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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
072r | Let $ABC$ be a triangle and $P$ be a point in its plane inside the region of the angle $BAC$, but lying outside triangle $ABC$.
a. Prove that any two of the following statements imply the third:
(i) the circumcentre of triangle $BPC$ lies on the ray $\vec{PA}$;
(ii) the circumcentre of triangle $CPA$ lies on the ray $... | [
"The problem may be rephrased as follows: line segments $PA$, $PB$, $PC$ are given. Let $l_1$, $l_2$, $l_3$ denote the perpendicular bisectors of $PA$, $PB$, $PC$ respectively. Given that $l_1$, $l_3$ meet on $PB$ and $l_2$, $l_3$ meet on $PA$, we have to show that $l_1$, $l_2$ intersect on $PC$.\n\nLet $l_1$, $l_3... | India | Indija TS 2006 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | null | proof only | null | |
04wp | Determine all pairs of integers $(x, y)$ satisfying the equation
$$
y(x + y) = x^3 - 7x^2 + 11x - 3.
$$ | [
"The considered equation is equivalent to\n$$\n(2y + x)^2 = 4x^3 - 27x^2 + 44x - 12 = (x - 2)(4x^2 - 19x + 6) = (x - 2)((x - 2)(4x - 11) - 16).\n$$\nThe expression above must be a perfect square. Therefore we have either $x = 2$ (and $y = -1$), or $(x - 2) = ks^2$, where $k \\in \\{-2, -1, 1, 2\\}$ and $s \\in \\ma... | Czech-Polish-Slovak Mathematical Match | Czech-Polish-Slovak Match | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | {(6, 3), (6, -9), (1, 1), (1, -2), (2, -1)} | |
00jh | In an arithmetic sequence, the differences between successive elements are all equal. We wish to consider integer sequences, in which the differences between successive elements are equal to the sum of all previous such differences. Which of these sequences with $a_0 = 2012$ and $1 \le d = a_1 - a_0 \le 43$ contain per... | [
"For $n \\ge 1$ we have\n$$\na_{n+1} - a_n = (a_n - a_{n-1}) + (a_{n-1} - a_{n-2}) + \\dots + (a_1 - a_0) = a_n - a_0,\n$$\nwhich yields $a_{n+1} = 2a_n - a_0$. The sequence can therefore be written in the form\n$$\na_0 = 2012,\\ a_1 = 2012 + d,\\ a_2 = 2012 + 2d,\\ \\dots,\\ a_n = 2012 + 2^{n-1}d,\\ \\dots\n$$\nFo... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | d = 13 and d = 26 | |
0ho0 | Problem:
Alice and Bob play the following game on the whiteboard. First, Alice writes an odd number in binary on the board. Then, beginning with Bob, the players take turns modifying the number in one of two ways: subtracting $1$ from it (preserving the binary notation), or erasing its last digit. When the whiteboard ... | [
"Solution:\n\nBob can win using the following strategy: Play so as to leave an empty whiteboard or a number with an odd number of terminal zeros (preceded by a $1$). We claim that Bob can fulfill this requirement at every move.\n\nOn his first move, or indeed any move in which Alice leaves an odd number on the boar... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Bob | |
0i02 | Problem:
What is the smallest square-free composite number that can divide a number of the form $4242 \ldots 42 \pm 1$ ? | [
"Solution:\nIt is easy to see that such a number can never be divisible by $2,3,5$, or $7$. They can be divisible by $11$, the smallest example being $4242424241 = 11 \\cdot 547 \\cdot 705073$.\n\nWhat makes this problem hard is finding the next prime that can divide such a number. Let $T_{n} = \\sum_{i=0}^{n} 42 \... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 319 | |
08zt | Consider a figure consisting of 15 circles and 20 line segments as shown below. Each circle is labeled with a 0, 1, or 2, and we define the *beauty* of a labeling as the number of line segments whose endpoints are labeled with numbers that differ by 1.
Let $M$ be the largest possible beauty that can be achieved with an... | [
"1920\nA line segment for which the difference between the numbers written in the two circles at its endpoints is 1 is called a good line segment, and one that does not satisfy this condition is called a bad line segment. Let us also name the six pentagons as $P_1$ to $P_6$, as shown in the figure below.\n![](attac... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 1920 | |
05bj | Let $n$ and $k$ be positive integers. On the board $k$ different positive integers not exceeding $n$ are written. Prove that the equation
$$
x + y = z + w
$$
has at least $\frac{k^4}{2n-1}$ solutions $(x, y, z, w)$, where $x, y, z, w$ are some (not necessarily different) numbers written on the board. | [
"For positive integer $s$ denote by $a_s$ the number of pairs $(x, y)$, where $x, y$ are numbers on the board and $x + y = s$. Since the sum of two numbers on the board cannot be less than $2$ or greater than $2n$, the number of solutions of equation $x + y = z + w$ is $a_2^2 + \\dots + a_{2n}^2$. Applying the AM-Q... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
09vz | Let a triangle $ABC$ such that $|AC| < |AB|$ be given, together with its circumcircle. Let $D$ be a varying point on the short arc $AC$. Let $E$ be the reflection of $A$ in the internal angular bisector of $\angle BDC$. Prove that the line $DE$ passes through a fixed point, independent of where $D$ lies. | [
"Let $M$ be the intersection of the internal angular bisector of $\\angle BDC$ with the circumcircle of $\\triangle ABC$. As $D$ lies on the short arc $AC$, we see that $M$ lies on the arc $BC$ not containing $A$. We have $\\angle BDM = \\angle MDC$ as $DM$ is the internal angular bisector of $\\angle BDC$, so arcs... | Netherlands | IMO Team Selection Test 3, June 2020 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0cwr | Let $ABCD$ be a quadrilateral having no two parallel sides, inscribed into the circle $\Omega$. Let $\omega_a, \omega_b, \omega_c, \omega_d$ be circles inscribed into triangles $DAB, ABC, BCD, CDA$, respectively. Let us draw common external tangents $t_1, t_2, t_3, t_4$ to pairs of circles $\omega_a$ and $\omega_b$, $\... | [
"Without loss of generality, let rays $AB$ and $DC$ intersect at point $P$, and rays $AD$ and $BC$ intersect at point $Q$. Denote the center of circle $\\omega_a$ as $I_a$, and define points $I_b, I_c, I_d$ similarly. Let the quadrilateral formed by the four tangent lines be $A'B'C'D'$ (line $A'B'$ is the common ex... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Coaxal circles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Russian | proof only | null | |
0jqo | Problem:
Let $D$ be a regular ten-sided polygon with edges of length 1. A triangle $T$ is defined by choosing three vertices of $D$ and connecting them with edges. How many different (non-congruent) triangles $T$ can be formed? | [
"Solution:\nAnswer: 8 The problem is equivalent to finding the number of ways to partition 10 into a sum of three (unordered) positive integers. These can be computed by hand to be $(1,1,8),(1,2,7),(1,3,6)$, $(1,4,5),(2,2,6),(2,3,5),(2,4,4),(3,3,4)$."
] | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 8 | |
0ebj | Problem:
a) Naj bo $a > 0$ in $b > 0$. Poenostavi izraz
$$
7\left(\frac{a^{5}}{2 b^{4}}\right)^{5} \cdot \left(\frac{2 b^{2}}{a^{4}}\right)^{5} - 3\left(\frac{a}{b^{2}}\right)^{3+2x} \cdot \left(\frac{a^{2}}{b^{4}}\right)^{1-x} + \left(4 a^{5} b\right)^{5} : \left(2 a^{4} b^{3}\right)^{5}
$$
b) Izračunaj vrednost izraz... | [] | Slovenia | 15. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | a) 36 a^5 / b^{10}; b) 45 | |
0dfc | Let $x(n)$ be the biggest prime divisor of $n$. Prove that there exist infinitely many number $n$ such that $x(n) < x(n + 1) < x(n + 2)$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
02rr | Find all surjective functions $f: (0, +\infty) \to (0, +\infty)$ such that
$$
2x \cdot f(f(x)) = (f(f(x)) + x) \cdot f(x)
$$
for all $x \in (0, +\infty)$.
A function $f: A \to B$ is said to be surjective iff the range of $f$ is $B$, that is, for all $y \in B$ there exists $x \in A$ such that $f(x) = y$. | [
"First let's prove that $f$ is injective: suppose that $f(a) = f(b)$. Then\n$$\n\\begin{aligned}\n\\frac{2a f(f(a))}{2b f(f(b))} &= \\frac{f(a)(a+f(f(a)))}{f(b)(b+f(f(b)))} \\\\\n&\\iff \\frac{a}{b} = \\frac{a+f(f(a))}{b+f(f(b))} = \\frac{f(f(a))}{f(f(b))} = 1 \\\\\n&\\iff a = b.\n\\end{aligned}\n$$\n\nThen $f$ is ... | Brazil | Brazilian Math Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | f(x) = x | |
0hro | Problem:
Let $P(x)$ be a polynomial with integer coefficients. Let $a_{0}=0$ and for $i \geq 0$ define $a_{i+1}=P\left(a_{i}\right)$. Prove that $\operatorname{gcd}\left(a_{m}, a_{n}\right)=a_{\gcd(m, n)}$ for any integers $m, n \geq 1$. | [
"Solution:\n\nIf $Q(x), R(x)$ are any two polynomials with integer coefficients, it is not hard to see that their composition $Q(R(x))$ is also such a polynomial. In particular, for any nonnegative integer $n$, $P^{n}(x)$, the function obtained by iterating $P$ $n$ times, is an integer polynomial. Note that $a_{n}=... | United States | Berkeley Math Circle Take-Home Contest #6 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0c3s | Problem:
Arătaţi că dacă într-un triunghi ortocentrul $H$, centrul de greutate $G$ şi centrul $I$ al cercului înscris sunt coliniare, atunci triunghiul este isoscel. | [
"Solution:\nDacă triunghiul este echilateral, concluzia este verificată.\nDacă triunghiul este dreptunghic, atunci o bisectoare este şi mediană, deci concluzia este valabilă.\nÎn caz contrar, deoarece $G$, $H$ şi centrul $O$ al cercului circumscris triunghiului sunt coliniare, deducem că $I$ este pe $OH$.\n\n![](at... | Romania | Olimpiada Naţională de Matematică Etapa Naţională | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
0l1p | For how many integer values of $x$ is $|2x| \le 7\pi$?
