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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
02xg | Problem:
Existem 100 números reais distintos arranjados ao redor de um círculo. Verifique que existem quatro números consecutivos ao redor do círculo de modo que a soma dos dois números do meio é estritamente menor que a soma dos outros dois números. | [
"Solution:\n\nSeja $a$ o menor número escrito no círculo e sejam $b$ e $c$ seus dois vizinhos, com $b < c$. Seja $d$ o outro vizinho de $b$. Assim, estarão escritos no círculo, em ordem, $d, b, a$ e $c$ ou $c, a, b$ e $d$. Em qualquer caso, como $a < d$ e $b < c$, temos $a + b < c + d$."
] | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
07os | Each square of an infinite square grid is to be coloured black or white in such a way that every $3 \times 4$ or $4 \times 3$ rectangle in the grid contains exactly $4$ black squares. In how many ways can this be done? | [
"The key to the solution is the following observation.\n\n(A) Each $1 \\times 3$ rectangle contains exactly one black square.\n\nConsider any $1 \\times 3$ rectangle and let $r$ be the number of black squares it contains, then $0 \\le r \\le 3$ and we want to show $r = 1$.\n\nObserve first that the two $3 \\times 3... | Ireland | Irska 2014 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 6 | |
092u | Problem:
A positive integer $n$ is called a Mozartian number if the numbers $1,2, \ldots, n$ together contain an even number of each digit (in base 10).
Prove:
(a) All Mozartian numbers are even.
(b) There are infinitely many Mozartian numbers. | [
"Solution:\n(a) Note that we need an even number of digits altogether if every digit occurs an even number of times. There is an odd number of numbers with one digit. For $k>1$, there are $9 \\cdot 10^{k-1}$ numbers with $k$ digits, which is an even number. Thus we need to end after a segment of odd length of numbe... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | null | proof only | null | |
0kad | Problem:
Let $A=\left\lceil 1 / H_{3}\right\rceil$, $B=\left\lceil H_{5} / 2\right\rceil$. How many ways are there to partition the set $\{1,2, \ldots, A+B\}$ into two sets $U$ and $V$ with size $A$ and $B$ respectively such that the probability that a number chosen from $U$ uniformly at random is greater than a numbe... | [
"Solution:\n\n$A=4$, $B=7$. There are $28$ total ways of choosing an element from $U$ and $V$, so there must be $14$ ways where $U$'s is larger. If we relabel the elements to be $0,1, \\cdots, 10$, then element $i$ is greater than exactly $i$ elements in the set. However, we overcount other elements in $U$, so the ... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 24 | |
04l1 | Find all rational numbers $a$ and $b$ such that
$$
sin 75^\circ \cdot \cos 15^\circ = a + \sqrt{b}.
$$
(Nikola Adžaga) | [] | Croatia | Mathematical competitions in Croatia | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | a = 1/2, b = 3/16 | |
0b3v | Problem:
A lattice point is a point $(x, y)$ where $x$ and $y$ are both integers. Find the number of lattice points that lie on the closed line segment whose endpoints are $(2002,2022)$ and $(2022,2202)$.
(a) 20
(b) 21
(c) 22
(d) 23 | [] | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | b | |
09cd | $2 < p \in \mathbb{P}$, $a_1, \dots, a_s$-нь $\mathbb{Z}_p$-ийн ялгаатай элементүүд, $b_1, \dots, b_s$ нь мөн $\mathbb{Z}_p$-ийн ялгаатай элементүүд бол $a_1 + b_{\sigma(1)}, \dots, a_s + b_{\sigma(s)}$ нь $\mathbb{Z}_p$-ийн ялгаатай элементүүд байх тийм сэлгэмэл $\sigma \in S_s$ олдохыг үзүүл. | [
"**ДБ-В3.** Энэ бодлогын бодолтыг хараахан хийж амжаагүй байна."
] | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | Mongolian | proof only | null | |
0584 | Natural numbers $1$ through $n$ are written on a blackboard. On each move, one erases from the blackboard $2$ or more numbers whose sum is divisible by any of the chosen numbers and writes their sum on the blackboard. Two players make moves by turns and the player who cannot move loses the game. Which player can win th... | [
"a. The first player can replace numbers $1$, $2$, $3$, $6$ with $12$. After that, the blackboard contains numbers $4$, $5$, $12$. In this state, the sum of no two or three numbers on the blackboard is divisible by all the added numbers. Thus the second player cannot move and the first player wins immediately.\n\nb... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | a: first player; b: first player | |
0exr | Problem:
The distance from $A$ to $B$ is $d$ kilometers. A plane $P$ is flying with constant speed, height and direction from $A$ to $B$. Over a period of 1 second the angle $PAB$ changes by $\alpha$ degrees and the angle $PBA$ by $\beta$ degrees. What is the minimal speed of the plane? | [
"Solution:\n\nAnswer: $20\\pi d\\sqrt{(\\alpha\\beta)}$ kilometers per hour.\n\nLet the plane be at height $h$ and a (horizontal) distance $y$ from $A$. Let the angle $PAB$ be $\\theta + \\alpha$ and the angle $PBA$ be $\\phi$. After 1 second, the angle $PAB$ is $\\theta$ and the angle $PBA$ is $\\phi + \\beta$. We... | Soviet Union | 6th ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 20π d √(αβ) kilometers per hour | |
00zp | Problem:
Prove that in every sequence of 79 consecutive positive integers written in the decimal system, there is a positive integer whose sum of digits is divisible by 13. | [
"Solution:\n\nAmong the first 40 numbers in the sequence, four are divisible by 10 and at least one of these has its second digit from the right less than or equal to 6. Let this number be $x$ and let $y$ be its sum of digits. Then the numbers $x, x+1, x+2, \\ldots, x+39$ all belong to the sequence, and each of $y,... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
00lj | Let $x_1, x_2, \dots, x_9$ be nonnegative real numbers satisfying
$$
x_1^2 + x_2^2 + \dots + x_9^2 \geq 25.
$$
*Prove that there exist three of these numbers with a sum of at least 5.* | [
"W.l.o.g. we may assume that $x_1 \\ge x_2 \\ge x_3 \\ge x_4 \\ge x_5 \\ge x_6 \\ge x_7 \\ge x_8 \\ge x_9 \\ge 0$. Then it follows that $x_1x_2 \\ge x_4^2 \\ge x_5^2$, $x_1x_3 \\ge x_6^2 \\ge x_7^2$ and $x_2x_3 \\ge x_8^2 \\ge x_9^2$. Hence we have\n$$\n(x_1+x_2+x_3)^2 = x_1^2 + x_2^2 + x_3^2 + 2x_1x_2 + 2x_1x_3 + ... | Austria | 48th Austrian Mathematical Olympiad Regional Competition (Qualifying Round) | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
04j7 | Determine the least positive integer $n$ such that in every set consisting of $n$ integers there are three pairwise distinct elements $a$, $b$ and $c$ such that $ab + bc + ca$ is divisible by $3$. (Ilko Brnetić) | [] | Croatia | Croatia Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof and answer | 6 | |
05h1 | Problem:
Trouver toutes les applications $f: \mathbb{R} \longrightarrow \mathbb{R}$ telles que pour tous $x, y$ dans $\mathbb{R}$,
$$
f(x+y)^2 - f\left(2x^2\right) = f(y+x) f(y-x) + 2x f(y)
$$ | [
"Solution:\nEn posant $x = y = 0$, on obtient que $f(0)^2 - f(0) = f(0)^2$, donc $f(0) = 0$.\n\nEn prenant $y = 0$, on obtient que\n$$\nf(x)^2 = f\\left(2x^2\\right) + f(x) f(-x)\n$$\ndonc $f^2$ est paire, i.e. $f(x)^2 = f(-x)^2$ pour tout $x$.\n\nEn remplaçant dans l'équation $x$ par $-x$, on obtient\n$$\nf(y-x)^2... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | f(x) = 0 for all x, or f(x) = x for all x | |
0jj1 | Problem:
Two fair octahedral dice, each with the numbers $1$ through $8$ on their faces, are rolled. Let $N$ be the remainder when the product of the numbers showing on the two dice is divided by $8$. Find the expected value of $N$. | [
"Solution:\n\nIf the first die is odd, which has probability $\\frac{1}{2}$, then $N$ can be any of $0, 1, 2, 3, 4, 5, 6, 7$ with equal probability, because multiplying each element of $\\{0, \\ldots, 7\\}$ with an odd number and taking modulo $8$ results in the same numbers, as all odd numbers are relatively prime... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | final answer only | 11/4 | |
0e9j | Problem:
Izračunaj vrednosti parametra $n$, tako da bo graf funkcije $f$ s predpisom $f(x)=2x+n$ tvoril s koordinatnima osema trikotnik s ploščino 25. | [
"Solution:\n\nGraf funkcije $f(x) = 2x + n$ seka $y$-os pri $x = 0$, torej v točki $(0, n)$.\n\nZa presečišče z $x$-osjo postavimo $f(x) = 0$:\n$$\n2x + n = 0 \\implies x = -\\frac{n}{2}\n$$\nTorej je presečišče z $x$-osjo v točki $\\left(-\\frac{n}{2}, 0\\right)$.\n\nTrikotnik, ki ga graf tvori s koordinatnima ose... | Slovenia | 14. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | n = 10 or n = -10 | |
0a2x | Xander draws five points and a number of infinitely long lines on an infinite sheet of paper. He does this in such a way that on each line there are at least two of those points and that the lines intersect only at points that Xander has drawn.
