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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bhl | Find all positive integers $n$ with at least two digits, $n$'s digits are pairwise distinct and $n$ equals the product of the sum of its digits with one of its digits. | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 24, 45, 48 | |
00r7 | Find all the pairs positive numbers $(x, y)$ with the following property:
If $\alpha, \beta$ are relatively prime and positive divisors of the number $x^3 + y^3$ then $\alpha + \beta - 1$ is a divisor of $x^3 + y^3$. | [
"We prove that $(x, y) = (2^n, 2^n)$ or $(x, y) = (3^n, 2 \\cdot 3^n)$ or $(x, y) = (2 \\cdot 3^n, 3^n)$ for a natural number $n$. We can easily check those solutions.\n\n**Step 1:** The number $x^3 + y^3$ has at most one odd prime divisor. Let $p_1 < p_2 < \\cdots < p_k$ be the odd prime divisors of $x^3 + y^3$ wi... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | null | proof and answer | (x, y) = (2^n, 2^n) or (x, y) = (3^n, 2·3^n) or (x, y) = (2·3^n, 3^n) for integers n ≥ 0 | |
0i64 | Problem:
A square and a regular hexagon are drawn with the same side length. If the area of the square is $\sqrt{3}$, what is the area of the hexagon? | [
"Solution:\n\nThe hexagon is composed of six equilateral triangles each of side length $\\sqrt[4]{3}$ (with base $b = \\sqrt[4]{3}$ and height $\\frac{\\sqrt{3}}{2} b$), so the total area is $\\frac{9}{2}$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | 9/2 | |
0ig3 | Problem:
If $a, b, c > 0$, what is the smallest possible value of $\left\lfloor\frac{a+b}{c}\right\rfloor + \left\lfloor\frac{b+c}{a}\right\rfloor + \left\lfloor\frac{c+a}{b}\right\rfloor$? (Note that $\lfloor x \rfloor$ denotes the greatest integer less than or equal to $x$.) | [
"Solution:\nSince $\\lfloor x \\rfloor > x - 1$ for all $x$, we have that\n$$\n\\begin{aligned}\n\\left\\lfloor\\frac{a+b}{c}\\right\\rfloor + \\left\\lfloor\\frac{b+c}{a}\\right\\rfloor + \\left\\lfloor\\frac{c+a}{b}\\right\\rfloor &> \\frac{a+b}{c} + \\frac{b+c}{a} + \\frac{c+a}{b} - 3 \\\\\n&= \\left(\\frac{a}{b... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 4 | |
0g9t | 已知正實數 $x, y, z$ 滿足 $x + y + z = 1$ 的正實數。試求使不等式
$$
\frac{x^2 y^2}{1-z} + \frac{y^2 z^2}{1-x} + \frac{z^2 x^2}{1-y} \le k - 3xyz
$$
恆成立的實數 $k$ 的最小值。
Let $x, y, z$ be positive real numbers satisfying $x + y + z = 1$. Find the smallest $k$ such that
$$
\frac{x^2 y^2}{1-z} + \frac{y^2 z^2}{1-x} + \frac{z^2 x^2}{1-y} \le k ... | [
"解：先令 $x = y = z = \\frac{1}{3}$，則 $k \\ge \\frac{1}{6}$。底下證明：\n$$\n\\frac{x^2 y^2}{1-z} + \\frac{y^2 z^2}{1-x} + \\frac{z^2 x^2}{1-y} \\le \\frac{1}{6} - 3xyz. \\quad (1)\n$$\n由 $x > 0, y > 0, z > 0$ 及 $x + y + z = 1$, 知不等式 (1) 等價於\n$$\n\\begin{align*}\n& \\frac{xy}{z(x+y)} + \\frac{yz}{x(y+z)} + \\frac{zx}{y(z+x)... | Taiwan | 二〇一六數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 1/6 | |
060z | Problem:
Déterminer tous les couples $(a, p)$ d'entiers strictement positifs, avec $p$ premier, tels que pour tout couple $(m, n)$ d'entiers strictement positifs, le reste de la division euclidienne de $a^{2^{n}}$ par $p^{n}$ est non nul, et est le même que celui de $a^{2^{m}}$ par $p^{m}$. | [
"Solution:\n\nEn prenant $n=1$ dans l'énoncé, on obtient que pour tout entier $m$ strictement positif, le reste de $a^{2^{m}}$ modulo $p^{m}$ vaut celui de $a^{2}$ modulo $p$, donc est constant. Notons $r$ ce reste : on a donc $r \\neq 0$ ainsi $r \\in \\{1, \\ldots, p-1\\}$. On a $a^{2} \\equiv r \\pmod{p}$ et $a^... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All pairs with the prime equal to two and the integer odd, and all pairs with the integer equal to one and any odd prime. | |
0hk7 | Problem:
Given 8 oranges on the table, 7 of them have exactly the same weight and the 8th is a little bit lighter. You are given a balance that can measure oranges against each other and you are allowed to use the balance at most twice! How can you determine which one of the oranges is lighter than the others? Explain... | [
"Solution:\n\nFirst we put 3 oranges on the left and 3 on the right-hand side of the balance. In the case that the balance shows equal weights, one of the remaining two oranges is lighter and in the second measurement we can easily see which one. However, if one side (say left) is lighter, then we know that the ora... | United States | Berkeley Math Circle Monthly Contest 3 | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Logic"
] | null | proof only | null | |
095j | Problem:
Fie $ABC$ un triunghi arbitrar, iar $A_{1}$, $B_{1}$ şi $C_{1}$ trei puncte, $A_{1} \in (BC)$, $B_{1} \in (AC)$, $C_{1} \in (AB)$, astfel încât $\frac{BA_{1}}{A_{1}C} = \frac{CB_{1}}{B_{1}A} = \frac{AC_{1}}{C_{1}B}$. Să se demonstreze că lungimile segmentelor $[AA_{1}]$, $[BB_{1}]$ şi $[CC_{1}]$ pot servi ca l... | [
"Solution:\nPrin punctul $C_{1}$ ducem o paralelă la $BB_{1}$, iar prin punctul $B_{1}$ o paralelă la $AB$. Notăm cu $M$ punctul de intersecţie a acestor paralele, iar $MB_{1} \\cap BC = \\{N\\}$. Fie $A_{1}P \\parallel AC$, $P \\in AB$.\n\nVom arăta că laturile triunghiului $MCC_{1}$ au lungimile egale cu lungimil... | Moldova | A 62-a OLIMPIADĂ DE MATEMATICĂ A REPUBLICII MOLDOVA | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
00sv | Let $P(x)$, $Q(x)$ be distinct polynomials of degree $2020$ with non-zero coefficients. Suppose that they have $r$ common real roots counting multiplicity and $s$ common coefficients. Determine the maximum possible value of $r + s$. | [
"We claim that the maximum possible value is $3029$.\nThe polynomials\n$$\nP(x) = (x^2 - 1)^{1009}(x^2 + 1) \\quad \\text{and} \\quad Q(x) = (x^2 - 1)^{1009}(x^2 + x + 1)\n$$\nsatisfy the conditions, have $2018$ common roots, and have $1011$ common coefficients (all coefficients of even powers). So $r + s \\ge 3029... | Balkan Mathematical Olympiad | BMO Short List | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Descartes' Rule of Signs"
] | English | proof and answer | 3029 | |
0l4t | Problem:
Let $ABCD$ be a convex quadrilateral. Define parabolas $\mathcal{P}_A$, $\mathcal{P}_B$, $\mathcal{P}_C$, and $\mathcal{P}_D$ to have directrices $BD$, $CA$, $DB$, and $AC$, and foci $A$, $B$, $C$, and $D$, respectively. Prove that no two of these parabolas intersect more than once.