(A) 16 (B) 17 (C) 19 (D) 20 (E) 21 | [
"**Answer (E):** From $3 < \\pi < 3.142$, it follows that\n$$\n21 = 7 \\cdot 3 < 7\\pi < 7 \\cdot 3.142 = 21.994 < 22.\n$$\nBecause $x$ is an integer, the values for $2x$ that make the inequality true are $-20, -18, -16, \\dots, -2, 0, 2, \\dots, 18$, and $20$. Each of these corresponds to a unique value of $x$. Th... | United States | 2024 AMC 12 B | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | E | |
08kz | Problem:
Let $n > 1$ be a positive integer and $p$ a prime number such that $n \mid (p-1)$ and $p \mid \left(n^{6}-1\right)$. Prove that at least one of the numbers $p-n$ and $p+n$ is a perfect square. | [
"Solution:\n\nSince $n \\mid p-1$, then $p = 1 + n a$, where $a \\geq 1$ is an integer. From the condition $p \\mid n^{6}-1$, it follows that $p \\mid n-1$, $p \\mid n+1$, $p \\mid n^{2}+n+1$ or $p \\mid n^{2}-n+1$.\n\n- Let $p \\mid n-1$. Then $n \\geq p+1 > n$ which is impossible.\n\n- Let $p \\mid n+1$. Then $n+... | JBMO | 2007 Shortlist JBMO | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof only | null | |
023y | Problem:
Três casais jantam todo sábado num mesmo restaurante, sempre à mesma mesa. A mesa é redonda e os casais combinaram que
a. jamais marido e mulher sentam à mesa como vizinhos; e
b. a disposição dos seis à mesa é diferente a cada sábado.
Desconsiderando rotações nas disposições à mesa, durante quantos sábados... | [
"Solution:\n\nPara simplificar, vamos denotar cada casal por um par de números, um número representando o marido e o outro a mulher. Temos, então, os três pares $(1,2)$, $(3,4)$, $(5,6)$, que não podem ser vizinhos. Podemos considerar o lugar do marido $1$ à mesa como sendo fixo, já que desconsideramos rotações na ... | Brazil | Nível 2 | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 16 | |
0cco | Find all positive integers $n$ so that the largest prime divisor of $n^2 + 2$ is equal to the largest prime divisor of $n^2 + 2n + 3$. | [
"A common divisor $d$ of the given numbers divides also $n^2 + 2n + 3 - (n^2 + 2) = 2n + 1$, $n \\cdot (2n + 1) = 2n^2 + n$, $2n^2 + n - 2 \\cdot (n^2 + 2) = n - 4$, $2n + 1 - 2 \\cdot (n - 4) = 9$.\nTherefore, if $d$ is a prime, then $d = 3$, and the only other prime factor contained by the given numbers is $2$, a... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1, 4 | |
0bz8 | Given an acute triangle $ABC$, erect triangles $ABD$ and $ACE$ externally, so that $\widehat{ADB} = \widehat{AEC} = 90^\circ$ and $\widehat{BAD} \equiv \widehat{CAE}$. Let $A_1 \in BC$, $B_1 \in AC$ and $C_1 \in AB$ be the feet of the altitudes of the triangle $ABC$, and let $K$ and $L$ be the midpoints of $[BC_1]$ and... | [
"Let $M$, $P$ and $Q$ be the midpoints of $[BC]$, $[CA]$ and $[AB]$, respectively.\n\nThe circumcircle of triangle $A_1B_1C_1$ is the Euler circle. Point $M$ lies on this circle.\n\nIt is enough to prove now that $[A_1M]$ is a common chord of the three circles, $(A_1B_1C_1)$, $(AKL)$ and $(DEA_1)$.\n\nThe segments ... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Circles > Coaxal circles",
"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 > A... | English | proof only | null | |
0994 | $p > 2$, $p \in \mathbb{R}$, $a_1, \ldots, a_k$ нь $\mathbb{Z}_p$-ийн ялгаатай элементүүд, $b_1, \ldots, b_k$ нь $\mathbb{Z}_p$-ийн элементүүд (зарим нь тэнцүү байж болно) бол $a_1 + b_{\sigma(1)}, \ldots, a_k + b_{\sigma(k)}$ нь $\mathbb{Z}_p$-ийн ялгаатай элементүүд байх тийм сэлгэмэл $\sigma \in S_k$ олдохыг үзүүл. | [
"$S_1 = \\dots = S_k = \\{a_1, \\dots, a_k\\}$ байг. $x_1 \\in S_1, \\ldots, x_k \\in S_k$ гэж авъя. $f \\Rightarrow \\prod_{1 \\le i < j \\le k} (x_j - x_i)(x_j + b_j - (x_i + b_i)) \\ne 0$ гэж харуулъя. Үүний тулд Теорем 2 (ГТАКТ-2) [MMK-II, х.105]-оор $A(f) = c x_1^{k-1} \\dots x_k^{k-1}$ гэвэл $c \\Leftrightarr... | Mongolia | ММО-46 | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Linear Algebra > Determinants"
] | Mongolian | proof only | null | |
03wv | Find all integers $n$ such that we can color all the edges and the diagonals of a convex $n$-polygon by $n$ given colors satisfying the following conditions;
(1) Each of the edges or the diagonals is colored by only one color;
(2) For any three distinct colors, there exists a triangle whose vertices are vertices of the... | [
"Answer: Any odd number $n > 1$.\n\nFirst of all, there are $\\binom{n}{3}$ ways to choose three among $n$ colors, and $\\binom{n}{3}$ ways to choose three vertices to form a triangle, so if the question's condition is fulfilled, all the triangles should have a different color combination (a 1-1 correspondence).\n\... | China | Chinese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | All odd integers greater than 1 | |
02ps | Problem:
Seja $n$ um número inteiro positivo. Para cada um dos inteiros $n+1$, \ldots, $2n$ considere o seu maior divisor ímpar. Prove que a soma de todos estes divisores é igual a $n^{2}$. | [
"Solution:\n\nChamemos de $S_{n}$ a soma dos maiores divisores ímpares dos números $n+1, \\ldots, 2n$. Por cálculo direto temos que $S_{1}=1$, $S_{2}=3+1=4=2^{2}$, $S_{3}=1+5+3=9=3^{2}$ e $S_{4}=5+3+7+1=16=4^{2}$.\n\nSe queremos calcular $S_{n+1}$, que é a soma dos maiores divisores ímpares dos números\n$$\nn+2, n+... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | null | proof only | null | |
08pv | Problem:
Show that there exist infinitely many positive integers $n$ such that
$$
\frac{4^{n}+2^{n}+1}{n^{2}+n+1}
$$
is an integer. | [
"Solution:\nLet $f(n) = n^{2} + n + 1$. Note that\n$$\nf\\left(n^{2}\\right) = n^{4} + n^{2} + 1 = \\left(n^{2} + n + 1\\right)\\left(n^{2} - n + 1\\right)\n$$\nThis means that $f(n) \\mid f\\left(n^{2}\\right)$ for every positive integer $n$. By induction on $k$, one can easily see that $f(n) \\mid f\\left(n^{2^{k... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0l7f | There are $n$ values of $x$ in the interval $0 < x < 2\pi$ where $f(x) = \sin(7\pi \cdot \sin(5x)) = 0$. For $t$ of these $n$ values of $x$, the graph of $y = f(x)$ is tangent to the $x$-axis. Find $n + t$. | [] | United States | AIME II | [
"Precalculus > Trigonometric functions",
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications"
] | null | final answer only | 150 | |
0bkg | Let $(G, \cdot)$ be a group with no elements of order $4$, and let $f : G \rightarrow G$ be a group morphism such that $f(x) \in \{x, x^{-1}\}$, for all $x \in G$. Prove that either $f(x) = x$ for all $x \in G$, or $f(x) = x^{-1}$ for all $x \in G$. | [
"Assume, by way of contradiction, that there exist $a, b \\in G$ such that $f(a) = a \\neq a^{-1}$ and $f(b) = b^{-1} \\neq b$. Then $f(ab) = f(a)f(b) = ab^{-1} \\neq ab$, therefore $f(ab) = (ab)^{-1} = b^{-1}a^{-1}$. It follows that $ab^{-1} = b^{-1}a^{-1}$, which implies $b^{-1} = ab^{-1}a$.\n\nNext, $f(ab^2) = f... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
00qz | A set of points in the plane is called *obtuse* when it contains no three collinear points, and every triangle with its vertices in this set has one angle $> 91^{\circ}$. Is it true that every finite obtuse set can be extended to an infinite obtuse set? | [
"It is true that every finite obtuse set can be thus extended.\nIt suffices to show that any finite obtuse set $\\{P_0, P_1, \\dots, P_n\\}$ can be enlarged to an obtuse set $\\{P_0, P_1, \\dots, P_n, Q\\}$. This is trivial for the empty set and a set with one element, so we may suppose $n \\ge 1$.\nNow we construc... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ape | Problem:
If $f(x+y)=f(x) \cdot f(y)$ for all positive integers $x, y$ and $f(1)=2$, find $f(2007)$. | [
"Solution:\n\n$2^{2007}$\n\nBy induction, it can be shown that $f(n) = [f(1)]^{n} = 2^{n}$ for all positive integers $n$. Thus, $f(2007) = 2^{2007}$."