What is the maximum number of lines Xander could have drawn?
A) 3 B) 4 ... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Combinatorial Geometry > Sylvester's theorem"
] | null | MCQ | B | |
006l | ¿Es posible colorear los puntos del plano que tienen coordenadas enteras con tres colores (deben usarse los tres colores) de manera que no haya ningún triángulo rectángulo con los tres vértices de colores diferentes? | [] | Argentina | XVII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | No | |
0b53 | Problem:
Fixăm un număr întreg $k \geq 2$. Determinaţi cel mai mic număr întreg $n$, astfel încât, printre oricare $n$ puncte în plan, să existe $k$ puncte între care fie toate distanţele sunt mai mici sau egale cu $2$, fie toate distanţele sunt strict mai mari decât $1$. | [] | Romania | TESTUL 1 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | (k - 1)^2 + 1 | |
0diy | For each $1 \le i \le 9$ and $T \in \mathcal{N}$, define $d_i(T)$ to be the total number of times the digit $i$ appears when all the multiples of $2023$ between $1$ and $T$ inclusive are written out in base $10$. Show that there are infinitely many $T \in \mathcal{N}$ such that there are precisely two distinct values a... | [
"Let $n = 2023$. First, we choose some $k$ such that $n \\mid 10^k - 1$. For instance, any multiple of $\\varphi(n)$ would work since $n$ is coprime to $10$. We still show that either $T = 10^k - 1$ or $T = 10^k - 2$ has the desired property, which completes the proof since $k$ can be taken to be arbitrarily large.... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof only | null | |
0c2s | Let $n$ be a positive integer and fix $2n$ distinct points on a circumference. Split these points into $n$ pairs and join the points in each pair by an arrow (i.e., an oriented line segment). The resulting configuration is good if no two arrows cross, and there are no arrows $\overrightarrow{AB}$ and $\overrightarrow{C... | [
"*First solution.* The required number is $\\binom{2n}{n}$. To prove this, trace the circumference counterclockwise to label the points $a_1, a_2, \\dots, a_{2n}$.\nLet $\\mathcal{C}$ be any good configuration and let $O(\\mathcal{C})$ be the set of all points from which arrows emerge. We claim that every $n$-eleme... | Romania | Eleventh Romanian Master of Mathematics | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / ... | null | proof and answer | \binom{2n}{n} | |
0e40 | Problem:
Ali obstaja celo število $n$, za katerega so vse ničle polinoma $p(x) = x^{4} - 2011 x^{2} + n$ cela števila? | [
"Solution:\n\nRecimo, da tako število obstaja. Iz $x^{4} - 2011 x^{2} + n = 0$ sledi\n$$\nx^{2} = \\frac{2011 \\pm \\sqrt{2011^{2} - 4 n}}{2}\n$$\nKer je to število celo, je $2011^{2} - 4 n$ popolni kvadrat. Zapišemo lahko $2011^{2} - 4 n = m^{2}$ za neko liho naravno število $m$ oziroma $n = \\frac{2011^{2} - m^{2... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No | |
0i0l | Problem:
Equilateral triangle $A B C$ is inscribed in a circle. Let $D$ be the midpoint of $A B$, and let $E$ be the midpoint of $A C$. The ray $\overrightarrow{D E}$ meets the circle at $P$. Prove that $D E^{2} = D P \cdot P E$. | [
"Solution:\n\nIf $Q$ is the other intersection of the circle with line $D E$, then one sees by symmetry that line $D E$ is parallel to $B C$ and $Q D = P E$. Then, by power of a point at $D$,\n$$\nP E \\cdot P D = Q D \\cdot P D = B D \\cdot A D = (A B / 2)^{2} = (B C / 2)^{2} = D E^{2}.\n$$\n(The last equality hol... | United States | Berkeley Math Circle | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0acv | The natural number $n > 1$ is "good" if for every natural numbers $b_1, b_2, \dots, b_{n-1}$ such that $1 \le b_1, b_2, \dots, b_{n-1} \le n-1$ we have there exist $I \subseteq \{1, 2, \dots, n-1\}$, such that $\sum_{k \in I} b_k \equiv i \pmod n$, for every $i \in \{0, 1, \dots, n-1\}$ (empty sum is equal to zero). Fi... | [
"We will prove that $n$ is a \"good\" number if and only if it is prime.\n\nFirst we will prove that if $n$ is not prime, then it is not \"good\". Let $n = rs$, $1 < r, s < n$. For $b_1 = b_2 = \\dots = b_{n-1} = r$ then\n$$\n\\{\\sum b_i \\pmod n \\mid I \\subseteq \\{1, 2, \\dots, n-1\\}\\} = \\{0, r, 2r, \\dots,... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | exactly the prime numbers | |
0kpm | Problem:
Three distinct vertices are randomly selected among the five vertices of a regular pentagon. Let $p$ be the probability that the triangle formed by the chosen vertices is acute. Compute $10 p$. | [
"Solution:\n\nThe only way for the three vertices to form an acute triangle is if they consist of two adjacent vertices and the vertex opposite their side. Since there are 5 ways to choose this and $\\binom{5}{3}=10$ ways to choose the three vertices, we have $p=\\frac{5}{10}=\\frac{1}{2}$."
] | United States | HMMT February | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 5 | |
0cs6 | Pete has put several tokens into some squares of a checkered $50 \times 50$ board (at most one token per square). Prove that Bazil can put at most 99 tokens into empty squares so that each row and each column contains an even number of tokens. | [
"11.8. См. решение задачи 10.8."
] | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
0ckm | Let $n \in \mathbb{N}$, $n \ge 2$. Consider the equation:
$$
\{x\} + \{2x\} + \dots + \{nx\} = [x] + [2x] + \dots + [2nx].
$$
a) Solve the equation in $\mathbb{R}$ for $n = 2$.