(A parabola with directri... | [
"Solution:\n\nLet $d(P, XY)$ be the distance from $P$ to line $XY$. We will first prove $\\mathcal{P}_A$ and $\\mathcal{P}_B$ intersect at most once.\n\nClaim 1. Let $\\ell_{AB}$ be the perpendicular bisector of $AB$. Then, $\\mathcal{P}_A$ is tangent to $\\ell_{AB}$.\n\nProof. Consider any point $X$ on $\\mathcal{... | United States | HMIC | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
04tv | Paul is filling the cells of a rectangular table alternately with crosses and circles (he starts with a cross). When the table is filled in completely, he determines his score as $X + O$ where $X$ is the number of rows containing more crosses than circles and $O$ is the number of columns containing more circles than cr... | [
"In total there are $2n(n+1)+1 < (2n+1)(n+1)$ crosses and $2n(n+1)$ circles. Hence the crosses can dominate in at most $2n$ rows and, similarly, circles can dominate in at most $2n$ columns for the total score $2n+2n=4n$.\n\nSuch a score can be achieved if, for example, Paul draws crosses in the left $n+1$ columns ... | Czech Republic | 67th Czech and Slovak Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 4n | |
069y | Determine all non-constant polynomials $P(x)$ and $Q(x)$ with real coefficients satisfying the equation
$$
P((Q(x))^3) = xP(x)(Q(x))^3.
$$ | [
"Let: $\\deg P(x) = m \\ge 1$, $\\deg Q(x) = n \\ge 1$. By taking the degrees of the two members of the given equation we get the equality:\n$$\n\\begin{aligned}\n3mn = 1 + m + 3n &\\Leftrightarrow (m-1)(3n-1) = 2 \\\\\n&\\Leftrightarrow m-1=1, 3n-1=2 \\quad \\text{or } m-1=2, 3n-1=1\\; (\\text{impossible in } \\ma... | Greece | 37th Hellenic Mathematical Olympiad 2020 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | P(x) = k x^2 with k ≠ 0 and Q(x) = x | |
0gp3 | Find all positive integers $n$ for which an integer that can be written as sum of squares of $n$ integers with each of them is divisible by $n$, can also be expressed as sum of squares of $n$ integers with none of them is divisible by $n$. | [
"The answer is all positive integers except $1$, $2$ and $4$.\n\nLet us call a positive integer *good* if it satisfies the condition given in the problem. We first show that if $n$ is good, so is any multiple of $n$.\n\nLet $m = nk$ and $x_1, x_2, \\dots, x_m$ be integers such that $m \\mid x_i$ for all $1 \\le i \... | Turkey | Team Selection Test | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | English | proof and answer | All positive integers except 1, 2, and 4. | |
07e0 | A table consisting of 5 columns and 32 rows, which are filled with zero and one numbers, called *varied*, if no two rows are filled in the same way.
On the exterior of a cylinder, a table with 32 rows and 16 columns is constructed. Is it possible to fill the numbers cells of the table with numbers zero and one, such th... | [
"**Answer.** Yes.\n\nImagine a $32 \\times 5$ table $T$ such that the set of its rows is equal to the set of all 5-tuples $(b_0, b_1, \\ldots, b_4)$ in which $\\forall\\ 0 \\le i \\le 4 : b_i \\in \\{0,1\\}$.\n\nNow consider a $32 \\times 16$ table $R$. Let $(c_i)$ be the $i$-th column and the $j$-th column $(c_j)$... | Iran | Iranian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | Yes | |
090h | How many tuples of positive integers $(a, b, c, d)$ satisfy $abcd = 2025$, such that each of $ab$, $bc$, $cd$, and $da$ is a perfect square? | [
"44\nSince $2025 = 3^4 \\times 5^2$, each of $a$, $b$, $c$, and $d$ has no prime factors other than $3$ and $5$. Therefore, we can write\n$$\na = 3^{x_1} 5^{y_1}, \\quad b = 3^{x_2} 5^{y_2}, \\quad c = 3^{x_3} 5^{y_3}, \\quad d = 3^{x_4} 5^{y_4}\n$$\nfor some non-negative integers $x_1, x_2, x_3, x_4, y_1, y_2, y_3... | Japan | The 35th Japanese Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 44 | |
0hge | Consider the circumscribed circle of an obtuse triangle $ABC$ with an obtuse angle $B$. Tangents to this circle at points $A$ and $B$ meet at point $P$, and the perpendicular to the line $BC$ at point $B$ intersects $AC$ at point $K$. Prove, that $PA = PK$.
(Danylo Khilko) | [
"First note, that as $\\angle ABC > 90^\\circ$, point $K$ lies on $AC$ (fig. 1). Also, it's clear that $PA = PB$. We will show that $K$ lies on the circle $\\omega$ with a center $P$ and radius $PA$. It's enough to prove, that $\\angle APB = 2(180^\\circ - \\angle AKB)$. Indeed, take on the larger arc of circle $\\... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
09d2 | Тус бүрдээ нэгээс их $a, b$ натураал тоонуудын хувьд $a^b$ хэлбэртэй бичигдэх тоо 1-1000 хүртэлх тоонууд дунд хичнээн байх вэ? | [
"$a > 1, b > 1$ гэдгээс $a = 2$ гэе.\n$$\n1 < 2^b < 1000 \\Rightarrow b = 2, 3, 4, 5, 6, 7, 8, 9 \\text{ гэсэн } 8 \\text{ боломжтой}\n$$\n$$\na = 3 \\text{ гэе. } 1 < 3^b < 1000 \\Rightarrow b = 2, 3, 4, 5, 6 \\text{ гэсэн } 5 \\text{ боломжтой}\n$$\n$$\na = 4 \\text{ гэе. } 1 < 4^b < 1000 \\Rightarrow b = 2, 3, 4... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Number Theory > Other"
] | Mongolian | proof and answer | 40 | |
049c | In the quadrilateral $ABCD$ the following holds:
$$
\angle ABC = \angle ADC = 90^\circ, \quad |AB| = |BC|, \quad |CD| + |DA| = m.
$$
Determine the area of $ABCD$ in terms of $m$. | [] | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | m^2/4 | |
0l11 | The Fibonacci numbers are defined by $F_1 = 1$, $F_2 = 1$, and $F_n = F_{n-1} + F_{n-2}$ for $n \ge 3$. What is
$$
\frac{F_2}{F_1} + \frac{F_4}{F_2} + \frac{F_6}{F_3} + \dots + \frac{F_{20}}{F_{10}}?
$$ | [
"The Fibonacci sequence starts out $1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, 987, 1597, 2584, 4181, 6765, \\ldots$\nso the given sum is\n$$\n\\frac{1}{1} + \\frac{3}{1} + \\frac{8}{2} + \\frac{21}{3} + \\frac{55}{5} + \\frac{144}{8} + \\frac{377}{13} + \\frac{987}{21} + \\frac{2584}{34} + \\frac{67... | United States | AMC 10 B | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | final answer only | 319 | |
06y1 | Decide whether for every sequence $\left(a_{n}\right)$ of positive real numbers,
$$
\frac{3^{a_{1}}+3^{a_{2}}+\cdots+3^{a_{n}}}{\left(2^{a_{1}}+2^{a_{2}}+\cdots+2^{a_{n}}\right)^{2}}<\frac{1}{2024}
$$
is true for at least one positive integer $n$. | [
"Solution 1. For every positive integer $n$, let $M_{n}=\\max \\left(a_{1}, a_{2}, \\ldots, a_{n}\\right)$. We first prove that\n$$\n\\frac{3^{a_{1}}+3^{a_{2}}+\\cdots+3^{a_{n}}}{\\left(2^{a_{1}}+2^{a_{2}}+\\cdots+2^{a_{n}}\\right)^{2}} \\leqslant\\left(\\frac{3}{4}\\right)^{M_{n}}\n$$\nFor $i=1,2, \\ldots, n$, fro... | IMO | IMO2024 Shortlisted Problems | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | Yes | |
03s3 | Let $T = \{0, 1, 2, 3, 4, 5, 6\}$ and $M = \{\frac{a_1}{7} + \frac{a_2}{7^2} + \frac{a_3}{7^3} + \frac{a_4}{7^4}; a_i \in T, i = 1, 2, 3, 4\}$. Arrange the numbers in $M$ in the descending order. Then the 2 005-th number is ( ).
(A) $\frac{5}{7} + \frac{5}{7^2} + \frac{6}{7^3} + \frac{3}{7^4}$
(B) $\frac{5}{7} + \frac... | [
"Let $[a_1 a_2 \\cdots a_k]_p$ be a number base $p$ with $k$ digits. Multiply each number in $M$ by $7^4$, and we get\n$$\nM' = \\{a_1 7^3 + a_2 7^2 + a_3 7 + a_4; a_i \\in T, i = 1, 2, 3, 4\\} = \\{[a_1 a_2 a_3 a_4]'_7 \\mid a_i \\in T, i = 1, 2, 3, 4\\}.\n$$\nThe maximum number in $M'$ is $[6666]_7 = [2400]_{10}$... | China | China Mathematical Competition (Jiangxi) | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Other"
] | English | MCQ | C | |
0el3 | Problem:
Katero je najmanjše naravno število $n$, za katero interval $\left[\frac{n+8}{2}, \frac{2 n+14}{3}\right]$ vsebuje vsaj 4 naravna števila?