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 2^2007 | |
0917 | Problem:
An acute-angled triangle $A B C$ is given. Let $E$ be a point such that $B$ and $E$ lie on different sides of the line $A C$, and let $D$ be an interior point of the segment $A E$. Suppose that $\angle A D B = \angle C D E$, $\angle B A D = \angle E C D$ and $\angle A C B = \angle E B A$. Prove that $B, C$ an... | [
"Solution:\n\nCondition $\\angle A D B = \\angle C D E$ motivates us to reflect $B$ over $A E$ to $B'$. (Fig. 3)\n\nThen $C, D$ and $B'$ are collinear and $\\angle E A B' = \\angle E A B = \\angle E C D = \\angle E C B'$, so $B' A C E$ is a cyclic quadrilateral. This implies that $\\angle E C A = \\pi - \\angle E B... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ktu | Problem:
Let $\mathbb{Q}^{+}$ denote the set of positive rational numbers. Find, with proof, all functions $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ such that, for all positive rational numbers $x$ and $y$, we have
$$
f(x)=f(x+y)+f\left(x+x^{2} f(y)\right)
$$ | [
"Solution:\nAnswer: $f(x)=\\frac{1}{x}$.\n\nIt is straightforward to check that $f(x)=\\frac{1}{x}$ works. We then focus on proving that there are no other solutions. Let $P(x, y)$ denote the given functional equation. First note that for all $x, y \\in \\mathbb{Q}^{+}$, $f(x)>f(x+y)$, so $f$ is strictly decreasing... | United States | HMIC 2023 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x)=1/x | |
014o | Problem:
Let $a, b, c, d, e, f$ be non-negative real numbers satisfying $a+b+c+d+e+f=6$. Find the maximal possible value of
$$
a b c+b c d+c d e+d e f+e f a+f a b
$$
and determine all 6-tuples $(a, b, c, d, e, f)$ for which this maximal value is achieved. | [
"Solution:\nIf we set $a=b=c=2$, $d=e=f=0$, then the given expression is equal to $8$. We will show that this is the maximal value.\n\nApplying the inequality between arithmetic and geometric mean we obtain\n$$\n\\begin{aligned}\n8 & =\\left(\\frac{(a+d)+(b+e)+(c+f)}{3}\\right)^{3} \\geq (a+d)(b+e)(c+f) \\\\\n& =(a... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Maximum value: 8. It is attained precisely for sextuples of the form (0, 0, t, 2, 2, 2 − t) with t between 0 and 2, and their cyclic permutations. | |
0k0r | Problem:
Triangle $G R T$ has $G R=5$, $R T=12$, and $G T=13$. The perpendicular bisector of $G T$ intersects the extension of $G R$ at $O$. Find $T O$. | [
"Solution:\n\nFirst, note that $T O = G O$ as $O$ lies on the perpendicular bisector of $G T$. Then if $M$ is the midpoint of $G T$, we have that $\\triangle G R T \\sim \\triangle G M O$, so we can compute\n\n$$\nT O = G O = G M \\cdot \\frac{G T}{G R} = \\frac{13}{2} \\cdot \\frac{13}{5} = \\frac{169}{10}.\n$$"
] | United States | HMMT February | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 169/10 | |
07rv | A paper rhombus $ABCD$, with side lengths $4$ and $\angle ABC = 60^\circ$, is folded so that $B$ coincides with the mid-point $H$ of $AD$ and creased. Prove that the length of the crease is $\frac{1}{2}\sqrt{21}$. | [
"Let $EF$ be the crease such that $E$ is on $AB$ and $F$ on $BC$. Join $H$ to $C$, $F$ and $E$.\n\n\n\nThe crease $EF$ lies along the perpendicular bisector of $HB$ and the triangles $\\triangle EHF$ and $\\triangle EBF$ are congruent. If we let $x = |AE|$ and $y = |CF|$, then $|EH| = |EB| ... | Ireland | Irish | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 7*sqrt(21)/10 | |
090i | Let $\triangle ABC$ be an acute triangle with circumcenter $O$. Let $O_1$ and $O_2$ be the circumcenters of triangles $ABO$ and $ACO$, respectively. Suppose that the circumcircle of triangle $AO_1O_2$ intersects line segment $BC$ at two distinct points $P$ and $Q$ (excluding the endpoints), with the four points $B, P, ... | [
"Let $Q'$ be the intersection of the perpendicular bisector of $AB$ with line $BC$. Then, we have\n$$\n\\angle AQ'O_1 = \\angle BQ'O_1 = 90^\\circ - \\angle ABC.\n$$\nMoreover, since $O_1$ and $O_2$ lie on the perpendicular bisector of segment $AO$ and $O_2$ is the circumcenter of triangle $ACO$, we have\n$$\n\\ang... | Japan | The 35th Japanese Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Concurrency ... | English | proof only | null | |
080y | Problem:
Sia $P_{1}$ un esagono regolare. Sia $P_{2}$ il nuovo esagono ottenuto congiungendo i punti medi dei lati consecutivi di $P_{1}$. Allo stesso modo si proceda a partire da $P_{2}$ ottenendo un nuovo esagono $P_{3}$. Quanto vale il rapporto tra l'area di $P_{3}$ e quella di $P_{1}$?
(A) $\frac{1}{4}$
(B) $\fra... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | C | |
0g2e | Problem:
Sei $n$ eine natürliche Zahl und $G$ die Menge der Punkte $(x, y)$ in der Ebene, sodass $x$ und $y$ ganze Zahlen mit $1 \leq x, y \leq n$ sind. Eine Teilmenge von $G$ heißt parallelogrammfrei, wenn sie keine vier nicht-kollineare Punkte enthält, die die Eckpunkte eines Parallelogramms sind. Wie viele Elemente... | [
"Solution:\n\nDie maximale Anzahl Elemente ist $2n-1$.\n\n- $\\geq 2n-1$ : Wir nehmen alle Punkte der Form $(0, y)$ und $(x, 0)$. Wenn wir nun 4 Punkte wählen, sind mindestens drei davon kollinear und können daher kein Parallelogramm bilden.\n\n- $\\leq 2n-1$ : Wir beweisen, dass wenn wir $2n$ Punkte wählen, dass w... | Switzerland | SMO - Finalrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof and answer | 2n - 1 | |
0edb | Let $ABCD$ be a convex quadrilateral such that $|AB| = |BC| = |CD|$. The lines $AB$ and $CD$ intersect at the point $E$, and the circumcircles of the triangles $ABC$ and $BDE$ intersect at $B$ and $F$. Let $P$ denote the intersection of the lines $AC$ and $BF$. Prove that $EP$ is the bisector of the angle $\angle AED$. | [
"**II/3.** Due to different configurations we will proceed using directed angles.\nLet $T$, distinct from $E$, be the intersection of the circumcircles of the triangles $BDE$ and $ACE$. The power line (radical axis) of the circumcircles of the triangles $ABC$ and $BDE$ is $BF$, the power line of the circumcircles o... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line,... | null | proof only | null | |
0if6 | Problem:
Find three real numbers $a < b < c$ satisfying:
$$
\begin{aligned}
a + b + c & = 21 / 4 \\
1 / a + 1 / b + 1 / c & = 21 / 4 \\
a b c & = 1
\end{aligned}
$$ | [
"Solution:\nBy inspection, one notices that if $b$ is a number such that $b + 1 / b = 17 / 4$, then $a = 1$, $c = 1 / b$ will work. Again by inspection (or by solving the quadratic $b^{2} - 17 b / 4 + 1 = 0$), one finds $b = 1 / 4$ or $4$, so the numbers are $1 / 4$, $1$, and $4$.\n\nAlternative Solution:\nNote tha... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 1/4, 1, 4 | |
0fy6 | Problem:
Bananen, Äpfel und Orangen sind irgendwie auf 100 Kisten verteilt. Beweise, dass man 51 Kisten auswählen kann, die zusammen mindestens die Hälfte der Früchte von jeder Sorte enthalten. | [
"Solution:\n\nWir beginnen mit folgendem\n\nLemma 2. Betrachte $2n$ Paare $(a_{1}, b_{1}), \\ldots, (a_{2n}, b_{2n})$ von nichtnegativen reellen Zahlen. Dann lassen sich diese immer in zwei Gruppen von $n$ Paaren aufteilen, sodass folgendes gilt: Die Summe aller $a_{k}$ in der ersten Gruppe und die Summe aller $a_{... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
091b | Problem:
Find all positive integers $n$ which satisfy the following two conditions:
(i) $n$ has at least four different positive divisors;
(ii) for any divisors $a$ and $b$ of $n$ satisfying $1 < a < b < n$, the number $b - a$ divides $n$. | [
"Solution:\nClearly primes, squares of primes and the number $1$ have the given property. We will exclude these numbers from further considerations.\n\nFirst assume that $n$ is even; thus $n = 2x$ for some integer $x$. Then $x - 2$ divides $n$. Any divisor of $n$ smaller than $x = n/2$ is at most $n/3$. Therefore, ... | Middle European Mathematical Olympiad (MEMO) | MEMO | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All such numbers are 1, all primes, all squares of primes, and 6, 8, 12. | |
0e6d | Problem:
Krt Črt ima v svojem brlogu 5 sob, oštevilčenih s številkami od $1$ do $5$. Med nekaterimi izmed njih je Črt izvrtal rove, tako da se iz vsake sobe lahko po nekaj rovih splazi v vsako drugo sobo. Nobena dva rova se ne sekata. Vsak rov se začne v neki sobi, konča pa v neki drugi (od začetne različni) sobi in v... | [
"Solution:\n\nGotovo sobi $5$ in $1$ nista sosednji, saj bi se Črt sicer lahko trikrat sprehodil po tem rovu iz sobe $5$ v sobo $1$, od tam nazaj v sobo $5$ in nato še enkrat v sobo $1$. S tem bi iz sobe $5$ prišel v sobo $1$ po natanko treh rovih, kar je protislovno z navodili naloge.\n\nPrav tako soba $5$ ni nepo... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | (1,4), (2,3), (2,4), (3,4), (4,5) | |
0kui | Problem:
Lucas writes two distinct positive integers on a whiteboard. He decreases the smaller number by $20$ and increases the larger number by $23$, only to discover the product of the two original numbers is equal to the product of the two altered numbers. Compute the minimum possible sum of the original two number... | [
"Solution:\n\nLet the original numbers be $m < n$. We know\n$$\nm n = (m - 20)(n + 23) = m n - 20 n + 23 m - 460 \\Longrightarrow 23 m - 20 n = 460\n$$\nFurthermore, $23 m < 23 n$ hence $460 < 3 n \\Longrightarrow n \\geq 154$. Furthermore, we must have $23 \\mid n$ hence the least possible value of $n$ is $161$ wh... | United States | HMMT November 2023 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 321 | |
0dnz | Problem:
Кружница уписана у $\triangle A B C$ има центар у тачки $I$ и додирује страницу $B C$ у тачки $D$. На дужима $B I$ и $C I$ одабране су тачке $P$ и $Q$, редом, такве да важи $\varangle B A C=2 \varangle P A Q$. Доказати: $\varangle P D Q=90^\circ$.