b) Prove that the equation has at most two real solutions for any $n \ge 2$. | [
"Since each fractional part $\\{kx\\} \\in [0, 1)$, the right-hand side is a sum of integers, hence nonnegative, so $x \\ge 0$.\n\na) For $n = 2$, the equation becomes $\\{x\\} + \\{2x\\} = [x] + [2x] + [3x] + [4x]$. The left side is in $[0, 2)$ and the right side is an integer number, so the right side is in $\\{0... | Romania | 75th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | x = 0 for n = 2 | |
031j | Problem:
In a group of $n$ people there are three that are familiar to each other and any of them is familiar with more than half of the people in the group. Find the minimum possible triples of familiar people? | [
"Solution:\n\nDenote by $A$, $B$ and $C$ the three familiar people in the group.\nLet $n=2k+1$ be an odd integer. Then any of $A$, $B$ and $C$ has at least $k+1$ familiar ( $k-1$ of them are not $A$, $B$ or $C$ ). Denote by $T$ the set of all people except $A$, $B$ and $C$ and let $a_{i}$, $i=0,1,2,3$, be the set o... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | For n = 2k + 1, the minimum is k; for n = 2k, the minimum is k + 1. | |
03ah | Planes through the points with integer coordinates in the three dimensional Euclidean space partition the space into unit cubes. Find all triples $(a, b, c)$, $a \le b \le c$, of positive integers such that the cubes can be colored in $abc$ colors in such a way that every parallelepiped of dimensions $a \times b \times... | [
"We shall prove that the solutions are the triples $(a, b, c)$ such that $a$ divides $b$ and $b$ divides $c$. We denote by $((x_0, y_0, z_0), p, q, r)$ the parallelepiped with a low right vertex $(x_0, y_0, z_0)$ and dimensions $p, q$ and $r$, at axes $Ox, Oy$ and $Oz$, respectively.\n\nLet us assume that $b$ is no... | Bulgaria | 58. National mathematical olympiad Final round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof and answer | (a, b, c) with a | b and b | c | |
0320 | Problem:
Prove that there are no positive integers $m$ and $n$ such that
$$
m(m+1)(m+2)(m+3)=n(n+1)^2(n+2)^3(n+3)^4
$$ | [
"Solution:\nWe shall use that if $\\left(\\sqrt{a^{2}-1}+1\\right)^{k}=x_{k} \\sqrt{a^{2}-1}+y_{k}$, then all the solutions of Pell's equation $\\left(a^{2}-1\\right) x^{2}+1=y^{2}$ are $\\left(x_{k}, y_{k}\\right)$. This implies that $\\left(a^{2}-1\\right) x^{2}+1$ is a perfect square if and only if $x$ is a term... | Bulgaria | Team selection test for 44. IMO | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof only | null | |
09yg | Problem:
Zij $p>10$ een priemgetal. Bewijs dat er positieve gehele getallen $m$ en $n$ met $m+n<p$ bestaan waarvoor $p$ een deler is van $5^{m} 7^{n}-1$. | [
"Solution:\n\nOplossing I. Wegens de kleine stelling van Fermat geldt $a^{p-1} \\equiv 1 \\bmod p$ voor alle $a$ met $p \\nmid a$. Omdat $p>10$ is $p$ oneven, dus $p-1$ is even. Er geldt\n$$\n\\left(a^{\\frac{p-1}{2}}-1\\right)\\left(a^{\\frac{p-1}{2}}+1\\right)=a^{p-1}-1 \\equiv 0 \\quad \\bmod p\n$$\nDus $p \\lef... | Netherlands | IMO-selectietoets III | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0cg0 | Let $ABCD$ be a square, $M$ be the midpoint of the side $AD$, $T$ be the common point of the lines $BM$ and $CD$, and $CP \perp BM$, $P \in MB$. The perpendicular line from $A$ to $AP$ intersects the line $BM$ in $Q$. Prove that:
a) $\angle APQ = \angle PCQ = 45^\circ$;
b) $PQ = QT = PC$. | [
"a) Let $F$ be the intersection of the lines $CP$ and $AB$ and $E$ be the foot of the perpendicular from $A$ to $BM$.\n\nSince $\\angle FCB = \\angle MBA = 90^\\circ - \\angle CBM$ and $CB = BA$, the right triangles $CBF$ and $BAM$ are congruent, whence $FB = MA = \\frac{AB}{2}$, so $F$ is ... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0b0q | Problem:
The circle $(x+3)^2+(y-4)^2=50$ and the line $y=2x-5$ intersect at two points. Determine the distance between these two points. | [
"Solution:\nFrom the given, we know $(x+3)^2+[(2x-5)-4]^2=50$.\n$$\n\\begin{aligned}\n(x+3)^2+(2x-9)^2 & =50 \\\\\n\\left(x^2+6x+9\\right)+\\left(4x^2-36x+81\\right) & =50 \\\\\n5x^2-30x+40 & =0 \\\\\n5(x-2)(x-4) & =0\n\\end{aligned}\n$$\nThus, $x=2$ or $x=4$. If $x=2$, then $y=-1$. If $x=4$, then $y=3$. The distan... | Philippines | Philippine Mathematical Olympiad, National Orals | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 2*sqrt(5) | |
004p | Sea $n > 2$ un entero par. En las casillas de un tablero de $n \times n$ se deben colocar fichas de modo que en cada columna la cantidad de fichas sea par y distinta de cero, y en cada fila la cantidad de fichas sea impar.
Determinar la menor cantidad de fichas que hay que colocar en el tablero para cumplir esta regla... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | Español | proof and answer | 2n | |
0af8 | Нека $a$, $b$, $c$ се страните а $\alpha$, $\beta$, $\gamma$ соодветните агли во триаголникот $ABC$ со плоштина $P$. Докажи дека важи равенството
$$
a^2(\sin 2\beta + \sin 2\gamma) + b^2(\sin 2\gamma + \sin 2\alpha) + c^2(\sin 2\alpha + \sin 2\beta) = 12P.
$$ | [
"Ќе ги прегрупираме собироците на левата страна од равенството во облик\n$$\n(a^2 \\sin 2\\beta + b^2 \\sin 2\\alpha) + (b^2 \\sin 2\\gamma + c^2 \\sin 2\\beta) + (c^2 \\sin 2\\alpha + a^2 \\sin 2\\gamma).\n$$\nСо примена на синусната теорема и формула за синус од двоен агол, ќе трансформираме збировите во заградит... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Macedonian, English | proof only | null | |
00os | A $9 \times 12$ rectangle is divided into unit squares. The centers of all the unit squares, except the four corner squares and the eight squares adjacent (by side) to them, are colored red. Is it possible to numerate the red centers by $C_{1}, C_{2}, \ldots, C_{96}$ so that the following two conditions are fulfilled:
... | [
"Place the given rectangle into the coordinate plane so that the center of the square at the intersection of $i$-th column and $j$-th row has the coordinates $(i, j)$. Suppose that a desired numeration of the red points exists; it corresponds to a path, i.e. a closed poligonal line consisting of 96 segments of leng... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Graph Theory",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | No | |
006x | En una circunferencia de centro $O$ sean $A$ y $B$ puntos de la circunferencia tales que $AB = 120^\circ$. El punto $C$ pertenece al menor arco $AB$ y el punto $D$ pertenece a la cuerda $AB$. Se sabe que $AD = 2$, $BD = 1$ y $CD = \sqrt{2}$. Calcular el área del triángulo $ABC$. | [] | Argentina | Argentina 2009 | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | Spanish | proof and answer | 3√2/4 | |
0emu | Given that
$$
\frac{1 + 3 + 5 + \cdots + (2n - 1)}{2 + 4 + 6 + \cdots + (2n)} = \frac{2011}{2012},
$$
determine $n$. | [
"Using the sum formula for arithmetic progressions, we obtain\n$$\n\\frac{1 + 3 + 5 + \\cdots + (2n - 1)}{2 + 4 + 6 + \\cdots + (2n)} = \\frac{n \\cdot \\frac{1+2n-1}{2}}{n \\cdot \\frac{2+2n}{2}} = \\frac{n}{n+1} = \\frac{2011}{2012},\n$$\nfrom which it follows that $n = 2011$."
] | South Africa | South African Mathematical Olympiad Third Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 2011 | |
0b8a | Let $A_1A_2A_3A_4$ be a convex quadrilateral with no pair of parallel sides. For each $i = 1, 2, 3, 4$, define $\omega_i$ to be the circle touching the quadrilateral externally, and which is tangent to the lines $A_{i-1}A_i$, $A_iA_{i+1}$ and $A_{i+1}A_{i+2}$ (indices are considered modulo 4, so $A_0 = A_4$, $A_5 = A_1... | [
"We start with a reformulation of a well-known statement on harmonic cyclic quadruples $(K_1, K_2, K_3, K_4)$, also provable by polar transformation (projective methods).\n**LEMMA.** Being given four pairwise non-parallel lines $\\ell_i$, $i = 1, 2, 3, 4$, tangent to a circle $\\omega$ at points $K_i$, and such tha... | Romania | Local Mathematical Competitions | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates"
] | English | proof only | null | |
0jvk | Problem:
Let $ABC$ be a triangle with $AB=13$, $BC=14$, $CA=15$. The altitude from $A$ intersects $BC$ at $D$. Let $\omega_{1}$ and $\omega_{2}$ be the incircles of $ABD$ and $ACD$, and let the common external tangent of $\omega_{1}$ and $\omega_{2}$ (other than $BC$) intersect $AD$ at $E$. Compute the length of $AE$. | [
"Solution:\n\nLet $I_{1}, I_{2}$ be the centers of $\\omega_{1}, \\omega_{2}$, respectively, $X_{1}, X_{2}$ be the tangency points of $\\omega_{1}, \\omega_{2}$ with $BC$, respectively, and $Y_{1}, Y_{2}$ be the tangency points of $\\omega_{1}, \\omega_{2}$ with $AD$, respectively. Let the two common external tange... | United States | HMMT November 2016 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 7 | |
0e74 | What is the greatest common divisor of the numbers $11n + 4$ and $7n + 2$, where $n$ is a positive integer? | [
"The greatest common divisor of $11n+4$ and $7n+2$ also divides $7(11n+4) - 11(7n+2) = 6$, so it can be at most $6$. If $n=4$, then $11n+4 = 48$ and $7n+2 = 30$, and the greatest common divisor of these two numbers is $6$. Hence, the answer is $6$."