(A) 16
(B) 17
(C) 18
(D) 19
(E) Nobeno izmed naštetih. | [
"Solution:\n\nDolžina intervala je enaka $\\frac{2 n+14}{3}-\\frac{n+8}{2}=\\frac{(4 n+28)-(3 n+24)}{6}=\\frac{n+4}{6}$. Ker mora interval vsebovati vsaj 4 naravna števila, mora biti njegova dolžina vsaj 3 (saj sta lahko obe krajišči naravni števili). Torej mora biti $\\frac{n+4}{6} \\geq 3$ oziroma $n \\geq 14$. Č... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | MCQ | E | |
052k | Inside a circle $c$ there are circles $c_1$, $c_2$ and $c_3$ which are tangent to $c$ at points $A$, $B$ and $C$ correspondingly, which are all different. Circles $c_2$ and $c_3$ have a common point $K$ in the segment $BC$, circles $c_3$ and $c_1$ have a common point $L$ in the segment $CA$, and circles $c_1$ and $c_2$... | [
"Take a point $X$ on the common tangent to the circles $c_1$ and $c$ which lies on the other side of the line $AB$ from the point $C$. Then $\\angle ALM = \\angle XAM = \\angle XAB = \\angle ACB$ (Fig. 4). Consequently $ML \\parallel BC$. Similarly $KM \\parallel CA$ and $LK \\parallel AB$. If $\\frac{|AM|}{|AB|} =... | Estonia | Open Contests | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
04w4 | Saba is trying to get from the ground floor of an infinitely tall skyscraper to the $n$th floor using a special elevator. The elevator has buttons $0$, $1$, $2$, ... and it moves alternately up/down (first it moves up). Moreover, upon pressing button $k$, the elevator moves by $2^k$ floors, and each subsequent button p... | [] | Czech Republic | First Round (take-home) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | English | proof only | null | |
0em0 | Prove that the equation $x^5 + y^3 = z^2$ has infinitely many solutions in the natural numbers. | [
"Put $x = k^6$, $y = 2k^{10}$ and $z = 3k^{15}$ for some natural number $k$. Then\n$$\nx^5 + y^3 = (k^6)^5 + (2k^{10})^3 = k^{30} + 8k^{30} = 9k^{30} = (3k^{15})^2 = z^2\n$$\nas required. Since there are infinitely many possible values for $k$, there are infinitely many triples $(x, y, z)$."
] | South Africa | South-Afrika 2011-2013 | [
"Number Theory > Diophantine Equations"
] | null | proof only | null | |
0asz | Problem:
Find all integers $n$ such that $5 n - 7$, $6 n + 1$, and $20 - 3 n$ are all prime numbers. | [
"Solution:\nOnly $n = 6$."
] | Philippines | Philippines Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | n = 6 | |
0awn | Problem:
Solve the following inequality.
$$
\log_{1/2} x - \sqrt{2 - \log_{4} x} + 1 \leq 0
$$ | [
"Solution:\nNote that $2 - \\log_{4} x \\geq 0 \\Longrightarrow 0 < x \\leq 16$. Let $t = \\sqrt{2 - \\log_{4} x}$. Then\n$$\n\\log_{4} x = 2 - t^2 \\Longrightarrow \\frac{\\log_{1/2} x}{\\log_{1/2} 4} = 2 - t^2 \\Longrightarrow \\log_{1/2} x = 2 t^2 - 4\n$$\nSubstituting back to the given inequality, we have\n$$\n... | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | [1/sqrt(2), 16] | |
0dfm | Find all non-constant functions $f: \mathbb{Q}^+ \to \mathbb{Q}^+$ satisfying the equation
$$
f(ab + bc + ca) = f(a)f(b) + f(b)f(c) + f(c)f(a)
$$
for all $a, b, c \in \mathbb{Q}^+$. | [
"Put $c = 1$ in the given condition, we have\n$$\nf(ab + a + b) = f(a)f(b) + f(a)f(1) + f(b)f(1); \\quad \\forall a, b \\in \\mathbb{Q}^{+}, \\quad (1)\n$$\nPut $b = 3$ into (1), we have\n$$\nf(4a + 3) = f(a)f(3) + f(a)f(1) + f(3)f(1); \\quad \\forall a \\in \\mathbb{Q}^{+}.\n$$\nPut $b = 1$ into (1), we have\n$$\n... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | f(x) = x for all positive rational x | |
0adl | One rectangle has side lengths $a$ cm and $b$ cm. If the side with length $a$ cm is enlarged by $b$ cm and the side with length $b$ cm is enlarged by $a$ cm then the resulting square has area of $100\,\mathrm{cm}^2$. Determine the rectangle that satisfies this condition with smallest area if its side lengths are positi... | [
"The area of the resulting square is $P_{sq} = (a + b)^2 = 100\\,\\mathrm{cm}^2$, hence $a + b = 10\\,\\mathrm{cm}$. Because $a$ and $b$ are positive integers we have the following cases: $a = 1$, $b = 9$; $a = 2$, $b = 8$; $a = 3$, $b = 7$; $a = 4$, $b = 6$; $a = 5$, $b = 5$; $a = 6$, $b = 4$; $a = 7$, $b = 3$; $a... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Sides 1 cm and 9 cm (or 9 cm and 1 cm); area 9 square centimeters. | |
07se | Let $\mathbb{N}$ denote the set of all non-negative integers (including zero). A function $f: \mathbb{N} \to \mathbb{N}$ satisfies for all $a, b \in \mathbb{N}$
$$
f(a + b^2) = f(a) + f(b^2) + ab^2.
$$
Given that $f(40) = 2020$ find $f(1)$. | [
"Putting $b = 1$ in the recurrence relation, we have: $f(a+1) = f(a) + f(1) + a$. It is then an easy inductive proof that for $a \\ge 1$\n$$\nf(a) = a \\left( f(1) + \\frac{a-1}{2} \\right).\n$$\nOn substitution, this satisfies the functional equation for any $a, b \\in \\mathbb{N}$ and so all solutions are of this... | Ireland | IRL_ABooklet_2020 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 31 | |
0iih | Problem:
Since $24=3+5+7+9$, the number $24$ can be written as the sum of at least two consecutive odd positive integers.
a. Can $2005$ be written as the sum of at least two consecutive odd positive integers? If yes, give an example of how it can be done. If no, provide a proof why not.
b. Can $2006$ be written as t... | [
"Solution:\n\nLet $N = (2k+1) + (2k+3) + \\cdots + (2k+2n-1)$ where $n$ and $k$ are integers, $n \\geq 2$, $k \\geq 0$. Then\n$$\nN = [(2k+1) + (2k+2n-1)] \\frac{n}{2} = (2k+n)n,\n$$\nwhich is a product of two integers with the same parity, since adding the even number $2k$ to the integer $n$ does not change its pa... | United States | Bay Area Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2005 = 397 + 399 + 401 + 403 + 405; 2006 cannot be written as a sum of at least two consecutive odd positive integers. | |
0cgc | We say that a simple periodic decimal fraction $f$ has the reduced length equal to $n$ (where $n$ is a positive integer) if $f$ has a $n$-digit period and $f$ cannot be represented as a simple periodic decimal fraction with a period having less than $n$ digits. For instance, $0.(223)$ has the reduced length 3, while $0... | [
"a) $f = 0.(2) \\cdot 0.(3) = \\frac{2}{9} \\cdot \\frac{3}{9} = \\frac{2}{27} = 0.(074)$ is a fraction of reduced length 3.\n\nb) Yes. For an example, $0.(3) \\cdot 0.(6) = 0.(2)$.\n\nc) Yes. For example, $0.(270) \\cdot 0.(370) = \\frac{270}{9 \\cdot 3 \\cdot 37} \\cdot \\frac{370}{999} = \\frac{100}{999} = 0.(10... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | a) 0.(2) × 0.(3) = 2/27 = 0.(074), which has reduced length 3.
b) Yes. Example: 0.(3) × 0.(6) = 0.(2), which has reduced length 1.