(Душан Ђукић) | [
"Solution:\n\nОзначимо са $E$ и $F$ редом подножја нормала из $P$ и $Q$ на праву $B C$, а са $M$ средиште дужи $P Q$.\n\nПосматрајмо тачку $X$ на страници $B C$ такву да је $\\varangle B A X=2 \\varangle B A P$. Тада је такође $\\varangle C A X=\\varangle B A C-2 \\varangle B A P=2 \\varangle C A Q$, па су $P$ и $Q... | Serbia | 12. СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА УЧЕНИКА СРЕДЊИХ ШКОЛА | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | null | proof only | null | |
0h88 | a) Determine if there exist positive integer numbers $x, y, z$ such that
$$
2016 = x^3 + y^3 + z^3?
$$
b) Determine if there exist positive integer numbers $x, y, z, t$ such that
$$
2016 = x^3 + y^3 + z^3 + t^3?
$$ | [
"b) It is enough to provide an example: $2016 = 1000 + 1000 + 8 + 8$.\n\na) Let us first note that $2016 = 2^5 \\cdot 3^2 \\cdot 7$. And consider remainders of $a^3, b^3, c^3$ modulo $7$. The remainders could equal to $0$ or $\\pm 1$. Thus, if $2016 = x^3 + y^3 + z^3$ then at least one of the numbers is divisible b... | Ukraine | UkraineMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) No.
b) Yes, for example x = 10, y = 10, z = 2, t = 2 (2016 = 10^3 + 10^3 + 2^3 + 2^3). | |
0j0l | Problem:
Let $a_{1}, a_{2}, \ldots$ be an infinite sequence of positive integers such that for integers $n > 2$, $a_{n} = 3 a_{n-1} - 2 a_{n-2}$. How many such sequences $\{a_{n}\}$ are there such that $a_{2010} \leq 2^{2012}$? | [
"Solution:\n\nAnswer: $36 \\cdot 2^{2009} + 36$\n\nConsider the characteristic polynomial for the recurrence $a_{n+2} - 3 a_{n+1} + 2 a_{n} = 0$, which is $x^{2} - 3x + 2$. The roots are at $2$ and $1$, so we know that numbers $a_{i}$ must be of the form $a 2^{i-1} + b$ for integers $a$ and $b$. Therefore $a_{2010}... | United States | Harvard-MIT November Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 36 * 2^{2009} + 36 | |
00bx | Write in the cells of a $4 \times 4$ table a different natural number such that the sums by rows are equal and the products by columns are also equal. | [
"Here is one way to construct such a table. Start by satisfying only the second part of the condition, that the products by columns are equal. This holds for the first table $T$ below. It has dimensions $4 \\times 4$ and all column products equal $120$. The second table has an additional column in which the row sum... | Argentina | Argentina_2018 | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 528 132 264 1056; 540 360 900 180; 198 990 297 495; 550 660 440 330 | |
0hp9 | Problem:
A triangular number is one of the numbers $1, 3, 6, 10, 15, \ldots$ of the form $T_{n} = 1 + 2 + \cdots + n$ or, equivalently, $T_{n} = \left(n^{2} + n\right) / 2$.
Find, with proof, all ways of writing $2015$ as the difference of two triangular numbers. | [
"Solution:\nSince\n$$\nT_{m} - T_{n} = \\frac{m^{2} + m}{2} - \\frac{n^{2} + n}{2} = \\frac{m^{2} - n^{2} + m - n}{2} = \\frac{(m + n + 1)(m - n)}{2},\n$$\nany expression of $2015$ as the difference of two triangular numbers yields a factorization $2 \\cdot 2015 = 4030 = (m + n + 1)(m - n)$. Conversely, assigning v... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series... | null | proof and answer | T_2015 − T_2014, T_1008 − T_1006, T_405 − T_400, T_206 − T_196, T_161 − T_148, T_90 − T_64, T_80 − T_49, T_63 − T_1 | |
0588 | Let $n$ be a natural number, $n \ge 2$. There are $n$ lamps on a circle. The lamps are labeled clockwise by natural numbers from $1$ to $n$. Each lamp can be either on or off. A switch between every two adjacent lamps enables one to change the state of both lamps simultaneously. In the beginning, all lamps are off. How... | [
"As the state of each lamp is determined by the parity of the number of switchings that influence this lamp, the result of every sequence of switchings is determined by the set of switches that have been touched an odd number of times. Hence all possible states can be achieved by sequences of switchings that touch ... | Estonia | Estonian Math Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | 2^{n-1} | |
08zw | Let $n \ge 2$ be an integer. Find all sets of real numbers $(a_1, a_2, \dots, a_n)$ such that $a_1 - 2a_2, a_2 - 2a_3, \dots, a_{n-1} - 2a_n, a_n - 2a_1$ is a permutation of $a_1, a_2, \dots, a_n$. Note that $a_1, a_2, \dots, a_n$ itself is also a permutation of $a_1, a_2, \dots, a_n$. | [
"Let $a_{n+1} = a_1$, $a_0 = a_n$, and let $M$ and $m$ denote the maximum and minimum values of $a_1, a_2, \\dots, a_n$, respectively. Remark that the maximum and minimum values of $a_1 - 2a_2, a_2 - 2a_3, \\dots, a_n - 2a_1$ are also $M$ and $m$, respectively. Take $s$ satisfying $a_s = m$. Then, from $M \\ge a_{s... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a1 = a2 = ... = an = 0 | |
0ief | Problem:
What is the smallest integer $x$ larger than $1$ such that $x^{2}$ ends in the same three digits as $x$ does? | [
"Solution:\n\nThe condition is that $1000 \\mid x^{2}-x = x(x-1)$. Since $1000 = 2^{3} \\cdot 5^{3}$, and $2$ cannot divide both $x$ and $x-1$, $2^{3} = 8$ must divide one of them. Similarly, $5^{3} = 125$ must divide either $x$ or $x-1$. We try successive values of $x$ that are congruent to $0$ or $1$ modulo $125$... | United States | Harvard-MIT Mathematics Tournament | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 376 | |
00ef | Mati is playing with some magic boxes and a machine. Each box has a value inside. When opening a box Mati sees its value, adds the value of the box to his score and it is destroyed (if the box's value is negative, Mati loses points). By placing a magic box with value $X$ in the machine, this box is destroyed and we obt... | [
"We say we *multiply a box* when we place it in the machine. We say we *multiply a box* 2 times if we multiply the box and then multiply each of the resulting boxes. Similarly, we say we *multiply a box* $N$ times if we do so $N-1$ times and then multiply each of the resulting boxes, obtaining a total of $2^N$ boxe... | Argentina | Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof and answer | Part (a): Yes. Part (b): Yes. | |
0fky | Problem:
Se pintan de rojo algunas de las aristas de un poliedro regular. Se dice que una coloración de este tipo es buena, si para cada vértice del poliedro, existe una arista que concurre en dicho vértice y no está pintada de rojo. Por otra parte, se dice que una coloración donde se pintan de rojo algunas de las ari... | [
"Solution:\n\nClaramente, las coloraciones completamente buenas son un subconjunto de las coloraciones buenas, con lo que si el máximo número de aristas que se pueden pintar de rojo para obtener una coloración buena se puede alcanzar con una coloración completamente buena, la pregunta del enunciado tiene respuesta ... | Spain | Fase Nacional de la XLV Olimpiada Matemática Española | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | tetrahedron, cube, dodecahedron | |
00l1 | Max has $2015$ jars labelled with the numbers $1$ to $2015$ and an unlimited supply of coins. Consider the following starting configurations:
a. All jars are empty.