] | Slovenia | National Math Olympiad 2013 - Final Round | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 6 | |
03a0 | All edges of a tetrahedron $ABCD$ are equal. Let $M$ be the midpoint of $DB$, $N$ be the point on the extension of $AB$ such that $2NA = NB$ and $P$ be a point on the altitude of $\triangle BCD$ through $D$. Find $\angle MPD$ if the intersection of the tetrahedron and the plane $(NMP)$ is a trapezoid.
Answer. $\angle M... | [] | Bulgaria | Fall Mathematical Competition | [
"Geometry > Solid Geometry > 3D Shapes",
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof and answer | 30° | |
09yx | Two circles $\Gamma_1$ and $\Gamma_2$ are given with centres $O_1$ and $O_2$ and common exterior tangents $\ell_1$ and $\ell_2$. The line $\ell_1$ intersects $\Gamma_1$ in $A$ and $\Gamma_2$ in $B$. Let $X$ be a point on segment $O_1O_2$, not lying on $\Gamma_1$ or $\Gamma_2$. The segment $AX$ intersects $\Gamma_1$ in ... | [
"\n\nWe consider the configuration in which $Y$ lies between $A$ and $X$; the other configurations are treated analogously. Let $C$ be the intersection of $\\ell_2$ and $\\Gamma_1$. Then $C$ is the reflection of $A$ in $O_1O_2$. We get\n$$\n\\begin{align*}\n\\angle O_1YX &= 180^\\circ - \\a... | Netherlands | IMO Team Selection Test 1 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00i8 | Show that $2010$ cannot be written as the difference of two squares. | [
"We assume that there are two integers $x, y$ with\n$$\n2010 = x^2 - y^2 = (x - y)(x + y).\n$$\nThe factors $(x - y)$ and $x + y = (x - y) + 2y$ have the same parity.\n\n• If both of them were odd, the product $2010$ would be odd which also gives a contradiction.\n\n• If both of them were even, the product $2010$ w... | Austria | Austria 2010 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
02w7 | Problem:
Um ponto $D$ é escolhido no lado $BC$ do triângulo $ABC$. A reta tangente aos incírculos dos triângulos $ABD$ e $ADC$ e diferente de $BC$ e $AD$ intersecta o segmento $AD$ em $T$. Se $AB=40\ \mathrm{cm}$, $AC=50\ \mathrm{cm}$ e $BC=60\ \mathrm{cm}$, determine o valor do comprimento de $AT$.
![](attached_imag... | [
"Solution:\n\nSabemos que os comprimentos dos segmentos tangentes traçados de um ponto externo a um círculo são congruentes e que $XY=HI$. Assim,\n$$\n\\begin{aligned}\n2AT &= (AF-TF)+(AJ-TJ) \\\\\n&= AG+AK-(TX+TY) \\\\\n&= (AB-BG)+(AC-CK)-XY \\\\\n&= AB+AC-(BH+HI+IC) \\\\\n&= AB+AC-BC \\\\\n&= 30\n\\end{aligned}\n... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
... | null | proof and answer | 30 cm | |
0has | Solve $\frac{2 \cos 2x}{6 - 3 \cos 3x} = \frac{\cos 2x + 1}{\cos 3x + 2}$ for $-\pi \le x \le \pi$. | [
"The equation can be written as:\n$$\n\\frac{2 \\cos 2x}{\\cos 2x + 1} = \\frac{6 - 3 \\cos 3x}{\\cos 3x + 2}.\n$$\nConsider the following two functions: $f(x) = \\frac{2x}{x+1}$ and $g(x) = \\frac{6-3x}{x+2}$, $x \\in (-1; 1]$. If $x \\in (-1; 1)$, then\n$$\nf(x) = \\frac{2x}{x+1} < 1 \\Leftrightarrow 2x < x+1 \\L... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 0 | |
0iq3 | Let $P$, $Q$, and $R$ be the points on sides $BC$, $CA$, and $AB$ of an acute triangle $ABC$ such that triangle $PQR$ is equilateral and has minimal area among all such equilateral triangles. Prove that the perpendiculars from $A$ to line $QR$, from $B$ to line $RP$, and from $C$ to line $PQ$ are concurrent. | [
"(By Zuming Feng) By Miquel's theorem (which can be shown by simple angle chasing), the circumcircles of triangles $AQR$, $BRP$, and $CPQ$ meet at a common point $X$. The key observation is that $XR \\perp AB$, $XP \\perp BC$, and $XQ \\perp CA$. Indeed, if $P_1Q_1R_1$ is an inscribed equilateral triangle, and the ... | United States | Team Selection Test | [
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
08no | Problem:
Along a round table are arranged 11 cards with the names (all distinct) of the 11 members of the $16^{ ext{th}}$ JBMO Problem Selection Committee. The distances between each two consecutive cards are equal. Assume that in the first meeting of the Committee none of its 11 members sits in front of the card with... | [
"Solution:\n\nYes it is: Rotating the table by the angles $\\frac{360^{\\circ}}{11}, 2 \\cdot \\frac{360^{\\circ}}{11}, 3 \\cdot \\frac{360^{\\circ}}{11}, \\ldots, 10 \\cdot \\frac{360^{\\circ}}{11}$, we obtain 10 new positions of the table. By the assumption, it is obvious that every one of the 11 members of the C... | JBMO | JBMO Shortlist | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | Yes | |
09n4 | Let $\{a_n\}_{n \ge 1}$ be the sequence of positive real numbers satisfying $a_1 = 1$ and
$$
a_{n+1} = a_n + \sqrt{a_n + a_{n+1}}
$$
for $n \ge 1$. Let $b_n = a_{n+1} - a_n$ for $n \ge 1$.
a. Show that $b_n \ge 1$.
b. Show that $a_n = b_n(b_n - 1)/2$.
c. Express $a_n$ in terms of $n$.
d. Find the sum $S = a_1 + a_2... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) b_n ≥ 1. b) a_n = b_n(b_n − 1)/2. c) a_n = n(n + 1)/2. d) S = 37820. | |
0j0p | Problem:
Triangle $A B C$ is given with $A B = 13$, $B C = 14$, $C A = 15$. Let $E$ and $F$ be the feet of the altitudes from $B$ and $C$, respectively. Let $G$ be the foot of the altitude from $A$ in triangle $A F E$. Find $A G$. | [
"Solution:\n\nBy Heron's formula we have $[A B C] = \\sqrt{21 \\cdot 8 \\cdot 7 \\cdot 6} = 84$.\n\nLet $D$ be the foot of the altitude from $A$ to $B C$; then $A D = 2 \\cdot \\frac{84}{14} = 12$.\n\nNotice that because $\\angle B F C = \\angle B E C$, $B F E C$ is cyclic, so $\\angle A F E = 90^\\circ - \\angle E... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 396/65 | |
0g5a | 令 $A_1A_2\cdots A_n$ 是一凸多邊形。點 $P$ 是此多邊形內一點且其在 $A_1A_2, \cdots, A_nA_1$ 之投影點分別為 $P_1, \cdots, P_n$, 其中 $P_1, \cdots, P_n$ 分別落在線段 $A_1A_2, \cdots, A_nA_1$ 內。試證: 對於任意分別在線段 $A_1A_2, \cdots, A_nA_1$ 內之點 $X_1, \cdots, X_n$, 滿足
$$
\max \left\{ \frac{X_1X_2}{P_1P_2} + \cdots + \frac{X_nX_1}{P_nP_1} \right\} \ge 1.