c) Yes. Example: 0.(270) × 0.(370) = 0.(100), which has reduced length 3. | |
0k5g | Let $ABC$ be an acute triangle with circumcircle $\omega$, and let $H$ be the foot of the altitude from $A$ to $\overline{BC}$. Let $P$ and $Q$ be the points on $\omega$ with $PA = PH$ and $QA = QH$. The tangent to $\omega$ at $P$ intersects lines $AC$ and $AB$ at $E_1$ and $F_1$ respectively; the tangent to $\omega$ a... | [
"Let $O$ be the center of $\\omega$, and let $M = \\overline{PQ} \\cap \\overline{AB}$ and $N = \\overline{PQ} \\cap \\overline{AC}$ be the midpoints of $\\overline{AB}$ and $\\overline{AC}$ respectively. Refer to the diagram below.\n\n\nThe main idea is to prove two key claims involving $O... | United States | USA TSTST | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Advanced Configurations > Polar... | null | proof only | null | |
0cqs | Ученик за одну неделю получил 17 оценок (каждая из них — $2$, $3$, $4$ или $5$). Среднее арифметическое этих 17 оценок — целое число. Докажите, что какую-то оценку он получил не более двух раз. | [
"Допустим противное. Тогда каждую из оценок $2$, $3$, $4$, $5$ ученик получил не меньше трёх раз. Возьмём по три оценки каждого вида; сумма 12 взятых оценок равна $42$. Так как каждая из оставшихся пяти оценок не меньше $2$ и не больше $5$, сумма всех $17$ оценок не меньше $42 + 5 \\cdot 2 = 52$ и не больше $42 + 5... | Russia | XL Russian mathematical olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof only | null | |
01nb | Two distinct points $A$ and $B$ are marked on the left half of the parabola $y = x^2$. Consider any pair of parallel lines which pass through $A$ and $B$ and intersect the right half of the parabola at points $C$ and $D$. Let $K$ be the intersection point of the diagonals $AC$ and $BD$ of the obtained trapezoid $ABCD$.... | [
"**Hint.** Show that if the given points have coordinates $A(a; a^2)$, $B(b; b^2)$, then $KM - KN = (b - a)^2$."
] | Belarus | Belorusija 2012 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
00pi | Let $ABCD$ be a square in the plane $P$. Find the minimum and the maximum values of the function $f: P \to \mathbb{R}$ defined by
$$
f(P) = \frac{PA + PB}{PC + PD}
$$
where $\mathbb{R}$ is the set of all real numbers. | [
"We have $f(A) = 1/(\\sqrt{2} + 1) = \\sqrt{2} - 1$. We will prove that this value is the minimum of function $f$, or in other words,\n$$\nPA + PB \\geq (\\sqrt{2} - 1)(PC + PD).\n$$\nfor all $P$.\nApplying Ptolemy's inequality for the points $P$, $A$, $B$, $C$, we have $PA + \\sqrt{2} PB \\geq PC$, that is\n$$\nPA... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"... | English | proof and answer | Minimum value: sqrt(2) - 1; Maximum value: sqrt(2) + 1 | |
01ts | A trapezoid $ABCD$ ($AB \parallel CD$) is inscribed into the parabola $y = x^2$. The line containing the midline $MN$ of this trapezoid meets the parabola at points $K$ and $L$.
Prove that the lengths of the segments $KM$ and $NL$ are equal.
(I. Gorodnin) | [
"Let $A(a, a^2)$, $B(b, b^2)$, $C(c, c^2)$, $D(d, d^2)$, $K(k, k^2)$, $L(l, l^2)$, $M(m, m^2)$, $N(n, n^2)$. Without loss of generality we can assume that the points are arranged as shown in the figure (all other cases are similar).\n\n\n\nIt is easy to write the equations of the straight l... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof only | null | |
0bn8 | Let $n$ be a positive integer and let $G_n$ be the set of all simple graphs on $n$ vertices. For each vertex $v$ of a graph in $G_n$, let $k(v)$ be the maximal cardinality of an independent set of neighbours of $v$. Determine $\max_{G \in \mathcal{G}_n} \sum_{v \in V(G)} k(v)$ and the graphs in $\mathcal{G}_n$ that ach... | [
"To prove this, let $G$ be a simple graph on $n$ vertices, and let $S$ be a maximal independent set of vertices of $G$. If $v$ is a member of $S$, then $k(v) \\le n - |S|$, since $v$ has at most $n - |S|$ neighbours. If a vertex $v$ is not in $S$, then $k(v) \\le |S|$, since $k(v)$ is the size of an independent set... | Romania | 2015 Thirteenth IMAR Mathematical Competition | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | The maximum is the floor of n squared over two, achieved uniquely by the complete bipartite graph with parts of sizes floor of n over two and ceiling of n over two. | |
0dqc | A subset $T$ of $\{7, 8, ..., 26\}$ does not contain three elements whose product is a perfect square. Determine the maximum number of elements in $T$. | [
"By checking the parity of the prime factorization of all the integers in $T$, we partition them into disjoint triples, whose products are all squares:\n$$\n\\{8, 13, 26\\}, \\{11, 18, 22\\}, \\{14, 21, 24\\}, \\{12, 15, 20\\}, \\{9, 16, 25\\}.\n$$\nIf $|T| \\ge 16$, at least one of these triples will be in $T$, a ... | Singapore | Singapore International Mathematical Olympiad Committee National Team Selection Test | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 15 | |
0g2g | Problem:
Das SMO-Land hat $1111$ Einwohner. Die elf Spieler der Liechtensteiner Nationalmannschaft verteilen Autogramme an alle Einwohner, wobei kein Einwohner ein Autogramm doppelt erhält (d.h. jeder Einwohner erhält von jedem Spieler entweder kein oder ein Autogramm).
a. Wie viele Möglichkeiten gibt es, welche Auto... | [
"Solution:\n\na. Jeder Einwohner kann von jedem Spieler entweder ein Autogramm erhalten oder kein Autogramm erhalten. Somit gibt es insgesamt $2^{11} = 2048$ verschiedene Möglichkeiten, Autogramme zu erhalten.\n\nb. Nun geht es darum, diese $2048$ Möglichkeiten auf die $1111$ Einwohner zu verteilen, sodass jeder Ei... | Switzerland | SMO - Vorrunde | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | a) 2048; b) Such a complementary pair of residents must exist. | |
0k3g | Problem:
An $n \times m$ maze is an $n \times m$ grid in which each cell is one of two things: a wall, or a blank. A maze is solvable if there exists a sequence of adjacent blank cells from the top left cell to the bottom right cell going through no walls. (In particular, the top left and bottom right cells must both ... | [
"Solution:\n\n```python\n# dfs that returns all paths with no adjacent vertices other than those consecutive in the path\ndef dfs(graph, start, end, path):\n if start == end:\n return [path]\n paths = []\n for child in graph[start]:\n skip = False\n if child in path:\n conti... | United States | HMMT November 2018 | [
"Discrete Mathematics > Algorithms"
] | null | final answer only | null | |
0bdm | Let $K$ be a convex quadrangle and let $\ell$ be a line through the point of intersection of the diagonals of $K$. Show that the length of the segment of intersection $\ell \cap K$ does not exceed the length of (at least) one of the diagonals of $K$.
D. Yu. Grigoriev, Kvant Magazine | [
"Consider a circular labelling, **A**, **B**, **C**, **D**, of the vertices of the quadrangle, and let the diagonals $AC$ and $BD$ meet at $O$. Without loss of generality, we may (and will) assume that the line $\\ell$ meets the opposite sides $AB$ and $CD$, say, at $X$ and $Y$, respectively. Further, let $\\alpha ... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
02g2 | Prove that there is at least one nonzero digit between the 1,000,000th and the 3,000,000th decimal digits of $\sqrt{2}$. | [
"Let us suppose that all digits between the 1,000,000th and the 3,000,000th decimal digits of $\\sqrt{2}$ are zeros. Then\n$$\n\\begin{aligned}\n\\sqrt{2} &= \\frac{n}{10^{10^6}} + \\epsilon, && n \\in \\mathbb{Z},\\ n < 2 \\cdot 10^{10^6} \\text{ and } 0 < \\epsilon < 10^{-3 \\cdot 10^6} \\\\\n\\Leftrightarrow 2 \... | Brazil | XXI OBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0a2c | Player Zero and player One play a game on an $n \times n$-board ($n \ge 1$). The columns of this $n \times n$-board are numbered by powers of two, so we have column 1, column 2, column 4 through column $2^{n-1}$. Alternately, players put their own number (i.e. Zero a 0 and One a 1) in one of the empty squares. Player Z... | [
"a) Note first that if $n = 1$, the unique box gets 0 because player Zero starts, so in this case it is possible.\nWe will now prove that for all other $n$ Zero can prevent One from winning. Since $4 \\mid 2^k$ for all $k \\ge 2$, only the first two columns determine who wins. If there is a row with two zeros at th... | Netherlands | IMO Team Selection Test 1 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Other"
] | English | proof and answer | a) n = 1. b) All n not congruent to 2 modulo 4; equivalently, n ≡ 0, 1, or 3 (mod 4). | |
0dhs | Five positive reals $a$, $b$, $c$, $d$ and $e$ having product equal to $1$ are given. Prove that
$$
\frac{a^2}{b^2} + \frac{b^2}{c^2} + \frac{c^2}{d^2} + \frac{d^2}{e^2} + \frac{e^2}{a^2} \geq a + b + c + d + e.