b. Jar $1$ contains $1$ coin, jar $2$ contains $2$ coins, and so on, up to jar $2015$ which contains $2015$ coins.
c. Jar $1$ contains $2015$ coins, jar... | [
"Max can achieve his goal in all three cases by the procedures described below. Let $N = 2015$ be the number of jars.\n\na. Let Max select jar $j$ exactly $\\left(\\frac{N!}{j}\\right)$ times. Then jar $j$ will contain\n$$\n\\sum_{k \\neq j} k \\cdot \\frac{N!}{k} = (N-1) \\cdot N!\n$$\ncoins which does not depend ... | Austria | Austrian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Discrete Mathematics > Algorithms"
] | null | proof and answer | Yes for all three starting configurations. | |
0jsf | Problem:
How many functions $f:\{0,1\}^{3} \rightarrow \{0,1\}$ satisfy the property that, for all ordered triples $(a_{1}, a_{2}, a_{3})$ and $(b_{1}, b_{2}, b_{3})$ such that $a_{i} \geq b_{i}$ for all $i$, $f(a_{1}, a_{2}, a_{3}) \geq f(b_{1}, b_{2}, b_{3})$? | [
"Solution:\nConsider the unit cube with vertices $\\{0,1\\}^{3}$. Let $O=(0,0,0)$, $A=(1,0,0)$, $B=(0,1,0)$, $C=(0,0,1)$, $D=(0,1,1)$, $E=(1,0,1)$, $F=(1,1,0)$, and $P=(1,1,1)$. We want to find a function $f$ on these vertices such that $f(1, y, z) \\geq f(0, y, z)$ (and symmetric representations). For instance, if... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 20 | |
0lfb | Consider functions $f: \mathbb{R} \to \mathbb{R}$ and $g: \mathbb{R} \to \mathbb{R}$ satisfying $f(0) = 2022$ and
$$
f(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x), \quad \forall x, y \in \mathbb{R}.
$$
a) Prove that $f$ is surjective and $g$ is injective.
b) Find all functions $f, g$ satisfying the given conditions. | [
"a. With $x, y \\in \\mathbb{R}$, $P(x, y)$ indicates the proposition containing the variable as follows\n$$\nf(x + g(y)) = x f(y) + (2023 - y) f(x) + g(x).\n$$\nFrom $P(0, y)$, we deduce\n$$\nf(g(y)) = (2023 - y) f(0) + g(0) = 2022(2023 - y) + g(0), \\quad \\forall y \\in \\mathbb{R}.\n$$\nNote that the right hand... | Vietnam | Vietnamese Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All solutions are the two affine pairs:
1) f(x) = ((-1 + sqrt(5)) / 2) x + 2022, g(x) = 1011(-1 - sqrt(5)) x + 2·1011^2·(-3 - sqrt(5));
2) f(x) = ((-1 - sqrt(5)) / 2) x + 2022, g(x) = 1011(-1 + sqrt(5)) x + 2·1011^2·(-3 + sqrt(5)). | |
0iz1 | Problem:
Suppose that $x$ and $y$ are positive reals such that
$$
x - y^{2} = 3, \quad x^{2} + y^{4} = 13
$$
Find $x$. | [
"Solution:\nSquaring both sides of $x - y^{2} = 3$ gives $x^{2} + y^{4} - 2x y^{2} = 9$.\n\nSubtract this equation from twice the second given to get\n$$\n2(x^{2} + y^{4}) - (x^{2} + y^{4} - 2x y^{2}) = 2 \\times 13 - 9\n$$\nwhich simplifies to\n$$\nx^{2} + 2x y^{2} + y^{4} = 17\n$$\nSo $(x + y^{2})^{2} = 17$, henc... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | (3 + sqrt(17)) / 2 | |
06r4 | Six stacks $S_{1}, \ldots, S_{6}$ of coins are standing in a row. In the beginning every stack contains a single coin. There are two types of allowed moves:
Move 1: If stack $S_{k}$ with $1 \leq k \leq 5$ contains at least one coin, you may remove one coin from $S_{k}$ and add two coins to $S_{k+1}$.
Move 2: If stack... | [
"Denote by $\\left(a_{1}, a_{2}, \\ldots, a_{n}\\right) \\rightarrow\\left(a_{1}^{\\prime}, a_{2}^{\\prime}, \\ldots, a_{n}^{\\prime}\\right)$ the following: if some consecutive stacks contain $a_{1}, \\ldots, a_{n}$ coins, then it is possible to perform several allowed moves such that the stacks contain $a_{1}^{\\... | IMO | 51st IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | Yes, it is possible. | |
0502 | A square $ABCD$ lies in the coordinate plane with its vertices $A$ and $C$ lying on different coordinate axes. Prove that one of the vertices $B$ or $D$ lies on the line $y = x$ and the other one on $y = -x$. | [
"Assume without loss of generality that $A$ is located on the $x$-axis and $C$ is located on the $y$-axis, let these points have coordinates of $A (a, 0)$ and $C (0, c)$. As the diagonals of a square bisect each other, we know that the intersection point $P$ of diagonal is also the mid-point of $AC$, i.e. $P(\\frac... | Estonia | Selected Problems from Open Contests | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Ve... | English | proof only | null | |
0hhx | Find all natural numbers $x, y, z$ that satisfy the equation: $2^x + 21^y = z^2$. | [
"Let's consider two cases for $x$.\n\nFor odd $x$, we have\n$$\nz^2 \\equiv 2^{2n+1} + 21^y \\equiv 2 + 0 \\equiv 2 \\pmod{3},\n$$\nwhich has no solutions.\n\nLet $x = 2n$. Then\n$$\n21^y = z^2 - 2^{2n} = (z - 2^n)(z + 2^n).\n$$\nSuppose that one of the two prime numbers $3$ or $7$ divides each of the two factors o... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (x, y, z) = (2, 1, 5) | |
0b5h | Given $a$, $b$ distinct positive integers, show that the system of equations
$$
xy + zw = a
$$
$$
xz + yw = b
$$
has only finitely many solutions in integers $x$, $y$, $z$, $w$. | [
"By adding and subtracting the equations we get $(x + w)(y + z) = a + b$ and $(x - w)(y - z) = a - b$, hence by multiplying, $0 < |(x^2 - w^2)(y^2 - z^2)| = |a^2 - b^2|$. Therefore $0 < |x^2 - w^2| \\le |a^2 - b^2|$ and $0 < |y^2 - z^2| \\le |a^2 - b^2|$. But an equation $0 < |A^2 - B^2| \\le |C|$ has only finitely... | Romania | Local Mathematical Competitions | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
00f4 | A set of $1990$ persons is divided into non-intersecting subsets in such a way that
(a) no one in a subset knows all the others in the subset;
(b) among any three persons in a subset, there are always at least two who do not know each other; and
(c) for any two persons in a subset who do not know each other, there is e... | [
"(i) Let $S$ be a subset of persons satisfying conditions (a), (b) and (c). Let $x \\in S$ be one who knows the maximum number of persons in $S$.\nAssume that $x$ knows $x_{1}, x_{2}, \\ldots, x_{n}$. By (b), $x_{i}$ and $x_{j}$ are strangers if $i \\neq j$. For each $x_{i}$, let $N_{i}$ be the set of persons in $S... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 398 | |
0hmg | Problem:
A line $l$ and two points $A$ and $B$ are given in a plane in such a way that $A$ belongs to $l$ but $B$ doesn't. Construct the circle $k$ that passes through $B$ and touches $l$ at the point $A$. | [
"Solution:\n\nLet $m$ be the bisector of the segment $A B$. The center of the circle $k$ has to belong to $m$. Similarly, since the circle has to be tangent to $l$ at the point $A$ its center has to be located on the line $a$ perpendicular to $l$ that passes through $A$.\n\nThe lines $m$ and $a$ are easy to constru... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0h8f | Three pairwise distinct numbers $a$, $b$, $c$ are such that their product is $80$.
