(x-y)+(z-y)|... | Netherlands | Dutch TST | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All triples of the form (y+1, y, y+1) and (y-1, y, y-1), where y is any real number. | |
04dc | For a positive integer $d$, let $f(d)$ be the smallest positive integer with exactly $d$ positive divisors. (E.g. $f(1) = 1$, $f(5) = 16$, $f(6) = 12$.)
Prove that for any positive integer $k$ the number $f(2^{k-1})$ divides $f(2^k)$. | [
"Let $n = p_1^{a_1} p_2^{a_2} \\cdots p_r^{a_r}, a_i \\ge 0$ be the prime factorization of the number $n$. Then the number of divisors of $n$ is $(a_1+1)(a_2+1)\\cdots(a_r+1)$.\n\nLet $k$ be a positive integer and let $f(2^k) = p_1^{a_1} p_2^{a_2} \\cdots p_r^{a_r}$. Then $a_i = 2^{b_i} - 1$ for some integers $b_i ... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0eub | Determine a polynomial $f(x)$ with integer coefficients which satisfies the following property: There are infinitely many relatively prime positive integers $a, b$ such that $a + b$ divides $f(a) + f(b)$. | [
"Let $f(x)$ be such a polynomial. Define\n$$\ng(x) := \\frac{f(x) - f(-x)}{2}, \\quad h(x) := \\frac{f(x) + f(-x)}{2}.\n$$\nThen $g(x)$ is a polynomial consisting of odd degree monomials of $f(x)$, $h(x)$ is a polynomial consisting of even degree monomials of $f(x)$ (including the constant term), and $f(x) = g(x) +... | South Korea | Korean Mathematical Olympiad Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Exactly those integer-coefficient polynomials whose even-degree part is either zero (so the polynomial has only odd-degree terms) or has at least two nonzero even-degree terms; equivalently, exclude polynomials whose even part consists of a single even-degree monomial. | |
0djm | Every rational number $\frac{p}{q}$ from the open interval $(0, 1)$ is covered by the closed interval
$$
\left[ \frac{p}{q} - \frac{1}{4q^2}, \frac{p}{q} + \frac{1}{4q^2} \right].
$$
Prove that the number $\frac{\sqrt{2}}{2}$ is not covered by any of these intervals. | [
"Suppose the contrary that means for some rational number $\\frac{p}{q}$, we have\n$$\n\\left| \\frac{\\sqrt{2}}{2} - \\frac{p}{q} \\right| \\le \\frac{1}{4q^2}.\n$$\nAs $\\frac{\\sqrt{2}}{2} + \\frac{p}{q} < 2$, multiplying the last two inequalities gives\n$$\n\\left| \\frac{1}{2} - \\frac{p^2}{q^2} \\right| < \\f... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
06d5 | Determine all primes of the form $n^2 + 1$, which are less than $10^{19}$ ($n$ is a positive integer). | [
"The only possible primes are $2$, $5$, $257$.\n\nIf $n = pm$ for some odd prime $p$, then we have\n$$\nn^n + 1 = (n^m)^p + 1 = (n^m + 1)(n^{m(p-1)} + n^{m(p-2)} + \\cdots + 1).\n$$\nClearly, both factors are larger than $1$, and hence $n^n+1$ is not a prime. Therefore, $n$ must be a power of $2$.\n\nNote that\n$$\... | Hong Kong | HKG TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | null | proof and answer | 2, 5, 257 | |
01a6 | Prove that there are at least 2013 different values of $m$, such that $m \times m$ square grid can be cut along grid lines into pieces $D_1, D_2, \dots, D_n$, such that $D_i$ consists of exactly $i$ cells. | [
"The sum of areas of $D_1, \\dots, D_n$ should be a perfect square:\n$$1 + 2 + \\dots + n = m^2 \\Rightarrow \\qquad (1)$$\n$$\\frac{n(n+1)}{2} = m^2 \\Rightarrow$$\n$$(2n+1)^2 = 8m^2 + 1$$\nIf we denote $x = 2n+1$ and $y = 2m$ then we have to find 2013 solutions to the equation $x^2 = 2y^2 + 1$ such that $x$ is od... | Baltic Way | Baltic Way 2013 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0cju | Determine all sequences $(a_n)_{n \ge 1}$ of positive rational numbers satisfying
$$
a_{k+1} = \frac{a_k^2 + a_{k+2}^2}{a_k + a_{k+2}} \quad \text{for all } k \ge 1. \quad (*)
$$ | [
"The required sequences are all constant sequences of positive rational numbers. Clearly, any such satisfies $(*)$.\n\nLet $(a_n)_{n \\ge 1}$ be a sequence of positive rational numbers satisfying $(*)$. Then so does $(r a_n)_{n \\ge 1}$, where $r$ is any positive rational number. Letting $r$ be the product of the d... | Romania | 75th NMO Selection Tests | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Prealgebra / Basic Algebra > Intege... | English | proof and answer | All constant sequences of positive rational numbers. | |
04lj | Sequences $(x_n)$ and $(y_n)$ are given recursively:
$x_1 = 3$, $y_1 = 1$,
$x_{n+1} = 3x_n + y_n$, for all $n \in \mathbb{N}$;
$y_{n+1} = x_n + 3y_n$, for all $n \in \mathbb{N}$.