$$ | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0jjt | Problem:
Let $f(x) = x^{2} + 6x + 7$. Determine the smallest possible value of $f(f(f(f(x))))$ over all real numbers $x$. | [
"Solution:\n\nConsider that $f(x) = x^{2} + 6x + 7 = (x+3)^{2} - 2$. So $f(x) \\geq -2$ for real numbers $x$. Also, $f$ is increasing on the interval $[-3, \\infty)$.\n\nTherefore\n$$\n\\begin{gathered}\nf(f(x)) \\geq f(-2) = -1, \\\\\nf(f(f(x))) \\geq f(-1) = 2,\n\\end{gathered}\n$$\nand\n$$\nf(f(f(f(x)))) \\geq f... | United States | HMMT November 2014 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 23 | |
0hvf | Problem:
Determine whether there exist two distinct finite subsets $A$ and $B$ of the reals such that for every polynomial $f$ of degree $2012$ with real coefficients,
$$
\sum_{x \in A} f(x) = \sum_{x \in B} f(x)
$$ | [
"Solution:\nWe will prove by induction on $n \\geq 0$ that there exist distinct sets $A_n$ and $B_n$ satisfying the desired equality\n$$\n\\sum_{x \\in A_n} f(x) = \\sum_{x \\in B_n} f(x)\n$$\nfor all polynomials $f$ of degree at most $n$.\n\nThe base case is $n=0$; we take $A = \\{0\\}$ and $B = \\{1\\}$.\n\nNow w... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0jhk | Problem:
Find all positive integers $n$ such that $n \mid 2^{n}-1$. | [
"Solution:\nThe only such $n$ is $n=1$. Clearly $1 \\mid 2^{1}-1$.\n\nSuppose that $n>1$ is a solution. Let $p$ be the smallest prime divisor of $n$. Note that $p>2$ (an even value of $n$ cannot divide $2^{n}-1$). By Fermat's little theorem, $2^{p-1} \\equiv 1 \\pmod{p}$. But we are also given $2^{n} \\equiv 1 \\pm... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 1 | |
09s5 | Problem:
Laat $a$ en $b$ twee positieve gehele getallen zijn die voldoen aan $\operatorname{ggd}(a, b)=1$. Beschouw een pion die op roosterpunt $(x, y)$ staat. Een stap van type A bestaat uit het verplaatsen van de pion naar één van de volgende velden: $(x+a, y+a), (x+a, y-a), (x-a, y+a)$ of $(x-a, y-a)$. Een stap van ... | [
"Solution:\nWe gaan bewijzen dat een roosterpunt $(x, y)$ bereikbaar is dan en slechts dan als $x+y \\equiv 0 \\bmod 2$.\n\nAls we de pion verplaatsen van $(x, y)$ naar $(x \\pm a, y \\pm a)$, dan wordt de som van de nieuwe coördinaten gelijk aan $x+y+2a$, $x+y$ of $x+y-2a$, dus modulo $2$ congruent aan de som van ... | Netherlands | IMO-selectietoets II | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All lattice points with x + y ≡ 0 (mod 2), i.e., those whose coordinate sum is even | |
06cc | Let $f$ be a function defined on the positive reals with the following properties:
(1) $f(1) = 1$,
(2) $f(x + 1) = x f(x)$, and
(3) $f(x) = 10^{g(x)}$,
where $g(x)$ is a function defined on the reals satisfying
$$
g(ty + (1-t)z) \leq t g(y) + (1-t) g(z)
$$
for all $y$ and $z$ and for $0 \leq t \leq 1$.
a. Prove that $... | [
"a.\nWe label the inequality as follows.\n$$\ng(ty + (1-t)z) \\leq t g(y) + (1-t) g(z) \\quad (1)\n$$\nPutting $y = n - 1$ and $z = n + c$ in (1), we obtain\n$$\ng(t(n-1) + (1-t)(n+c)) \\leq t g(n-1) + (1-t) g(n+c).\n$$\nSuppose $c \\in [0, 1]$. Replacing $t$ by $\\frac{c}{1+c} \\in [0, 1]$, this becomes\n$$\ng\\le... | Hong Kong | CHKMO | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 4/3 <= f(1/2) <= (4/3) sqrt(2) | |
0gcf | 給定一正整數 $n \ge 3$. 我們稱一實數 $n$-序對 $(x_1, x_2, \cdots, x_n)$ 為閃耀的, 若對 $x_1, x_2, \cdots, x_n$ 的每一個排列 $y_1, y_2, \cdots, y_n$, 皆滿足
$$
\sum_{i=1}^{n-1} y_i y_{i+1} = y_1 y_2 + y_2 y_3 + y_3 y_4 + \cdots + y_{n-1} y_n \ge -1.
$$
試求最大的數 $K = K(n)$ 使得對每一個閃耀的 $n$-序對 $(x_1, x_2, \cdots, x_n)$,
$$
\sum_{1 \le i < j \le n} x_i x_j... | [
"First of all, we show that we may not take a larger constant $K$. Let $t$ be a positive number, and take\n$$\nx_2 = x_3 = \\cdots = t \\text{ and } x_1 = -\\frac{1}{2t}.\n$$\nThen, every product $x_i x_j$ ($i \\ne j$) is equal to either $t^2$ or $(-1/2)$. Hence, for every permutation $y_i$ of the $x_i$, we have\n$... | Taiwan | 2018 數學奧林匹亞競賽第二階段選訓營, 模擬競賽(二) | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | -(n-1)/2 | |
0013 | Se tiene una sucesión infinita que utiliza los dígitos $1$, $2$, $\dots$, $9$. Consideramos cada tramo de dígitos consecutivos de la sucesión como un entero positivo escrito en base $10$.
Demostre que para cualquier entero $n \ge 2$ al menos una de las siguientes dos afirmaciones es verdadera:
(i) Se pueden encontrar... | [] | Argentina | XII Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | español | proof only | null | |
0dln | $$
x^2 + a x + 1 = 0 \text{ and } x^2 + b x + c = 0
$$
have a common real root. In addition, the equations
$$
x^2 + x + a = 0 \text{ and } x^2 + c x + b = 0
$$
have a common real root. What values can the sum $a + b + c$ take? | [] | Saudi Arabia | Saudi Booklet | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | (-∞, -3] | |
0096 | Let $a \ge 2$ and $n \ge 3$ be integers. Show that one of the numbers
$$
a^n + 1, a^{n+1} + 1, \dots, a^{2n-2} + 1,
$$
does not share any odd divisor greater than $1$ with any other number in the set. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | English | proof only | null | |
0bbc | a) Show that if $a, b > 1$ are distinct real numbers then
$$
\log_a(\log_a b) > \log_b(\log_a b).
$$
b) Let $a_1 > a_2 > \dots > a_n > 1$ be real numbers, $n \ge 2$. Prove that
$$
\log_{a_1}(\log_{a_1}(a_2) + \log_{a_2}(\log_{a_2}(a_3) + \dots + \log_{a_{n-1}}(\log_{a_{n-1}}(a_n) + \log_{a_n}(\log_{a_n}(a_1)))) > 0.