Determine the least possible prime sum of these numbers. | [
"Clearly, their sum is greater than $2$, thus their sum is odd. Since all three numbers cannot be odd (since their product is $80$), then two of the numbers are even and one is odd. There are only two odd divisors of $80 = 2^4 \\cdot 5$: $1$ and $5$. Consider these cases.\n\n$c=1$, the following is possible:\n\n$a ... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 19 | |
04xq | Let $ABCD$ be a cyclic quadrilateral and $\omega$ its circumcircle. Let $I$, $J$, and $K$ be the incenters of the triangles $ABC$, $ACD$, and $ABD$, respectively. Also, let $E$ be the midpoint of the arc $DB$ of $\omega$ containing $A$. The line $EK$ intersects $\omega$ at $F$ ($F \neq E$). Prove that the points $C$, $... | [
"Let $M$ and $N$ be the midpoints of the arcs $AB$ and $AD$ (not containing any other vertices of $ABCD$), respectively. Then the incenter $I$ lies on $CM$, the incenter $J$ lies on $CN$, and the incenter $K$ lies on $BN$. Moreover, we have\n$$\nMI = MA = MB \\quad \\text{and} \\quad NJ = NA = ND = NK \\quad (1)\n$... | Czech-Polish-Slovak Mathematical Match | 12th Czech-Polish-Slovak Mathematics Competition | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscel... | English | proof only | null | |
0hzh | Problem:
$f$ is a continuous real-valued function such that $f(x+y)=f(x) f(y)$ for all real $x, y$. If $f(2)=5$, find $f(5)$. | [
"Solution:\n\nSince $f(n x)=f(x)^{n}$ for all integers $n$, $f(5)=f(1)^{5}$ and $f(2)=f(1)^{2}$, so $f(5)=f(2)^{5 / 2}=25 \\sqrt{5}$.",
"Solution:\n\nMore generally, since $f(n x)=f(x)^{n}$ for all integers $n$, $f(1)=c=f(1 / n)^{n}$ for some constant $c$ and all integers $n$. Thus $f(k / n)=f(1 / n)^{k}=f(1)^{k ... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | final answer only | 25 sqrt(5) | |
047j | Let $m$, $n$, $r$ be positive integers satisfying $n > m$, where both $m^2 + r$ and $n^2 + r$ are powers of $2$. Prove that:
$$
n > \frac{2m^2}{r}.
$$ | [
"*Proof.* Let $m^2 + r = 2^k$ and $n^2 + r = 2^l$ with $k, l \\in \\mathbb{Z}_{>0}$.\nFirst, we analyze the 2-adic valuations:\n* Since $v_2(m^2) = v_2(2^k - r) = v_2(r)$, we conclude $v_2(r)$ must be even.\n* Let $v_2(r) = 2t$ where $t \\in \\mathbb{Z}_{\\ge 0}$, and write $r = 2^{2t}r_1$, $m = 2^t m_1$.\n* Simila... | China | 2024 CGMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0a8t | Problem:
Let $x_{11}, x_{21}, \ldots, x_{n1}$, $n>2$, be a sequence of integers. We assume that all of the numbers $x_{i1}$ are not equal. Assuming that the numbers $x_{1k}, x_{2k}, \ldots, x_{nk}$ have been defined, we set
$$
\begin{aligned}
& x_{i, k+1}=\frac{1}{2}\left(x_{ik}+x_{i+1, k}\right), \quad i=1,2, \ldots,... | [
"Solution:\n\nWe compute the first index modulo $n$, i.e. $x_{1k}=x_{n+1, k}$. Let $M_{k}=\\max_{j} x_{jk}$ and $m_{k}=\\min_{j} x_{jk}$. Evidently $(M_{k})$ is a non-increasing and $(m_{k})$ a non-decreasing sequence, and $M_{k+1}=M_{k}$ is possible only if $x_{jk}=x_{j+1, k}=M_{k}$ for some $j$. If exactly $p$ co... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 18 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | For odd n, some term becomes non-integer. For even n, not necessarily: for example, starting with alternating zeros and twos yields all ones from the next step onward, so all terms remain integers. | |
0jyx | Problem:
Triangle $A B C$ has $A B=10$, $B C=17$, and $C A=21$. Point $P$ lies on the circle with diameter $A B$. What is the greatest possible area of $A P C$? | [
"Solution:\n\nTo maximize $[A P C]$, point $P$ should be the farthest point on the circle from $A C$. Let $M$ be the midpoint of $A B$ and $Q$ be the projection of $M$ onto $A C$. Then $P Q = P M + M Q = \\frac{1}{2} A B + \\frac{1}{2} h_{B}$, where $h_{B}$ is the length of the altitude from $B$ to $A C$.\n\nBy Her... | United States | HMMT November 2017 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 189/2 | |
0jj3 | Problem:
Sindy writes down the positive integers less than $200$ in increasing order, but skips the multiples of $10$. She then alternately places $+$ and $-$ signs before each of the integers, yielding an expression $+1-2+3-4+5-6+7-8+9-11+12-\cdots-199$. What is the value of the resulting expression? | [
"Solution:\n\nGroup the numbers into $(1-2+3-4+\\ldots+18-19)+(21-22+\\ldots+38-39)+\\ldots+(181-182+\\ldots+198-199)$. We can easily show that each group is equal to $-10$, and so the answer is $-100$."
] | United States | HMMT November 2014 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | -100 | |
0fna | Resuelve la ecuación exponencial
$$
2^x \cdot 3^{5-x} + \frac{3^{5x}}{2^x} = 6
$$ | [
"Aplicando la desigualdad de las medias aritmética y geométrica y, después, una de sus más conocidas consecuencias (la suma de un número real positivo y su inverso es siempre mayor o igual que $2$, y la igualdad sólo se da para el número $1$) tenemos,\n$$\n6 = 2^x 3^{5-x} + 2^{-x} 3^{5x} \\geq 2\\sqrt{2^x 3^{5-x} 2... | Spain | Olimpiada Matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | Spanish | proof and answer | x = 0 | |
08oq | Problem:
A board $n \times n$ ($n \geq 3$) is divided into $n^{2}$ unit squares. Integers from $0$ to $n$ inclusive are written down: one integer in each unit square, in such a way that the sums of integers in each $2 \times 2$ square of the board are different. Find all $n$ for which such boards exist. | [
"Solution:\nThe number of the $2 \\times 2$ squares in a board $n \\times n$ is equal to $(n-1)^{2}$. All possible sums of the numbers in such squares are $0, 1, \\ldots, 4n$. A necessary condition for the existence of a board with the required property is $4n + 1 \\geq (n-1)^{2}$ and consequently $n(n-6) \\leq 0$.... | JBMO | Junior Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | n = 3, 4, 5, 6 | |
04uj | Let $ABC$ be an acute triangle and $I$ its incenter. Rays $BI$, $CI$ meet the circumcircle of triangle $ABC$ again at $S \neq B$, $T \neq C$, respectively. The segment $ST$ meets the sides $AB$, $AC$ at $K$, $L$, respectively. Prove that $AKIL$ is a rhombus. (Josef Tkadlec) | [] | Czech Republic | First Round | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0bk3 | Let $A$ be an invertible matrix in $M_4(\mathbb{R})$, such that $\operatorname{tr} A = \operatorname{tr} A^* \neq 0$, where $A^*$ is the adjugate of $A$. Prove that the matrix $A^2 + I_4$ is singular if and only if there exists a nonzero matrix $B$ in $M_4(\mathbb{R})$, so that $AB = -BA$.
Mariean Andronache | [
"We show first that if $A$ is a matrix from $M_n(\\mathbb{R})$ such that $A^2 + I_n$ is singular, then there exists a matrix $B$ in $M_n(\\mathbb{R})$, such that $AB = -BA$.\nAs $A^2 + I_n$ is singular, $i$ is a proper value of $A$, and $-i$ is a proper value of the transpose $A^\\tau$. There exist two proper vecto... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0fp4 | Encontrar la solución entera más pequeña de la ecuación
$$
\left\lfloor \frac{x}{8} \right\rfloor - \left\lfloor \frac{x}{40} \right\rfloor + \left\lfloor \frac{x}{240} \right\rfloor = 210.
$$
(Si $x$ es un número real, $\lfloor x \rfloor$ es la parte entera de $x$, esto es, el mayor número entero menor o igual que $x$... | [
"Sea $x$ una solución entera de la ecuación. Dividiendo, primero por $240$, luego el resto $r_1$ por $40$ y el nuevo resto $r_2$ por $8$, resulta\n$$\nx = 240c_1 + r_1 = 240c_1 + 40c_2 + r_2 = 240c_1 + 40c_2 + 8c_3 + r_3,\n$$\ndonde $0 \\leq c_2 < 6$, $0 \\leq c_3 < 5$ y $0 \\leq r_3 < 8$ (las desigualdades para $c... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Spanish | proof and answer | 2016 | |
0bjg | Let $ABC$ be a triangle with $\angle A < 90^\circ$ and $AB \neq AC$. Denote by $H$ the ortocenter of triangle $ABC$, by $N$ the midpoint of $[AH]$, by $M$ the midpoint of the side $[BC]$ and by $D$ the intersection point of the angle bisector of $\angle BAC$ with $[MN]$. Prove that $\angle ADH = 90^\circ$. | [
"Let $O$ be the circumcircle of triangle $ABC$. Rays ($AO$ and ($AH$ are isogonal. We deduce that $\\angle NAD \\equiv \\angle OAD$. (1)\n\nBut $OM \\parallel AH$ and $OM = \\frac{1}{2}AH = AN$, which means that $MOAN$ is a parallelogram. It follows that $MN \\parallel OA$. (2)\n\nCombining (1) and (2) gives $\\ang... | Romania | THE 2014 DANUBE MATHEMATICAL COMPETITION | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0ayt | Problem:
Let $\alpha$ be the unique positive root of the equation
$$
x^{2018} - 11x - 24 = 0
$$
Find $\left\lfloor \alpha^{2018} \right\rfloor$. (Here, $\lfloor x\rfloor$ denotes the greatest integer less than or equal to $x$.) | [
"Solution:\nLet $\\alpha$ be the unique positive root of $x^{2018} - 11x - 24 = 0$.\n\nSo,\n$$\n\\alpha^{2018} = 11\\alpha + 24\n$$\nWe are to find $\\left\\lfloor \\alpha^{2018} \\right\\rfloor$.\n\nLet us estimate $\\alpha$.\n\nSuppose $\\alpha > 1$. Try $\\alpha = 2$:\n$$\n2^{2018} - 11 \\times 2 - 24 = 2^{2018}... | Philippines | 21st PMO Area Stage | [
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 35 | |
03eg | Solve the equation
$$
(x + 1)\sqrt{x^2 + 2x + 2} + x\sqrt{x^2 + 1} = 0.