Prove that $x_{2017}^2 - y_{2017}^2 = 8^{2017}$. | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0j59 | Problem:
Denote $\{1,2, \ldots, n\}$ by $[n]$, and let $S$ be the set of all permutations of $[n]$. Call a subset $T$ of $S$ good if every permutation $\sigma$ in $S$ may be written as $t_{1} t_{2}$ for elements $t_{1}$ and $t_{2}$ of $T$, where the product of two permutations is defined to be their composition. Call ... | [
"Solution:\n\nCall an element $t \\in S$ an involution if and only if $t^{2}$ is the identity permutation. We claim that the set of all involutions in $S$ constitutes a good subset of $S$. The proof is simple. Let $s$ be an arbitrary permutation in $S$. Note that $s$ may be decomposed into a product of disjoint cyc... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
041z | A non-empty set $A \subseteq \{1, 2, 3, \dots, n\}$ is called a *good set* of degree $n$ if $|A| \le \min_{x \in A} x$. Denote by $a_n$ the number of good sets of degree $n$. Prove that $a_{n+2} = a_{n+1} + a_n + 1$ for any positive integer $n$. (posed by Li Weigu) | [
"Let $A$ be a good set of degree $n$, and $|A| = k$, then $\\min_{x \\in A} x \\ge k$, so $A \\subseteq \\{k, k+1, \\dots, n\\}$. Hence the number of good sets of degree $n$ with $k$ elements is $C_{n-k+1}^k$. It follows that $a_n = \\sum_{k=1}^{\\lfloor \\frac{n+1}{2} \\rfloor} C_{n-k+1}^k = C_n^1 + C_{n-1}^2 + C_... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0bkt | Prove that if $41$ divides $\overline{abcde}$, then $41$ divides $\overline{eabcd}$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof only | null | |
0b8f | Find all functions $f : \mathbb{N} \to \mathbb{N}$ such that
$$
3f(f(f(n))) + 2f(f(n)) + f(n) = 6n, \text{ for all } n \in \mathbb{N}.