$$ | [
"記 $P_{n+1} = P_1$, $X_{n+1} = X_1$, $A_{n+1} = A_1$.\n\n引理：令 $Q$ 為 $A_1A_2\\cdots A_n$ 內一點。則 $Q$ 必落在三角形 $X_1A_2X_2, \\cdots, X_nA_1X_1$ 之外接圓之其中一個。\n\n證明：若 $Q$ 在三角形 $X_1A_2X_2, \\cdots, X_nA_1X_1$ 之其中一個，則顯然成立。否則 $Q$ 在多邊形 $A_1A_2\\cdots A_n$ 內（如圖1）。則\n$$\n\\begin{aligned}\n& (\\angle X_1 A_2 X_2 + \\angle X_1 Q X_2)... | Taiwan | 二〇一一數學奧林匹亞競賽第二階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0hjr | Problem:
Given are $n+1$ real linear equations in $n$ variables (of the form $a_{1} x_{1}+a_{2} x_{2}+\cdots+a_{n} x_{n}=a$). Prove that each $=$ sign can be replaced with either $\leq$ or $\geq$ so that the resulting $n+1$ inequalities have the following property: for every choice of real numbers $x_{1}, x_{2}, \ldot... | [
"Solution:\n\nHenceforth, we assume that every equation has some $a_{i} \\neq 0$, since otherwise it is of the form $0=a$, and then we can change it to either $0 \\leq a$ or $0 \\geq a$ to form an inequality which is always true. We use induction on $n$.\n\nIf $n=1$, we have the two equations $a x=b$, $c x=d$. We c... | United States | Berkeley Math Circle Take-Home Contest | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Linear Algebra > Vectors",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
00tf | There is a population $P$ of $10000$ bacteria, some of which are friends (friendship is mutual), so that each bacterion has at least one friend and if we wish to assign to each bacterion a coloured membrane so that no two friends have the same colour, then there is a way to do it with $2021$ colours, but not with $2020... | [
"We will use the terminology of graph theory. Here the vertices of our main graph $G$ are the bacteria and there is an edge between two precisely when they are friends. The degree $d(v)$ of a vertex $v$ of $G$ is the number of neighbours of $v$. The minimum degree $\\delta(G)$ of $G$ is the smallest amongst all $d(... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | Yes | |
06s8 | Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=1$, and
$$
u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text{ for } k=1, \ldots, n-1 .$$
Prove tha... | [
"We prove by induction on $k$ that\n$$\n\\begin{equation*}\nu_{k}=\\sum_{\\substack{0<i_{1}<\\ldots<i_{t}<k \\\\ i_{j+1}-i_{j} \\geqslant 2}} a_{i_{1}} \\ldots a_{i_{t}} \\tag{1}\\end{equation*}\n$$\nNote that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product i... | IMO | International Mathematical Olympiad Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Matrices",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
00ax | Find the angles of a convex quadrilateral $ABCD$ such that $\hat{ABD} = 29^\circ$, $\hat{ADB} = 41^\circ$, $\hat{ACB} = 82^\circ$ and $\hat{ACD} = 58^\circ$. | [
"We have $\\hat{BAD} = 180^\\circ - (29^\\circ + 41^\\circ) = 110^\\circ$, $\\hat{BCD} = 82^\\circ + 58^\\circ = 140^\\circ$. Consider the circumcircle $\\gamma$ of $\\triangle BCD$. Since $\\hat{BAD} + \\hat{BCD} > 180^\\circ$, point $A$ is interior to $\\gamma$.\n\nExtend $CA$ beyond $A$ to meet $\\gamma$ at $E$.... | Argentina | Argentina_2017 | [
"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"
] | English | proof and answer | ∠A = 110°, ∠B = 49°, ∠C = 140°, ∠D = 61° | |
0i9s | Problem:
We wish to color the integers $1,2,3, \ldots, 10$ in red, green, and blue, so that no two numbers $a$ and $b$, with $a-b$ odd, have the same color. (We do not require that all three colors be used.) In how many ways can this be done? | [
"Solution:\n\nThe condition is equivalent to never having an odd number and an even number in the same color. We can choose one of the three colors for the odd numbers and distribute the other two colors freely among the 5 even numbers; this can be done in $3 \\cdot 2^{5}=96$ ways. We can also choose one color for ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 186 | |
0aue | Problem:
What is the greatest common factor of all integers of the form $p^{4}-1$, where $p$ is a prime number greater than $5$? | [
"Solution:\n\nLet $f(p) = p^{4} - 1 = (p-1)(p+1)(p^{2}+1)$. Note that $f(7) = 2^{5} \\cdot 3 \\cdot 5^{2}$ and $f(11) = 2^{4} \\cdot 3 \\cdot 5 \\cdot 61$. We now show that their greatest common factor, $2^{4} \\cdot 3 \\cdot 5$, is actually the greatest common factor of all numbers $p^{4}-1$ so described.\n\n- Sin... | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 240 | |
0h7q | Let $AM$ be a median in an acute triangle $ABC$. Its extension intersect the circumcircle $w$ of $ABC$ at $P$. Let $AH_1$ be an altitude of $\triangle ABC$, $H$ - its orthocenter. The rays $MH$ and $PH_1$ intersect $w$ at $K$ and $T$ respectively. Prove that the circumcircle of $\triangle AKTH_1$ is tangent to $BC$.
(... | [
"It suffices to show that $\\angle TKH_1 = \\angle TH_1B$ (Fig. 14). Let us extend $KH_1$ and intersect if with $w$ at $S$. Then\n$$\n\\angle TKS = \\angle TAB + \\angle BAS, \\quad \\angle TH_1B = \\angle TAB + \\angle PAC,\n$$\nSo it is sufficient to show that $\\angle PAC = \\angle BAS$.\nDenote by $A_1$ the poi... | Ukraine | UkraineMO | [
"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... | null | proof only | null | |
0jm2 | Problem:
We have a calculator with two buttons that displays an integer $x$. Pressing the first button replaces $x$ by $\left\lfloor\frac{x}{2}\right\rfloor$, and pressing the second button replaces $x$ by $4x+1$. Initially, the calculator displays $0$. How many integers less than or equal to $2014$ can be achieved th... | [
"Solution:\n\nWe consider the integers from this process written in binary. The first operation truncates the rightmost digit, while the second operation appends $01$ to the right.\n\nWe cannot have a number with a substring $11$. For simplicity, call a string valid if it has no consecutive $1$'s. Note that any num... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 233 | |
06qj | For an integer $m \geq 1$, we consider partitions of a $2^{m} \times 2^{m}$ chessboard into rectangles consisting of cells of the chessboard, in which each of the $2^{m}$ cells along one diagonal forms a separate rectangle of side length 1. Determine the smallest possible sum of rectangle perimeters in such a partition... | [
"Solution 1. For a $k \\times k$ chessboard, we introduce in a standard way coordinates of the vertices of the cells and assume that the cell $C_{ij}$ in row $i$ and column $j$ has vertices $(i-1, j-1), (i-1, j), (i, j-1), (i, j)$, where $i, j \\in \\{1, \\ldots, k\\}$. Without loss of generality assume that the ce... | IMO | IMO Problem Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | (m+1) 2^{m+2} | |
0d08 | Let $M = \{1, 2, 4, 5, 7, 8, \dots\}$ be the set of all positive integers not divisible by $3$. The sum of $2n$ consecutive elements of $M$ is $300$. Determine the possible values of $n$. | [
"Let $S_i$ be the sum of the first $i$ numbers in the set $M$. We have\n$$\n\\begin{aligned}\nS_{2k} &= (1+2) + (4+5) + \\dots + (3k-2+3k-1) \\\\\n&= (6 \\cdot 1 - 3) + (6 \\cdot 2 - 3) + \\dots + (6k-3) \\\\\n&= 3k(k+1) - 3k = 3k^2.\n\\end{aligned}\n$$\nThen $S_{2k+1} = S_{2k} + 3k + 1 = 3k^2 + 3k + 1$.\n\n**Case ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | {1, 2, 4, 5, 10} | |
0hd8 | Prove that there are no two distinct positive integers $a$ and $b$, such that $\{\frac{a}{b}\} + \{\frac{b}{a}\} = 0$.
Here $\{x\}$ denotes the difference between $x$ and the greatest integer that does not exceed $x$, for example, $\{\frac{7}{5}\} = \frac{2}{5}$, $\{\frac{2019}{3}\} = 0$ and $\{\frac{2020}{3}\} = \frac... | [
"Suppose such two numbers $a$ and $b$ exist. Without loss of generality, one can assume they are relatively prime. The condition of the problem implies that the number $\\frac{a+b}{a} = \\frac{a^2+b^2}{ab}$ is an integer. But then $a^2 + b^2$ has to be divisible by both $a$ and $b$. Hence $b^2 \\mid a$ and $a^2 \\m... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
01n4 | For any point $X$ inside an acute-angled triangle $ABC$ we define
$$
f(X) = \frac{AX}{A_1X} \cdot \frac{BX}{B_1X} \cdot \frac{CX}{C_1X}
$$
where $A_1, B_1$, and $C_1$ are the intersection points of the lines $AX, BX$, and $CX$ with the sides $BC, AC$, and $AB$, respectively.