$$ | [
"Let $a = \\sqrt{x^2 + 2x + 2} > 0$ and $b = \\sqrt{x^2 + 1} > 0$. Therefore\n$$ x = \\frac{(x^2 + 2x + 2) - (x^2 + 1) - 1}{2} = \\frac{a^2 - b^2 - 1}{2} $$\n$$ x + 1 = \\frac{a^2 - b^2 + 1}{2}. $$\nThe equation is equivalent to:\n$$ \\frac{a^2 - b^2 + 1}{2} \\cdot a + \\frac{a^2 - b^2 - 1}{2} \\cdot b = 0 $$\n$$(a... | Bulgaria | Autumn tournament | [
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | -1/2 | |
0emm | In triangle $XYZ$, $L$ is a variable point on a fixed line passing through $X$. $LZ$ meets $XY$ at $P$ and $LY$ meets $XZ$ at $Q$. Show that $PQ$ passes through a fixed point on $YZ$. | [
"Let $R$ be the intersection of $YZ$ and $PQ$. We wish to prove that $R$ is constant. Let $R'$ be the intersection of $LQ$ and $XR$. Then it is known that $L, Y, R', Q$ form a harmonic range, i.e. $(L, Y; R', Q) = -1$. The cross ratio depends on the angles $\\angle LXY, \\angle YXR', \\angle R'XZ$. Since $XL, XY$ a... | South Africa | South-Afrika 2011-2013 | [
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0c5a | Let $ABCD A'B'C'D'$ be a cuboid so that $ABB'A'$ is a square, and let the points $M \in (AC)$, $N \in (BC')$ be such that $AM = BN$ and $MN \perp BD'$.
a) Prove that $M$ and $N$ are the midpoints of the segments $AC$, respectively $BC'$.
b) If, moreover, $d(MN, BD') = \frac{1}{6}AN$ and $AB \ge BC$, then $ABCD A'B'C'... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Algebra > Linear Algebra > Vectors"
] | English | proof only | null | |
0jhq | Problem:
Find the largest real number $\lambda$ such that $a^{2}+b^{2}+c^{2}+d^{2} \geq a b+\lambda b c+c d$ for all real numbers $a, b, c, d$. | [
"Solution:\n\nLet $f(a, b, c, d) = (a^{2}+b^{2}+c^{2}+d^{2})-(a b+\\lambda b c+c d)$. For fixed $(b, c, d)$, $f$ is minimized at $a=\\frac{b}{2}$, and for fixed $(a, b, c)$, $f$ is minimized at $d=\\frac{c}{2}$, so simply we want the largest $\\lambda$ such that $f\\left(\\frac{b}{2}, b, c, \\frac{c}{2}\\right)=\\f... | United States | HMMT November 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 3/2 | |
01tf | Two points $A$ and $B$ are marked on the right branch of the hyperbola $y = 1/x$ ($x > 0$). The straight line $\ell$ passing through the origin $O$ is perpendicular to the line $AB$ and meets $AB$ and the given branch of the hyperbola at points $D$ and $C$, respectively. The circle $S$ passes through the points $A, B, ... | [
"Let $A(a; 1/a)$, $B(b; 1/b)$, $C(c; 1/c)$. It is easy to see that $y = -\\frac{x}{ab} + \\frac{1}{a} + \\frac{1}{b}$ is the equation of the line $AB$ and $y = abx$ is the equation of the line $OD$ because $OD$ is perpendicular to $AB$. Then we easily calculate the coordinates of $C$: $C = (\\frac{1}{\\sqrt{ab}}, \... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | English | proof and answer | 1/2 | |
06ef | Suppose $M$ is a point on the side $AB$ of $\triangle ABC$ such that the incircles of $\triangle AMC$ and $\triangle BMC$ have the same radius. The two incircles, centered at $O_1$ and $O_2$, meet $AB$ at $P$ and $Q$ respectively. It is known that the area of $\triangle ABC$ is six times the area of the quadrilateral $... | [
"The ratio can be $\\frac{5}{3}$ or $\\frac{5}{4}$.\nLet $a = BC$, $b = CA$, $c = AB$, $d = a+b$, $s = \\frac{a+b+c}{2}$, $x = MA$, $y = MB$ and $z = MC$. Also, let $r$ be the inradius of $\\triangle ABC$, and let $r'$ be the inradius of $\\triangle AMC$.\nFirstly, by considering $\\triangle AMC$, we obtain $r' = \... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | proof and answer | 5/3 or 5/4 | |
0gpu | Some unit squares of the grid $2013 \times 2013$ are marked so that any sub-square $19 \times 19$ of the grid consisting of unit squares has at least 21 marked unit squares. What is the minimal possible number of marked unit squares? | [
"The answer is $233625$.\n\nSuppose that the centers of unit squares have coordinates $(i,j)$, where $i = 1,2,\\ldots,2013$; $j = 1,2,\\ldots,2013$. The unit square with center at $(i,j)$ will be denoted by $u(i,j)$. Let the marked unit squares be:\n$u(19k, 19l + 1)$, $u(19k, 19l + 2)$, where $1 \\le k \\le 105$, $... | Turkey | Team Selection Test for IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 233625 | |
08di | Problem:
È data una sequenza di 2019 numeri $a_{1}, a_{2}, a_{3}, \ldots, a_{2019}$. Si sa che scelti qualsiasi 4 termini consecutivi della sequenza, la loro somma è costante. Similmente, presi due numeri consecutivi, la loro differenza in valore assoluto è costante (cioè $\left|a_{1}-a_{2}\right|=\left|a_{2}-a_{3}\ri... | [
"Solution:\n\nLa risposta è (D). Dato che la somma di 4 termini consecutivi è costante, per ogni $n \\geq 1$ abbiamo $a_{n}+a_{n+1}+a_{n+2}+a_{n+3}=a_{n+1}+a_{n+2}+a_{n+3}+a_{n+4}$, da cui $a_{n}=a_{n+4}$, cioè la sequenza è periodica e si ripete dopo 4 termini.\n\nPoiché vale $a_{1}<a_{2}<a_{3}$, le differenze $a_... | Italy | Gara di Febbraio | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | MCQ | D | |
0juq | Problem:
Let $a, b, c, d, e, f$ be integers selected from the set $\{1,2, \ldots, 100\}$, uniformly and at random with replacement. Set
$$
M = a + 2b + 4c + 8d + 16e + 32f
$$
What is the expected value of the remainder when $M$ is divided by $64$?
Proposed by: Evan Chen | [
"Solution:\nConsider $M$ in binary. Assume we start with $M = 0$, then add $a$ to $M$, then add $2b$ to $M$, then add $4c$ to $M$, and so on. After the first addition, the first bit (defined as the rightmost bit) of $M$ is toggled with probability $\\frac{1}{2}$. After the second addition, the second bit of $M$ is ... | United States | HMMT February 2016 | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 63/2 | |
09fz | Find all $n$ such that any convex $n$-gon in the plane can be divided into finite number of triangles satisfying the following conditions.
(i) No vertex is added to the sides of the $n$-gon. However, any number of vertices can be added to the interior of the $n$-gon.
(ii) All triangles have exactly three bounding edge... | [
"This is essentially Exercise 6.4.10 from Invitation to Discrete Mathematics, 2nd edition, Oxford University Press, by Jiří Matoušek and Jaroslav Nešetřil."
] | Mongolia | 2015 Mongolian IMO Team Selection Tests | [
"Discrete Mathematics > Graph Theory > Euler characteristic: V-E+F",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | All n that are multiples of 3 | |
01e8 | Let $b_i$, $c_i$, $0 \le i \le 100$ be two sequences of positive integers with two exceptions: $c_0 = 0$, $b_{100} = 0$. Several villages are connected by roads, each road connects two villages which are called neighbours and has length $1$ km. Roads do not intersect each other, but can pass over/under each other. The ... | [
"Let $z$ be an arbitrary village, $S_i(z)$ be the set of villages at distance $i$ from $z$, and $k_i$ be the number of elements in $S_i(z)$. Then the sequence $k_i$ does not depend on $z$!\nWe prove this statement by induction on $i$. We will show also that $k_{i+1} = k_i \\cdot \\frac{b_i}{c_i}$. Clearly, $k_0 = 1... | Baltic Way | Baltic Way shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
0gxy | The integers $a, b, c$ satisfy the following condition: $\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = 0$. Prove that the number $a^2 + b^2 + c^2$ is a perfect square of some integer. | [
"We note the following facts: $abc \\neq 0$ and $ab + bc + ca = 0$, then\n$$(a + b + c)^2 = a^2 + b^2 + c^2 + 2ab + 2bc + 2ca = a^2 + b^2 + c^2$$\nas required."
] | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
05r2 | Problem:
Montrer que, parmi $2048$ entiers, on peut toujours en trouver $1024$ dont la somme est divisible par $1024$. | [
"Solution:\n\nOn va montrer par récurrence sur $n \\geqslant 0$ la propriété $P(n)$ selon laquelle, parmi $2^{n+1}$ entiers, on peut toujours en trouver $2^{n}$ dont la somme est divisible par $2^{n}$. Tout d'abord, pour $n=0$, le résultat est évident, puisque tout entier est divisible par $2^{0}=1$. La propriété $... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0aeg | Let $O$ be the circumcentre of the triangle $ABC$. Let $K$ and $L$ be the intersection points of the circumcircles of the triangles $BOC$ and $AOC$ with the bisectors of the angles at $A$ and $B$ correspondently. Let $P$ be the midpoint of $\overline{KL}$, $M$ symmetrical to $O$ relatively $P$ and $N$ symmetrical to $O... | [
"The angles $LCA$ and $LOA$ are equal as inscribed angles over a same arc. The angle $LOA$ is equal to the sum of the angles $OAB$ and $OBA$ (because $LOA$ is an exterior angle of the triangle $ABO$). Hence:\n$$\n\\begin{aligned} \\angle LCO &= \\angle LCA + \\angle OCA = \\angle LOA + \\angle OCA = \\angle OAB + \... | North Macedonia | Sixteenth Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | Macedonian, English | proof only | null | |
0eyo | Problem:
In a group of students, $50$ speak English, $50$ speak French and $50$ speak Spanish. Some students speak more than one language. Prove it is possible to divide the students into $5$ groups (not necessarily equal), so that in each group $10$ speak English, $10$ speak French and $10$ speak Spanish. | [
"Solution:\n\nLet $EF$ denote the number of students speaking English and French. Similarly define $ES$, $FS$, $E$, $F$, $S$, $EFS$. Then $ES + EF + E + EFS = 50$, $EF + FS + F + EFS = 50$. Subtracting: $ES - F = FS - E$. Similarly, $ES - F = EF - S$.\n\nPair off members of $FS$ with members of $E$. Similarly, memb... | Soviet Union | 2nd ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0d0j | Find all positive real numbers $x$, $y$, $z$ if
$$
\frac{1}{x^2+1} + \frac{1}{y^2+4} + \frac{1}{z^2+9} = \frac{7}{12} \sqrt{\frac{1}{x^2} + \frac{1}{y^2} + \frac{1}{z^2}}.
$$ | [
"We have $x^2+1 \\ge 2x$, $y^2+4 \\ge 4y$, $z^2+9 \\ge 6z$, hence\n$$\n\\frac{1}{x^2+1} + \\frac{1}{y^2+4} + \\frac{1}{z^2+9} \\le \\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}. \\quad (1)\n$$\nUsing Cauchy-Schwarz inequality it follows\n$$\n\\left(\\frac{1}{2x} + \\frac{1}{4y} + \\frac{1}{6z}\\right)^2 \\le \\left... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | x = 1, y = 2, z = 3 | |
09xa | We have two integers consisting of two digits, and both numbers do not start with a $0$. If you add these numbers, you get the number $S$. If you interchange the two digits of both numbers and add the new numbers, you get $4S$.
Determine all possible pairs of two-digit numbers satisfying these constraints. Make sure to... | [
"$\\{14, 19\\}$, $\\{15, 18\\}$, and $\\{16, 17\\}$"
] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | {14, 19}, {15, 18}, {16, 17} | |
0jrb | Problem:
Evaluate
$$
sin (\arcsin (0.4)+\arcsin (0.5)) \cdot \sin (\arcsin (0.5)-\arcsin (0.4))
$$
where for $x \in[-1,1]$, $\arcsin (x)$ denotes the unique real number $y \in[-\pi, \pi]$ such that $\sin (y)=x$. | [
"Solution:\n\n$0.09$ OR $\\frac{9}{100}$\n\nUse the difference of squares identity $\\sin(a-b)\\sin(a+b)=\\sin^2(a)-\\sin^2(b)$ to get $0.5^2-0.4^2=0.3^2=0.09=\\frac{9}{100}$."
] | United States | HMMT February | [
"Precalculus > Trigonometric functions",
"Precalculus > Functions"
] | null | final answer only | 9/100 | |
02uy | Problem:
Um triângulo $AB C$ tem lados de comprimentos $AB = 50~\mathrm{cm}$, $BC = 20~\mathrm{cm}$ e $AC = 40~\mathrm{cm}$. Sejam $M$ e $N$ pontos no lado $AB$ tais que $CM$ é a bissetriz relativa ao ângulo $\angle ACB$ e $CN$ é a altura relativa ao lado $AB$. Qual a medida, em centímetros, de $MN$? | [
"Solution:\n\nPela Lei dos Cossenos aplicado ao triângulo $\\triangle ABC$, temos\n$$\n\\begin{aligned}\nAC^2 & = AB^2 + BC^2 - 2 \\cdot AB \\cdot BC \\cdot \\cos \\angle ABC \\\\\n1600 & = 2500 + 400 - 2 \\cdot 50 \\cdot 20 \\cdot \\cos \\angle ABC \\\\\n\\cos \\angle ABC & = \\frac{13}{20}\n\\end{aligned}\n$$\nAl... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 11/3 | |
0i58 | Problem:
Suppose that a positive integer $n$ has the property that $n, 2 n, 3 n, \ldots, 9 n$ are all palindromes. Prove that the decimal digits of $n$ are all zeros or ones. | [
"Solution:\nFirst consider the ones digit $a$ of $n$; we claim that $a=1$. Certainly $a$ cannot be even, for then $5 n$ would be divisible by $10$. If $a$ is $5$, $7$, or $9$, then $2 n$ has an even ones digit, while its most significant digit is $1$. If $a$ is $3$, then $4 n$ has an even ones digit but most signif... | United States | HMMT 2002 | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
02a5 | Problem:
Interseção entre circunferências - Com centros nos vértices do triângulo equilátero $\triangle ABC$ de lado $a$, foram desenhadas três circunferências de raio $r$.
Se $r < a$ e $2r > a$, estas três circunferências são duas a duas concorrentes nos pontos $X, Y$ e $Z$, exteriores ao triângulo $\triangle ABC$. M... | [
"Solution:\n\nSeja $G$ o baricentro (encontro das medianas) do triângulo $ABC$. Como a figura é invariante por rotações de $60^\\circ$ ao redor do ponto $G$, temos que o triângulo $\\triangle XYZ$ é equilátero, e que $G$ também é o seu baricentro.\n\nVamos calcular o comprimento $L$ do seu lado.\n\nSeja $CM$ a altu... | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a/2 + sqrt(3)*sqrt(r^2 - a^2/4) | |
06yj | Find all positive integers $n$ with the following property: for all positive divisors $d$ of $n$, we have that $d+1 \mid n$ or $d+1$ is prime. | [
"It is easy to verify that $n=1,2,4,12$ all work. We must show they are the only possibilities. We write $n=2^{k} m$, where $k$ is a nonnegative integer and $m$ is odd. Since $m \\mid n$, either $m+1$ is prime or $m+1 \\mid n$.\n\nIn the former case, since $m+1$ is even it must be $2$, so $n=2^{k}$. If $k \\geqslan... | IMO | IMO2024 Shortlisted Problems | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 1, 2, 4, 12 | |
0g0e | Problem:
Sei $n$ eine natürliche Zahl. Wir nennen ein Zahlenpaar unverträglich, falls ihr größter gemeinsamer Teiler gleich $1$ ist. Wie viele unverträgliche Paare treten mindestens auf, wenn man die Zahlen $\{1, 2, \ldots, 2n\}$ in $n$ Paare aufteilt? | [
"Solution:\n\nWir nennen ein nicht unverträgliches Paar verträglich. Es ist klar, dass Primzahlen strikt größer als $n$ Teil eines unverträglichen Paars sind, und $1$ ebenfalls Teil eines unverträglichen Paars ist. Also gibt es wegen diesen Zahlen mindestens\n$$\n\\left\\lceil \\frac{(\\text{Anzahl Primzahlen zwisc... | Switzerland | IMO-Selektion | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ceil((π(2n) − π(n) + 1)/2) | |
01lu | Point $P$ inside an acute-angled triangle $A_1A_2A_3$ is chosen so that its projections $P_1, P_2, P_3$ onto the sides $A_1A_2$, $A_2A_3$, $A_3A_1$ respectively lie on the sides of the triangle.
Prove that for points $X_1$, $X_2$, $X_3$ on the sides $A_1A_2$, $A_2A_3$, $A_3A_1$ respectively, $$\max \left\{ \frac{X_1X_2... | [
"a), b) (Solution of M. Mankevich, A. Nekrashevich, A. Tanana.) The statement from a) and b) simply follow from the general\n\n**Lemma.** If there is a point $X$ such that $X_1$, $X_2$, $X_3$ are the orthogonal projections of $X$ onto the sides $A_2A_3$, $A_3A_1$, $A_1A_2$ respectively, then the statement of the pr... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > ... | English | proof only | null |
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