$... | [
"a) For $a < b$, $\\log_a(\\log_a b) = (\\log_a b)(\\log_b(\\log_a b)) > \\log_b(\\log_a b)$ because $\\log_b(\\log_a b) > 0$ and $\\log_a b > 1$.\nFor $a > b$, the claim is reached from $\\log_a b < 1$ and $\\log_b(\\log_a b) < 0$.\n\nb) Induct on $n$. For $n = 2$,\n$$\n\\log_{a_1}(\\log_{a_1}(a_2) + \\log_{a_2}(\... | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
007l | Is there a number which is the sum of $2345$ positive integers that have the same digit sum, and also the sum of $5678$ positive integers that have the same digit sum? If the answer is *yes*, find the least such number. If not, explain why. | [
"Such numbers exist. The least one is $11725$.\nLet $N$ be the sum of $2345$ positive integers with digit sum $R$, and let $R \\equiv r \\pmod{9}$, $r \\in [1, 9]$. Then $N \\equiv 2345r \\equiv 5r \\pmod{9}$ as each summand is congruent to $r$ modulo $9$. Similarly if $N$ is the sum of $5678$ numbers with digit su... | Argentina | National Olympiad of Argentina | [
"Number Theory > Modular Arithmetic",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 11725 | |
00qx | In the acute angled triangle $ABC$ consider the altitudes $BB'$ and $CC'$. The half-line $C'B'$ intersects the circumcircle of triangle at $B''$ and denote by $\alpha_A$ the angle $\widehat{ABB''}$. In a similar way define the angles $\alpha_B$ and $\alpha_C$. Prove the inequality
$$
\sin \alpha_A \sin \alpha_B \sin \a... | [
"Let $C''$ be the second intersection point of the line $B'C'$ with the circumcircle of $\\triangle ABC$.\n\nConsider the intersection point $A_1$ of the diameter $AA_2$ with the line $B'C'$. Since $BC'B'C$ cyclic, we have $\\widehat{ABC} = \\widehat{AA_2C} = \\frac{\\pi}{2} - \\widehat{A_1... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry >... | null | proof and answer | 3√6/32 | |
01te | Let $D$, $E$, $F$ denote the tangent points of the incircle of $ABC$ with the sides $BC$, $AC$, $AB$, respectively. Let $M$ be the midpoint of the segment $EF$. Let $L$ be the intersection point of the circle passing through $D$, $M$, $F$ and the segment $AB$, $K$ be the intersection point of the circle passing through... | [
"Let *I* be the incenter of the triangle *ABC*. Let $\\Gamma$ denote the circle passing through $A$, $K$, $L$. We have\n$$\n\\begin{align*}\n\\angle AKD = \\angle EKD &= 180^\\circ - \\angle DME = \\angle DMF = 180^\\circ - \\angle FLD = \\\\\n&= 180^\\circ - \\angle ALD.\n\\end{align*}\n$$\nTherefore, $D$ lies on ... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0fk7 | Problem:
Hallar todas las soluciones reales de la ecuación
$$
3^{x^{2}-x-y}+3^{y^{2}-y-z}+3^{z^{2}-z-x}=1
$$ | [
"Solution:\nAplicando la desigualdad entre las medias aritmética y geométrica resulta\n$$\n\\begin{aligned}\n& 3^{x^{2}-x-y}+3^{y^{2}-y-z}+3^{z^{2}-z-x} \\geq 3 \\sqrt[3]{3^{x^{2}+y^{2}+z^{2}-2x-2y-2z}} = 3^{\\frac{1}{3}\\left(x^{2}+y^{2}+z^{2}-2x-2y-2z\\right)+1} \\\\\n& = 3^{\\frac{1}{3}\\left[(x-1)^{2}+(y-1)^{2}... | Spain | Spanish Mathematical Olympiad - Local Stage | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | (1,1,1) | |
0eg3 | Problem:
Poišči vse pare realnih števil $x$ in $y$, ki rešijo sistem enačb
$$
\begin{aligned}
\frac{3}{x-4y} + \frac{2}{x+y-5} &= 0 \\
\frac{2}{x^2-4y^2} + \frac{1}{x^2+y^2-5} &= 0
\end{aligned}
$$ | [
"Solution:\n\nOdpravimo ulomke, da dobimo\n$$\n\\begin{aligned}\n3(x+y-5) + 2(x-4y) &= 0 \\\\\n2\\left(x^2 + y^2 - 5\\right) + \\left(x^2 - 4y^2\\right) &= 0\n\\end{aligned}\n$$\nin obe enačbi poenostavimo do\n$$\n\\begin{array}{r}\n5x - 5y - 15 = 0 \\\\\n3x^2 - 2y^2 - 10 = 0\n\\end{array}\n$$\nIz prve enačbe izraz... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | x = -14, y = -17 | |
0f5r | Problem:
Starting with the polynomial $x^2 + 10x + 20$, a move is to change the coefficient of $x$ by $1$ or to change the coefficient of $x^0$ by $1$ (but not both). After a series of moves the polynomial is changed to $x^2 + 20x + 10$. Is it true that at some intermediate point the polynomial had integer roots? | [
"Solution:\n\nWe have $x^2 + (n + 1)x + n = (x + n)(x + 1)$, so $x^2 + a x + b$ has integer roots if $a = b + 1$ (and $a$ and $b$ are integers). But initially $a - b$ is $-10$ and it ends up as $+10$. Each move changes $a - b$ by $\\pm 1$, so it must pass through all values between $-10$ and $+10$."
] | Soviet Union | 18th ASU | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes | |
09mb | Let $ABC$ be an isosceles triangle with $AB = BC$. Let $M$ and $N$ be midpoints of $AC$ and $BM$, respectively. $P$ is the foot of the altitude from $A$ to $AN$ of triangle $AMN$. Prove that triangles $APM$ and $CPB$ are similar.
(Khulan Tumenbayar) | [
"Let us denote $\\angle PNM = \\alpha$. Then $\\angle PMC = \\angle PNB = 180^\\circ - \\alpha$ and $\\angle PAC = \\angle PMB$. Hence $\\triangle APM \\sim \\triangle MPN$. As we have $AM = MC$, $MN = NB$ it yields $\\triangle APC \\sim \\triangle MPB$. It implies $\\angle APC = \\angle MPB$ and $\\angle MPC = \\a... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0kog | Problem:
There are 21 competitors with distinct skill levels numbered $1,2, \ldots, 21$. They participate in a pingpong tournament as follows. First, a random competitor is chosen to be "active", while the rest are "inactive." Every round, a random inactive competitor is chosen to play against the current active one. ... | [
"Solution:\n\nInsert a player with skill level 0, who will be the first active player (and lose their first game).\nIf Alice plays after any of the players with skill level $12,13, \\ldots, 21$, which happens with probability $\\frac{10}{11}$, then she will play exactly 1 game.\nIf Alice is the first of the players... | United States | HMMT November 2022 | [
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 47/42 | |
0c4v | Let $x$, $y$, $z$, $t$ be a permutation of the real numbers $a$, $b$, $c$, $d$. It is known that $x \le 2a - b$, $y \le 2b - c$, $z \le 2c - d$, $t \le 2d - a$. Prove that $a = b = c = d$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0dpd | In countries of the Silk Road there are finite number of cities and some of them are connected by directed roads (one pair of cities can have several roads, moreover they can have an opposite direction). It is known that every two paths by these roads form the city $A$ to the city $B$ use at least one same road. Prove ... | [] | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut"
] | English | proof only | null | |
00y4 | Problem:
How many positive integers satisfy the following three conditions:
(i) All digits of the number are from the set $\{1,2,3,4,5\}$;
(ii) The absolute value of the difference between any two consecutive digits is $1$;
(iii) The integer has $1994$ digits? | [
"Solution:\n\nConsider all positive integers with $2n$ digits satisfying conditions $(i)$ and $(ii)$ of the problem. Let the number of such integers beginning with $1,2,3,4$ and $5$ be $a_{n}, b_{n}, c_{n}, d_{n}$ and $e_{n}$, respectively. Then, for $n=1$ we have $a_{1}=1$ (integer $12$), $b_{1}=2$ (integers $21$ ... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 8*3^996 | |
0ful | Problem:
Sei $A$ eine $m \times m$-Matrix. Sei $X_{i}$ die Menge der Einträge in der $i$-ten Zeile und $Y_{j}$ die Menge der Einträge in der $j$-ten Spalte, $1 \leq i, j \leq m$. $A$ heisst cool, wenn die Mengen $X_{1}, \ldots, X_{m}, Y_{1}, \ldots, Y_{m}$ alle verschieden sind. Bestimme den kleinsten Wert für $n$, so... | [
"Solution:\n\nSei $\\mathcal{X}$ die Menge der $X_{i}$ und $\\mathcal{Y}$ die Menge der $Y_{j}$. Es muss gelten $2^{n} \\geq|\\mathcal{X} \\cup \\mathcal{Y}|=2 \\cdot 2005$, also $n \\geq 12$. Nehme an, $n=12$ wäre möglich. Genau $2^{12}-2 \\cdot 2005=86$ Teilmengen von $\\{1,2, \\ldots, 12\\}$ liegen nicht in $\\m... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 13 | |
0cer | A positive integer is *square full* if it is divisible by the square of each of its prime divisors. Prove that $n$ and $n + 1$ are both square full for infinitely many positive integers $n$. | [
"Note that $8 = 2^3$ and $8 + 1 = 9 = 3^2$ are square full. Now, if $n$ and $n + 1$ are both square full, then so are $4n(n + 1)$ and $4n(n + 1) + 1 = (2n + 1)^2$. As $4n(n + 1) > n + 1$, the conclusion follows."