$$
Vasile Pop | [
"We firstly notice that the function is injective. Indeed, if $f(n) = f(m)$, then $3f(f(f(n))) + 2f(f(n)) + f(n) = 3f(f(f(m))) + 2f(f(m)) + f(m)$, hence $6n = 6m$, so $n = m$.\n\nPlugging $n = 0$ yields $3f(f(f(0))) + 2f(f(0)) + f(0) = 0$, so $f(0) = 0$. We prove, using induction, that $f(n) = n$, for every $n \\in... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(n) = n for all n ∈ ℕ | |
05kk | Problem:
Déterminer tous les nombres réels $t$ tels que si $a, b, c$ sont les longueurs des côtés d'un triangle, alors il en est de même pour $a^{2}+b c t, b^{2}+c a t, c^{2}+a b t$. | [
"Solution:\n\nPar symétrie, nous pouvons ordonner $0<a \\leq b \\leq c$. De plus, les conditions de l'énoncé étant homogènes en $a, b, c$, nous pouvons fixer $a=1$. La seule inégalité triangulaire non triviale à exploiter sera donc $c<1+b$. Comme nous allons le voir, le fait d'ordonner $a, b, c$ permet de rendre de... | France | EGMO 2014, 1ère journée | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | [2/3, 2] | |
02u9 | Problem:
Sejam $p$, $q$ e $r$ três números primos maiores que $3$. Sabe-se que o número $p+q+r$ também é primo. Mostre que $p+q$, $p+r$ ou $q+r$ é um múltiplo de $6$. | [
"Solution:\n\nPrimeiramente vejamos que, se um número $n$ é maior que $3$ e deixa resto $0$, $2$, $3$ ou $4$ na divisão por $6$, então esse número não pode ser primo. Para isso, basta mostrar que o número pode ser escrito como o produto de dois números maiores que $1$:\n$$\n\\begin{aligned}\n& n=6m+0=2 \\cdot 3 \\c... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
0g48 | Problem:
Let $k_{1}$ be a circle centred at $M$ and $\ell$ a line tangent to $k_{1}$ at $A$. Let $k_{2}$ be a circle inside $k_{1}$ also tangent to $\ell$ at $A$. Let $P$ be a point on $\ell$ different from $A$. The second tangent to $k_{1}$ through $P$ touches $k_{1}$ at $T$. Let $B$ be the second intersection of $A T... | [
"Solution:\nFirst, observe that\n(1) $\\angle P A M = \\angle P T M = 90^{\\circ}$ since lines $P A$ and $P T$ are tangent to circle $k_{1}$.\n(2) $P A$ is tangent to both circles $k_{1}$ and $k_{2}$ by the problem condition.\n(3) $P A = P T$ since both lines $P A$ and $P T$ are tangent to circle $k_{1}$. Hence tri... | Switzerland | Second round 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
026s | Problem:
A soma dos primos de 1 até 1000 é no máximo quanto?
Há muitos anos, um professor que não queria dar aula, ordenou que seus alunos calculassem a soma dos números de 1 até 100. Um aluno muito esperto, chamado Gauss, descobriu um jeito muito simples de realizar a tarefa descobrindo a fórmula:
$$
1+2+3+\ldots+n=... | [
"Solution:\n\na. Seja $p$ um primo maior que 3. Este primo, como qualquer número inteiro positivo, só pode deixar restos de 0 a 5 na divisão por 6. Veja que se ele deixasse os restos $0, 2, 3$ ou $4$, poderíamos fatorá-lo:\n$$\n\\begin{aligned}\n& p=6q+0=2 \\cdot 3q \\\\\n& p=6q+2=2 \\cdot (3q+1) \\\\\n& p=6q+3=3 \... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | < 166000 | |
08vz | For a triangle $ABC$ the line tangent to its circum-circle at $A$ and the line $BC$ intersects at a point $P$. Let $Q$ and $R$ be the points which are symmetrically located from $P$ with respect to the lines $AB$ and $AC$, respectively. Prove that the lines $BC$ and $QR$ intersect perpendicularly. | [
"Without loss of generality, we may assume that the points $B$, $C$, $P$ lie on the same straight line in this order. Let us denote by $Q'$ the point of intersection of the lines $AB$ and $PQ$ and by $R'$ the point of intersection of the lines $AC$ and $PR$.\nIf $\\angle ACB \\neq 90^\\circ$, we see from $PQ' \\per... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0btc | The positive integers $m$ and $n$ are such that $m^{2016} + m + n^2$ is divisible with $mn$.
a) Give an example of such $m$ and $n$, with $m > n$.
b) Prove that $m$ is a perfect square. | [
"a) For instance, $m = 4$, $n = 2$.\n\nb) Let $d$ be the greatest common divisor of $m$ and $n$, and $a, b \\in \\mathbb{N}^*$ be such that $m = d a$, $n = d b$, with $(a, b) = 1$.\nThe initial condition becomes: $d^{2016} a^{2016} + d a + d^2 b^2$ is divisible with $d^2 a b$. So $d a b$ divides $d^{2015} a^{2016} ... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | Example: m = 4, n = 2; Conclusion: m is a perfect square. |
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