Let $H, I$, and $G$ be the orthocenter, the ... | [
"Let $a = BC$, $b = AC$, $c = AB$, and $\\angle A = \\alpha$, $\\angle B = \\beta$, $\\angle C = \\gamma$. Let $AH_1$, $BH_2$, $CH_3$ be altitudes, $AL_1$, $BL_2$, $CL_3$ be angle bisectors, and $AM_1$, $BM_2$, $CM_3$ be medians of $\\triangle ABC$.\nThen\n$$\n\\frac{AI}{L_1I} = \\frac{b+c}{a}, \\quad \\frac{BI}{L_... | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequa... | English | proof only | null | |
0k9w | Problem:
There are 36 students at the Multiples Obfuscation Program, including a singleton, a pair of identical twins, a set of identical triplets, a set of identical quadruplets, and so on, up to a set of identical octuplets. Two students look the same if and only if they are from the same identical multiple. Nithya ... | [
"Solution:\n\nLet $X$ and $Y$ be the students Nithya encounters during the day. The number of pairs $(X, Y)$ for which $X$ and $Y$ look the same is $1 \\cdot 1 + 2 \\cdot 2 + \\ldots + 8 \\cdot 8 = 204$, and these pairs include all the ones in which $X$ and $Y$ are identical. As $X$ and $Y$ are chosen uniformly and... | United States | HMMT November 2019 | [
"Statistics > Probability > Counting Methods > Other",
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 3/17 | |
0i78 | Problem:
Find all values of $x$ with $0 \leq x < 2\pi$ that satisfy $\sin x + \cos x = \sqrt{2}$. | [
"Solution:\nSquaring both sides gives $\\sin^2 x + \\cos^2 x + 2 \\sin x \\cos x = 1 + \\sin 2x = 2$, so $x = \\frac{\\pi}{4}, \\frac{5\\pi}{4}$."
] | United States | Harvard-MIT Math Tournament | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | π/4 | |
0i48 | Problem:
Eight knights are randomly placed on a chessboard (not necessarily on distinct squares). A knight on a given square attacks all the squares that can be reached by moving either (1) two squares up or down followed by one square left or right, or (2) two squares left or right followed by one square up or down. ... | [
"Solution:\n\n$0$. Since every knight attacks at most eight squares, the event can only occur if every knight attacks exactly eight squares. However, each corner square must be attacked, and some experimentation readily finds that it is impossible to place a knight so as to attack a corner and seven other squares a... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 0 | |
0jqh | Problem:
Marcus and four of his relatives are at a party. Each pair of the five people are either friends or enemies. For any two enemies, there is no person that they are both friends with. In how many ways is this possible? | [
"Solution:\n\nDenote friendship between two people $a$ and $b$ by $a \\sim b$. Then, assuming everyone is friends with themselves, the following conditions are satisfied:\n- $a \\sim a$\n- If $a \\sim b$, then $b \\sim a$\n- If $a \\sim b$ and $b \\sim c$, then $a \\sim c$\n\nThus we can separate the five people in... | United States | HMMT November 2015 | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 52 | |
0iv8 | Problem:
A knight begins on the lower-left square of a standard chessboard. How many squares could the knight end up at after exactly 2009 legal knight's moves? (A knight's move is 2 squares either horizontally or vertically, followed by 1 square in a direction perpendicular to the first.) | [
"Solution:\n\nAnswer: 32\n\nThe knight goes from a black square to a white square on every move, or vice versa, so after 2009 moves he must be on a square whose color is opposite of what he started on. So he can only land on half the squares after 2009 moves. Note that he can access any of the 32 squares (there are... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 32 | |
0ezn | Problem:
$A$ and $B$ are adjacent vertices of a $12$-gon. Vertex $A$ is marked $-$ and the other vertices are marked $+$. You are allowed to change the sign of any $n$ adjacent vertices. Show that by a succession of moves of this type with $n = 6$ you cannot get $B$ marked $-$ and the other vertices marked $+$. Show t... | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0cve | An interstellar hotel has 100 rooms, their capacities are $101$, $102$, $\ldots$, $200$ people. These rooms are occupied by $n$ people in total. Now a VIP guest is about to arrive, and the owner wants to provide him with a personal room. For that purpose, the owner wants to choose two rooms, $A$ and $B$, and move all g... | [
"Предположим, что при $8824$ постояльцах директор не может осуществить переселение. Разобьём комнаты на пары по вместимости: $101$ – $200$, $102$ – $199$, $\\ldots$, $150$ – $151$. Отметим, что для каждой пары комнат суммарное количество человек, живущих в двух комнатах, больше, чем вместимость большей комнаты из п... | Russia | Final round | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | 8824 | |
09pp | Problem:
Zij $n \geq 1$. Vind alle permutaties $(a_1, a_2, \ldots, a_n)$ van $(1,2, \ldots, n)$ waarvoor geldt
$$
\frac{a_k^2}{a_{k+1}} \leq k+2 \quad \text{voor } k=1,2, \ldots, n-1
$$ | [] | Netherlands | TOETS TRAININGSKAMP | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a_k = k for all k | |
04d6 | Along the coast of an island there are 20 villages. Each village has 20 fighters. Every fighter fights all the fighters from all the other villages. No two fighters have equal strength and the stronger fighter wins the fight. We say that the village $A$ is stronger than the village $B$ if in at least $k$ fights among t... | [
"We first show that $k > 290$ is impossible. In every village we rank the fighters (from 1 to 20) according to their strength and we consider the tenth fighter in every village. Let the weakest of these considered fighters comes from the village $A$. Then in any other village $B$ (and in particular in the neighbour... | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 290 | |
094y | Problem:
Let $k$ be a positive integer and $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers such that
$$
a_{i} a_{i+1} \mid k-a_{i}^{2}
$$
for all integers $i \geq 1$. Prove that there exists a positive integer $M$ such that $a_{n}=a_{n+1}$ for all integers $n \geq M$. | [
"Solution:\n\nNote that $a_{i} \\mid k$ for all $i \\geq 1$. Furthermore, we have $a_{i+1} \\mid a_{i}^{2}$, so there are only finitely many primes that divide any element of the sequence.\nFor a prime number $p$ and a positive integer $n$, let $\\nu_{p}(n)$ denote the exponent of $p$ in the prime factorization of ... | Middle European Mathematical Olympiad (MEMO) | MEMO Szeged | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
04k2 | In a lake, there are two kinds of fish: red and yellow. Of all the fish in the lake, two fifths are yellow, while the others are red. Three quarters of all the yellow fish are female. If the total number of female fish equals the total number of male fish, what is the percentage of red male fish in the lake? (The Nethe... | [] | Croatia | Mathematical competitions in Croatia | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | 40% | |
09uu | Which of the following numbers is the largest number you can get by separating the numbers $1$, $2$, $3$, $4$, and $5$ by using each of the operations $+$, $-$, $:$, and $\times$ exactly once, where you may use parentheses to indicate the order in which the operations should be executed? For example: $(5 - 3) \times (4... | [
"E) $35$"
] | Netherlands | First Round, January 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | MCQ | E | |
0226 | Problem:
As duas partículas, $A$ e $B$, percorrem uma circunferência de $120\ m$ de comprimento. A partícula $A$ gasta 3 segundos menos que $B$, por estar animada com uma velocidade maior de $2$ metros por segundo. Qual é a velocidade de cada partícula? | [