] | Romania | Eighteenth STARS OF MATHEMATICS Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Other"
] | English | proof only | null | |
07r2 | An equilateral triangle of integer side length $n \ge 1$ is subdivided into small triangles of unit side length, as illustrated in the figure below for the case $n=5$. In this diagram, a sub-triangle is a triangle of any size which is formed by connecting vertices of the small triangles along the grid-lines.
It is desi... | [
"The answer is $f(1) = 6$, $f(2) = 18$, $f(3) = 36$ and $f(n) = 0$ for $n \\ge 4$.\n\nFirst consider the case $n=1$. All $2^3 = 8$ possible colourings are valid, except for the two monochromatic colourings. So $f(1) = 6$.\n\nNext consider the case $n=2$ as shown in the Figure below. The 'inner' triangle cannot be m... | Ireland | Ireland_2017 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | f(1) = 6, f(2) = 18, f(3) = 36, and f(n) = 0 for all n ≥ 4 | |
02d5 | The vertex $C$ of the triangle $ABC$ is allowed to vary along a line parallel to $AB$. Find the locus of the orthocenter. | [
"Take axes so that $A = (-a, 0)$, $B = (a, 0)$ and $C = (k, b)$. Then the orthocenter lies on the line $x = k$. The line $AC$ has gradient $\\frac{b}{k+a}$, so the perpendicular has gradient $-\\frac{k+a}{b}$. Hence the altitude from $B$ has equation $y + \\frac{(x-a)(k+a)}{b} = 0$. So the intersection is $x = k, y... | Brazil | I OBM | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof and answer | The locus is the entire parabola b y = a^2 − x^2, in coordinates with A = (−a, 0), B = (a, 0), and C moving along y = b. | |
02sj | Problem:
Sobre um tabuleiro de $5 \times 5$ casas foram distribuídos os números $1,2,3, \ldots, 25$ de modo tal que cada casa seja ocupada por um único número e que dois números consecutivos sempre estejam colocados em casas vizinhas. A figura a seguir mostra um exemplo de como distribuir esses números.
| 3 | 4 | 5 |... | [
"Solution:\n\na) Vamos colorir as casas do tabuleiro de branco e preto como indica a figura seguinte. Desse modo, as casas correspondentes a dois números consecutivos estarão sempre pintadas de cores distintas.\n\n\n\nEm particular, se a casa com o número $1$ está pintada com a cor $A$ (bra... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | 33 | |
00fa | In a circle $C$ with centre $O$ and radius $r$, let $C_{1}, C_{2}$ be two circles with centres $O_{1}, O_{2}$ and radii $r_{1}, r_{2}$ respectively, so that each circle $C_{i}$ is internally tangent to $C$ at $A_{i}$ and so that $C_{1}, C_{2}$ are externally tangent to each other at $A$.
Prove that the three lines $O A... | [
"Because of the tangencies, the following triples of points (two centers and a tangency point) are collinear:\n$$\nO_{1}, O_{2}, A, \\quad O, O_{1}, A_{1}, \\quad O, O_{2}, A_{2}.\n$$\n\nBecause of that we can ignore the circles and only draw their centers and tangency points.\n\n\nNow the ... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0i5k | Problem:
In the $x$-$y$ plane, draw a circle of radius $2$ centered at $(0,0)$. Color the circle red above the line $y=1$, color the circle blue below the line $y=-1$, and color the rest of the circle white. Now consider an arbitrary straight line at distance $1$ from the circle. We color each point $P$ of the line wi... | [
"Solution:\n\nLet $O = (0,0)$, $P = (1,0)$, and $H$ the foot of the perpendicular from $O$ to the line. If $\\angle P O H$ (as measured counterclockwise) lies between $\\pi / 3$ and $2\\pi / 3$, the line will fail to contain blue points; if it lies between $4\\pi / 3$ and $5\\pi / 3$, the line will fail to contain ... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 2/3 | |
0fed | Problem:
¿Existe algún triángulo tal que las medidas de sus lados son tres números consecutivos y el ángulo mayor es el doble que el menor? Si existe, determinad sus medidas. | [
"Solution:\n\nNotamos que el triángulo debe ser acutángulo. Si $a, b, c$ son sus lados, hay que probar, gracias al teorema del coseno, que $a^{2} + b^{2} > c^{2}$, $a^{2} + c^{2} > b^{2}$, $b^{2} + c^{2} > a^{2}$. Si tomamos $a \\leq b \\leq c$, basta probar $a^{2} + b^{2} > c^{2}$. Sea $a = x - 1$, $b = x$, $c = x... | Spain | null | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | null | proof and answer | 4, 5, 6 | |
0cgg | Let $ABCD$ be a convex quadrilateral in which the diagonals intersect at $O$. Given that $\overrightarrow{AB} + \overrightarrow{AD} + \overrightarrow{AO} = \overrightarrow{BC} + \overrightarrow{DC} + \overrightarrow{OC}$, prove that $ABCD$ is a parallelogram.
Marius Dolcan | [
"We obtain that the points $M$, $N$ and $O$ are collinear, so $M = N = O$. Thus point $O$ is the midpoint of each of the diagonals, therefore $ABCD$ is a parallelogram."
] | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | English | proof only | null | |
0hod | Problem:
The number $2011$ is written on a blackboard. It is permitted to transform the numbers on it by two types of moves:
a) Given a number $n$, we can erase $n$ and write two nonnegative integers $a$ and $b$ such that $a+b=n$.
b) Given two numbers $a$ and $b$, we can erase them and write their difference $a-b$, ... | [
"Solution:\n\nThe answer is no.\n\nWe claim that the sum of the numbers on the blackboard is always odd. Indeed, the initial sum, $2011$, is odd, and when a move of type (a) is performed, the sum does not change. When a move of type (b) is performed, the sum decreases by\n$$\n(a+b)-(a-b)=2b,\n$$\nan even number. Si... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | No | |
0kmy | Problem:
Paul and Sara are playing a game with integers on a whiteboard, with Paul going first. When it is Paul's turn, he can pick any two integers on the board and replace them with their product; when it is Sara's turn, she can pick any two integers on the board and replace them with their sum. Play continues until... | [
"Solution:\n\nWe claim that Paul wins if and only if there are exactly 1 or 2 odd integers on the board at the start. Assuming this, the answer is $\\frac{2021+\\left(\\frac{2021}{2}\\right)}{2^{2021}}$. Since the numerator is odd, this fraction is reduced. Now, $m+n \\equiv 2^{2021}+21+2021 \\cdot 1010 \\equiv 231... | United States | HMMT November 2021 Team Round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Chinese remainder theorem"
] | null | proof and answer | 383 | |
0gq7 | Find the greatest real number $M$ satisfying
$$
a^3 + b^3 + c^3 - 3abc \geq M(ab^2 + bc^2 + ca^2 - 3abc)
$$
for all positive real numbers $a, b, c$. | [
"Since the inequality is cyclic, w.l.o.g. let $\\min\\{a, b, c\\} = c$. Then $a = c + x$, $b = c + y$ for some nonnegative $x$ and $y$. After substitution $a = c + x$ and $b = c + y$ we get\n$$\n\\begin{align*}\na^3 + b^3 + c^3 - 3abc &= (c + x)^3 + (c + y)^3 + c^3 - 3(c + x)(c + y)c \\\\\n&= (3c + x + y)(x^2 - xy ... | Turkey | 21st Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 3 / 4^(1/3) | |
02et | Show that there is a number of the form $199\dots 91$ (with $n$ 9s) with $n > 2$ which is divisible by $1991$. | [
"There are many ways to solve the problem, using, for instance, Euler-Fermat theorem. But one student, André Reys Leal, obtained a clever, simple solution: consider all numbers of the form $1999\\dots 91$ with more than two nines. If one of them is multiple of $1991$ we are done. If not, since there are infinite of... | Brazil | XIII OBM | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | English | proof only | null | |
0j89 | Problem:
Let $ABC$ be an equilateral triangle with $AB = 3$. Circle $\omega$ with diameter $1$ is drawn inside the triangle such that it is tangent to sides $AB$ and $AC$. Let $P$ be a point on $\omega$ and $Q$ be a point on segment $BC$. Find the minimum possible length of the segment $PQ$. | [