"Solution:\n\nSeja $v_A$ a velocidade da partícula $A$ e $v_B$ a velocidade da partícula $B$.\n\nSabemos que:\n- $v_A = v_B + 2$\n- O tempo gasto por $A$ é $t_A = \\dfrac{120}{v_A}$\n- O tempo gasto por $B$ é $t_B = \\dfrac{120}{v_B}$\n- $A$ gasta 3 segundos a menos: $t_B = t_A + 3$\n\nSubstituindo as expressões do... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | Particle A: 10 m/s; Particle B: 8 m/s | |
03fz | Given is a triangle $ABC$ and a circle $\omega$ with center $I$ that touches $AB$, $AC$ and meets $BC$ at $X$, $Y$. The line through $I$ perpendicular to $BC$ meets the line through $A$ parallel to $BC$ at $Z$. Show that the circumcircles of $\triangle XYZ$ and $\triangle ABC$ are tangent to each other. | [
"Let $W$ be the midpoint of the major arc $BAC$, $A' \\in (ABC)$ be such that $AA' \\parallel BC$, $T$ be the intersection of the circle with diameter $AI$ and $(ABC)$ and let $\\omega$ touch $AC$, $AB$ at $E$, $F$. We claim the two circles touch at $T$.\n\nFirstly, observe that $Z \\in (AEF)$, so\n$$\n\\begin{alig... | Bulgaria | 5 Bulgarian National Olympiad - Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Ho... | English | proof only | null | |
08m1 | Problem:
Let $x$, $y$, $z$ be positive real numbers. Prove that:
$$
\left(x^{2}+y+1\right)\left(x^{2}+z+1\right)\left(y^{2}+z+1\right)\left(y^{2}+x+1\right)\left(z^{2}+x+1\right)\left(z^{2}+y+1\right) \geq (x+y+z)^{6}
$$ | [
"Solution:\nApplying Cauchy-Schwarz's inequality:\n$$\n\\left(x^{2}+y+1\\right)\\left(z^{2}+y+1\\right)=\\left(x^{2}+y+1\\right)\\left(1+y+z^{2}\\right) \\geq (x+y+z)^{2}\n$$\nUsing the same reasoning we deduce:\n$$\n\\left(x^{2}+z+1\\right)\\left(y^{2}+z+1\\right) \\geq (x+y+z)^{2}\n$$\nand\n$$\n\\left(y^{2}+x+1\\... | JBMO | 2009 Shortlist JBMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0a8o | Problem:
Find all triples of integers $(x, y, z)$ satisfying
$$
x^{3}+y^{3}+z^{3}-3 x y z=2003
$$ | [
"Solution:\nIt is a well-known fact (which can be rediscovered e.g. by noticing that the left hand side is a polynomial in $x$ having $-(y+z)$ as a zero) that\n$$\n\\begin{aligned}\n& x^{3}+y^{3}+z^{3}-3 x y z=(x+y+z)\\left(x^{2}+y^{2}+z^{2}-x y-y z-z x\\right) \\\\\n&=(x+y+z) \\frac{(x-y)^{2}+(y-z)^{2}+(z-x)^{2}}{... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 17 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All permutations of (668, 668, 667) | |
031d | Problem:
Aleksander writes a positive integer as a coefficient of a polynomial of degree four, then Elitza writes a positive integer as another coefficient of the same polynomial and so on till all the five coefficients of the polynomial are filled in. Aleksander wins if the polynomial obtained has an integer root; ot... | [
"Solution:\n\nWe shall prove that Elitza has a winning strategy. If the polynomial is $a_{0} x^{4} + a_{1} x^{3} + a_{2} x^{2} + a_{3} x + a_{4}$ and Aleksander writes $a_{0}$, $a_{1}$, $a_{2}$ or $a_{3}$, then Elitza writes respectively $a_{1} = a_{0}$, $a_{0} = a_{1}$, $a_{3} = a_{2}$ or $a_{2} = a_{3}$; if he wr... | Bulgaria | Bulgarian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Elitza | |
0l6u | Circle $\omega_1$ with radius $6$ centered at point $A$ is internally tangent at point $B$ to circle $\omega_2$ with radius $15$. Points $C$ and $D$ lie on $\omega_2$ such that $\overline{BC}$ is a diameter of $\omega_2$ and $\overline{BC} \perp \overline{AD}$. The rectangle $EFGH$ is inscribed in $\omega_1$ such that ... | [] | United States | AIME II | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 293 | |
0gjd | For positive integers $n$ and $k \ge 2$ define $E_k(n)$ as the greatest exponent $r$ such that $k^r$ divides $n!$. Prove that there are infinitely many $n$ such that $E_{10}(n) > E_9(n)$ and infinitely many $m$ such that $E_{10}(m) < E_9(m)$. | [
"Set $n = 5^{2l-1}$, we have\n$$\nE_{10}(n) = v_5(n!) = 5^{2l-2} + 5^{2l-3} + \\dots + 5 + 1 = \\frac{5^{2l-1}-1}{4} = \\frac{n-1}{4}.\n$$\nSince $n \\equiv 2 \\pmod 4$, so\n$$\nE_9(n) = \\frac{1}{2}v_3(n!) < \\frac{1}{2}\\left(\\frac{n-2}{3} + \\frac{n}{3^2} + \\dots\\right) = \\frac{n}{4} - \\frac{1}{3}.\n$$\nThu... | Taiwan | IMO 2J, Independent Study 1 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | Chinese; English | proof only | null | |
01o7 | A quadratic trinomial $x^2 + p x + q$ with integer coefficients $p$ and $q$ is said to be *irrational* if it has irrational roots $\alpha_1$ and $\alpha_2$.
Find the smallest value of the sum $|\alpha_1| + |\alpha_2|$ among all irrational trinomials. | [
"Answer: $\\sqrt{5}$.\nBy condition, the trinomial $x^2 + p x + q$ has the roots, so its discriminant $D = p^2 - 4q \\ge 0$. Since $p^2 = 4q + D$ and $p$ and $q$ are integer, we have $D \\ne 2$ and $D \\ne 3$ because the square of the integer number is congruent neither to 2 nor to 3 modulo 4.\nMoreover, the trinom... | Belarus | Belorusija 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | sqrt(5) | |
0ibs | Problem:
Find all ordered pairs of integers $(x, y)$ such that $3^{x} 4^{y} = 2^{x+y} + 2^{2(x+y)-1}$. | [
"Solution:\nThe right side is $2^{x+y} (1 + 2^{x+y-1})$. If the second factor is odd, it needs to be a power of $3$, so the only options are $x+y=2$ and $x+y=4$. This leads to two solutions, namely $(1,1)$ and $(2,2)$. The second factor can also be even, if $x+y-1=0$. Then $x+y=1$ and $3^{x} 4^{y} = 2 + 2$, giving ... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1,1), (2,2), (0,1) | |
0iqq | Problem:
Let $f(x)=\frac{1}{1-x}$. Let $f^{k+1}(x)=f\left(f^{k}(x)\right)$, with $f^{1}(x)=f(x)$. What is $f^{2008}(2008)$? | [
"Solution:\n\n$\\boxed{\\frac{-1}{2007}}$ Notice that, if $x \\neq 0,1$, then $f^{2}(x)=\\frac{1}{1-\\frac{1}{1-x}}=\\frac{x-1}{x}$, which means that $f^{3}(x)=\\frac{1}{1-\\frac{x-1}{x}}=x$. So $f^{n}$ is periodic with period $n=3$, which means that $f^{2007}(x)=x$ so $f^{2008}(2008)=f(2008)=\\frac{-1}{2007}$."
] | United States | 1st Annual Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | -1/2007 | |
0j0h | Problem:
George, Jeff, Brian, and Travis decide to play a game of hot potato. They begin by arranging themselves clockwise in a circle in that order. George and Jeff both start with a hot potato. On his turn, a player gives a hot potato (if he has one) to a randomly chosen player among the other three (if a player has... | [
"Solution:\n\nNotice that Travis can only have the hot potato at the end if he has two potatoes before his turn. A little bit of casework shows that this can only happen when\n\nCase 1: George gives Travis his potato, while Jeff gives Brian his potato, which then goes to Travis. The probability of this occurring is... | United States | Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | final answer only | 5/27 | |
002q | Alrededor de una circunferencia se han escrito cierta cantidad de ceros y la misma cantidad de unos. Se sabe que hay exactamente $99$ ternas de números consecutivos que contienen dos o tres ceros. Determinar el mínimo número de ternas de números consecutivos que contienen dos o tres unos. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Español | proof and answer | 35 | |
04n7 | Determine all real numbers $x$ such that
$$
1 \le \frac{1 + \sin x}{1 - \sin x} \le 3.
$$ | [] | Croatia | Croatia_2018 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | All real x with sin x in [0, 1/2], i.e., for any integer k: x in [2kπ, 2kπ + π/6] ∪ [2kπ + 5π/6, 2kπ + π]. | |
08br | Problem:
Dato un triangolo $ABC$ di lati $AB = 13$, $BC = 14$ e $AC = 15$, sia $H$ il piede dell'altezza relativa al lato $BC$, $M$ il punto medio di $BC$ e $N$ il punto medio di $AM$. Quanto vale la lunghezza di $HN$?
(A) $2 + 2\sqrt{3}$
(B) $6$
(C) $\sqrt{37}$
(D) $4 + \sqrt{7}$
(E) $\sqrt{42}$ | [
"Solution:\n\nLa risposta è (C). Il triangolo $AHM$ è rettangolo, quindi abbiamo $\\overline{HN} = \\overline{AM} / 2$ perché $HN$ è la mediana relativa all'ipotenusa. Ponendo $\\overline{BH} = x$ abbiamo $\\overline{HC} = \\overline{BC} - \\overline{BH} = 14 - x$ e, per il teorema di Pitagora, $\\overline{AH}^2 = ... | Italy | Gara di Febbraio | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | C | |
0ha0 | There are numbers $1, 2, 3, \ldots, 100$ on the board. Each is written exactly once. Petryk and Ivasyk are playing in the following game (Petryk starts): each player erases one number from the board. If after player's turn the sum of all erased numbers (by both players) cannot be represented as a difference of squares ... | [
"Notice that $n$ cannot be represented as a difference of squares of integers iff $n \\equiv 2 \\pmod{4}$. Indeed, if $n \\equiv 2 \\pmod{4}$, then suppose $n = (x + y)(x - y)$. If $x, y$ are either both odd or both even, then $n \\equiv 0 \\pmod{4}$, otherwise $n \\equiv \\pm 1 \\pmod{4}$. If $n = 4k$, then let $x... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Petryk | |
09gf | Let $ABC$ be a scalene triangle and $F$ be midpoint of arc $BC$ of circumcircle of triangle $ABC$ containing $A$. The circle passing through $A$ and $F$ intersected to the rays $[BA]$ and $[CA]$ at $M$ and $N$, respectively. Let $P$ be intersection of $MC$ and $NB$. Let $S$ be the point such that the quadrilateral $ABS... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0g35 | Problem:
Bestimme alle natürlichen Zahlen $n \geq 2$, die eine Darstellung der Form
$$
n = k^{2} + d^{2}
$$
haben, wobei $k$ der kleinste Teiler von $n$ grösser als 1 und $d$ ein beliebiger Teiler von $n$ ist. | [
"Solution:\n\nFalls $n$ ungerade ist, sind auch alle Teiler von $n$, also insbesondere auch $k$ und $d$, ungerade. Dann ist aber $k^{2} + d^{2}$ gerade, was nicht möglich ist. Somit ist $n$ gerade und der kleinste Teiler einer geraden Zahl grösser als 1 ist 2, also ist $k = 2$. Da $n$ und $k$ gerade sind, muss also... | Switzerland | Vorrunde 2019 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 8 and 20 | |
0f7n | Problem:
A convex pentagon is cut along all its diagonals to give 11 pieces. Show that the pieces cannot all have equal areas. | [] | Soviet Union | 21st ASU | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0kcn | Problem:
Let $T$ be a trapezoid with two right angles and side lengths $4, 4, 5$, and $\sqrt{17}$. Two line segments are drawn, connecting the midpoints of opposite sides of $T$ and dividing $T$ into 4 regions. If the difference between the areas of the largest and smallest of these regions is $d$, compute $240 d$. | [
"Solution:\n\n\n\nBy checking all the possibilities, one can show that $T$ has height $4$ and base lengths $4$ and $5$. Orient $T$ so that the shorter base is on the top.\n\nThen, the length of the cut parallel to the bases is $\\frac{4+5}{2}=\\frac{9}{2}$. Thus, the top two pieces are trap... | United States | HMMO | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 120 | |
0bzf | Determine the smallest positive integer $n$ such that, for any coloring of the elements of the set $\{2, 3, \dots, n\}$ with two colors, the equation $x + y = z$ has a monochrome solution with $x \neq y$. (We say that the equation $x + y = z$ has a monochrome solution if there exist $a, b, c$ distinct, having the same ... | [
"For $n = 12$ there exists a coloring of the numbers from $\\{2, 3, \\dots, 12\\}$ with two colors such that the equation $x + y = z$ has no monochrome solution: we color the numbers from $A = \\{2, 3, 4, 11, 12\\}$ with one color, and the elements of $B = \\{5, 6, 7, 8, 9, 10\\}$ with the second color.\nFor $n < 1... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 13 | |
0esf | Let $k$ and $m$ be integers with $1 < k < m$. For a positive integer $i$, let $L_i$ be the least common multiple of $1, 2, \dots, i$. Prove that $k$ is a divisor of $L_i \cdot \left[\binom{m}{i} - \binom{m-k}{i}\right]$ for all $i \ge 1$. [Here, $\binom{n}{i} = \frac{n!}{i!(n-i)!}$ denotes a binomial coefficient. Note ... | [
"We prove the statement by induction on $m$. When $m = k$, we have\n$$\nL_i \\left[ \\binom{k}{i} - \\binom{0}{i} \\right] = L_i \\binom{k}{i} = L_i \\cdot \\frac{k!}{i!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\frac{(k-1)!}{(i-1)!(k-i)!} = k \\cdot \\frac{L_i}{i} \\cdot \\binom{k-1}{i-1}.\n$$\nSince $\\frac{L_i}{i... | South Africa | South African Mathematics Olympiad Third Round | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0abg | If $a, b, c \in \mathbb{R}^+$ such that $(a+b)(b+c)(c+a)=8$, then prove that
$$
\frac{a+b+c}{3} \geq \sqrt[27]{\frac{a^3+b^3+c^3}{3}}.
$$ | [
"$$\n\\begin{align*}\n(a+b+c)^3 &= a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a) = a^3 + b^3 + c^3 + 24 \\\\\n&= a^3 + b^3 + c^3 + \\underbrace{3 + \\dots + 3}_{8} \\geq 9\\sqrt[9]{(a^3 + b^3 + c^3)^3}\n\\end{align*}\n$$\n\\text{i.e. } $\\left(\\frac{a+b+c}{3}\\right)^3 \\geq \\sqrt[9]{\\frac{a^3+b^3+c^3}{3}}$, $\\frac{a+b+c}... | North Macedonia | Macedonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0kji | Each of the $5$ sides and the $5$ diagonals of a regular pentagon are randomly and independently colored red or blue with equal probability. What is the probability that there will be a triangle whose vertices are among the vertices of the pentagon such that all of its sides have the same color? | [
"It will be easier to compute the probability that no monochromatic triangles exist. Suppose one of the vertices, say $A$, has $3$ segments of the same color connecting it to $3$ other vertices, say $B$, $C$, and $D$. If one of the edges of $\\triangle BCD$ has the same color as edges $\\overline{AB}$, $\\overline{... | United States | Fall 2021 AMC 10 B | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | final answer only | 253/256 | |
09fb | A sequence of length $n$, consisting of positive integers not exceeding $n - 1$, is given. If there exists exactly one subsequence with sum divisible by $n$, then show that the sequence is constant. | [
"For $n = 2$, there is nothing to prove. Suppose $n \\ge 3$.\nLet $a_1, \\dots, a_n$ denote the sequence and let $I \\subseteq \\{1, \\dots, n\\}$ denote the index set of the non-empty subsequence whose sum is divisible by $n$, that is $\\sum_{i \\in I} a_i \\equiv 0 \\pmod{n}$ and no other non-empty subset of $\\{... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
04ed | Determine all primes $p$ for which there exist positive integers $x$ and $y$ such that
$$
\begin{cases}
p + 1 = 2x^2 \\
p^2 + 1 = 2y^2.
\end{cases}
$$ | [
"Subtracting the given equations we get $p(p-1) = 2(y-x)(y+x)$.\nFrom this we conclude\n$$\np \\mid y + x,\n$$\nbecause otherwise $p$ would be a divisor of $y - x$, and $p - 1$ would be a multiple of number $y + x$, which is impossible (we would have $p - 1 \\ge y + x > y - x \\ge p$ then).\nSince $p > y$ (from the... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 7 | |
00m7 | Anna und Berta spielen ein Spiel, bei dem sie abwechselnd Murmeln vom Tisch nehmen. Anna macht den ersten Zug. Wenn zu Beginn eines Zuges $n \ge 1$ Murmeln am Tisch sind, dann nimmt die Spielerin, die am Zug ist, $k$ Murmeln weg, wobei $k \ge 1$ entweder eine gerade Zahl mit $k \le \frac{n}{2}$ oder eine ungerade Zahl ... | [
"Behauptung: Die Verlustsituationen sind jene Situationen mit $n = 2^a - 2$ Murmeln am Tisch für alle ganzen Zahlen $a \\ge 2$. Alle anderen Situationen sind Gewinnsituationen.\nBeweis: Mit Induktion über $n \\ge 1$. Für $n = 1$ gewinnt man, indem man die einzige verbleibende Murmel nimmt. Für $n = 2$ kann man nur ... | Austria | 48. Österreichische Mathematik-Olympiade Bundeswettbewerb für Fortgeschrittene, Teil 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | German | proof and answer | 131070 |
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