"Solution:\n\nThe minimum possible length is $\\frac{3\\sqrt{3} - 3}{2}$.\n\nLet $P$, $Q$ be the points which minimize the distance. We see that we want both to lie on the altitude from $A$ to $BC$. Hence, $Q$ is the foot of the altitude from $A$ to $BC$ and $AQ = \\frac{3\\sqrt{3}}{2}$.\n\nLet $O$, which must also... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (3*sqrt(3) - 3)/2 | |
09du | $x_0 = a$, $x_1 = 2$, $x_n = 2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1$. Find all integers $a$ such that $2x_{3n}$ are perfect squares for $\forall n \in \mathbb{N}$. | [
"If we substitute $a_n = 2x_n - 1$ in the $2x_n - 1 = 2(2x_{n-1}x_{n-2} - x_{n-1} - x_{n-2} + 1) - 1 = (2x_{n-1} - 1)(2x_{n-2} - 1)$ we get $a_n = a_{n-1}a_{n-2}$. Initial conditions of the recurrence relation are $a_1 = 2 \\cdot 2 - 1 = 3$, $a_0 = 2a - 1$. It is required that $a_0 = 2a - 1$ must be a perfect squar... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | all integers a of the form 2k^2 + 2k + 1 for integer k | |
0a00 | Problem:
Zij $n>1$ een geheel getal. Op een rij staan $n$ dozen, en we hebben $n+1$ identieke stenen. Een verdeling is een manier om de stenen over de dozen te verdelen, waarbij elke steen in precies één doos zit. We zeggen dat twee van zulke verdelingen zich op een steenworp afstand van elkaar bevinden als we de ene ... | [
"Solution:\n\nOplossing I. We tellen het aantal, $N_{k}$, verdelingen dat precies $k$ lege dozen bevat met $0 \\leq k \\leq n-1$ (aangezien niet alle dozen leeg kunnen zijn). Er zijn $\\binom{n}{k}$ manieren om de lege dozen te kiezen en wegens het paaseierenprincipe $\\binom{(k+1)+(n-k)-1}{k+1}=\\binom{n}{k+1}$ ma... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | (n^2 - 1)/2 | |
0aks | Find all polynomials $P(x)$ with integer coefficients such that for all real numbers $s$ and $t$, if $P(s)$ and $P(t)$ are both integers, then $P(st)$ is also an integer. | [] | North Macedonia | Asian-Pacific Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | All such polynomials are of the form P(x) = ±x^n + c for some integer n ≥ 0 and integer c. | |
0ewu | Problem:
Given a triangle $ABC$. Let the line through $C$ parallel to the angle bisector of $B$ meet the angle bisector of $A$ at $D$, and let the line through $C$ parallel to the angle bisector of $A$ meet the angle bisector of $B$ at $E$. Prove that if $DE$ is parallel to $AB$, then $CA = CB$. | [
"Solution:\nThe idea is to find an expression for the perpendicular distance $h$ from $D$ to $AB$. Let $\\gamma = \\angle ACB$, $\\alpha = \\frac{1}{2} \\angle CAB$, and $\\beta = \\frac{1}{2} \\angle ABC$. We have $h = AP \\sin \\alpha$.\n\nUsing the sine rule on $APC$, we have $AP = AC \\frac{\\sin (\\gamma + \\b... | Soviet Union | 3rd ASU | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
0j24 | Let $q = \frac{3p-5}{2}$ where $p$ is an odd prime, and let
$$
S_q = \frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7} + \dots + \frac{1}{q(q+1)(q+2)}.
$$
Prove that if $\frac{1}{p} - 2S_q = \frac{m}{n}$ for integers $m$ and $n$, then $m-n$ is divisible by $p$. | [
"**Solution** (by Titu Andreescu). We have\n$$\n\\begin{aligned}\n\\frac{2}{k(k+1)(k+2)} &= \\frac{(k+2)-k}{k(k+1)(k+2)} \\\\\n&= \\frac{1}{k(k+1)} - \\frac{1}{(k+1)(k+2)} \\\\\n&= \\left( \\frac{1}{k} - \\frac{1}{k+1} \\right) - \\left( \\frac{1}{k+1} - \\frac{1}{k+2} \\right) \\\\\n&= \\frac{1}{k} + \\frac{1}{k+1... | United States | USAMO 2010 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof only | null | |
0i4e | Problem:
Determine the value of
$$
2002+\frac{1}{2}\left(2001+\frac{1}{2}\left(2000+\cdots+\frac{1}{2}\left(3+\frac{1}{2} \cdot 2\right)\right) \cdots\right) .
$$ | [
"Solution:\n4002. We can show by induction that\n$$\nn+\\frac{1}{2}\\left([n-1]+\\frac{1}{2}\\left(\\cdots+\\frac{1}{2} \\cdot 2\\right) \\cdots\\right)=2(n-1).\n$$\nFor $n=3$ we have $3+\\frac{1}{2} \\cdot 2=4$, giving the base case, and if the result holds for $n$, then\n$$(n+1)+\\frac{1}{2} 2(n-1)=2 n=2(n+1)-2.$... | United States | Harvard-MIT Math Tournament | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | final answer only | 4002 | |
00q6 | The sequence $a_1, a_2, a_3, \dots$ is defined by $a_1 = a_2 = 1$, $a_{2n+1} = 2a_{2n} - a_n$ and $a_{2n+2} = 2a_{2n+1}$ for $n \in \mathbb{N}$. Prove that if $n > 3$ and $n - 3$ is divisible by 8 then $a_n$ is divisible by 5. | [
"First, for $k \\in \\mathbb{N}$, $k \\ge 2$, we have\n$$\na_{2k+1} + a_{2k-1} = 2a_{2k} - a_k + a_{2k-1} = 5a_{2k-1} - a_k \\equiv -a_k \\pmod{5}. \\quad (1)\n$$\nWe prove the assertion of the problem by induction on $k$, where $n = 8k+3$. For the base case $k=1$, we compute $a_3 = 1$, $a_4 = 2$, $a_5 = 3$, $a_6 =... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlist | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0hk2 | Problem:
Calculate, with proof, the last digit of
$$
3^{3^{3^{3^{3}}}}
$$ | [
"Solution:\nWhen $3$ is raised to the successive powers $1, 2, 3, 4, \\ldots$, the units digits are $3, 9, 7, 1, \\ldots$ From then on, since the digit $1$ has been reached, the units digits will repeat in this cycle of four elements. So it is necessary to find the remainder when\n$$\nn_{4} = 3^{3^{3^{3}}}\n$$\nis ... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Number Theory > Modular Arithmetic",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | 7 | |
090d | As shown in the figure, seven regular hexagonal cells form a hexagonal pattern. We write one integer from $1$ to $7$ in each cell without repetition. For any two cells that share an edge, the sum of the integers written in those cells must be at most $10$. How many ways are there to write the integers under these condi... | [
"$72$\n\nSince the sum of the integers in any two adjacent cells must be at most $10$, the only possible integers that can appear in the neighbors of the cell containing $7$ are $1$, $2$, or $3$. Therefore, the cell labeled $7$ cannot be the central cell, and there are exactly $6$ possible cells in which to place t... | Japan | The 35th Japanese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | 72 | |
09n0 | Consider a regular polygon with $360$ vertices labeled clockwise from $1$ to $360$. Starting at vertex $1$, count clockwise and remove the $359$th vertex. Continue counting from the next vertex, and remove the subsequent $358$th vertex. Repeat this process, removing the subsequent $357$th vertex, and so on, until only ... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Algorithms"
] | English | proof and answer | 152 | |
0gjb | 令 $ABC$ 為一三角形, 其外心為 $O$。圓 $\Gamma$ 分別與 $OB$ 和 $OC$ 相切於 $B$ 和 $C$。令 $D$ 為 $\Gamma$ 上異於 $B$ 的一點, 使得 $CB = CD$。令 $E$ 為 $DO$ 和 $\Gamma$ 異於 $D$ 的交點, 而 $F$ 為 $EA$ 和 $\Gamma$ 異於 $D$ 的交點。令 $X$ 為 $AC$ 上一點, 使得 $XB \perp BD$。證明 $\angle ADF$ 的一半等於 $\angle BDX$ 或 $\angle BXD$。
Let $ABC$ be a triangle with $O$ as its circumcenter. ... | [
"**解.** Since $\\angle ADF$ is the angle between $\\Gamma$ and $\\odot(ADE)$, consider the circle $\\odot(ADE)$ and its second intersection $A'$ with $\\odot(ABC)$. Also let $DA'$ intersects $BX$ at $X'$. Denote $\\angle ADF$ by $\\theta$. Now we first show the following two lemmas.\n\n**Lemma 1.** $(A, A'; B, C) =... | Taiwan | IMO 2J, Independent Study 1 | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, E... | Chinese; English | proof only | null | |
0hot | Problem:
Let $P(x)$ be a nonzero polynomial with real coefficients such that
$$
P(x) = P(0) + P(1) x + P(2) x^{2}
$$
holds for all $x$. What are the roots of $P(x)$? | [
"Solution.\n\nLet $c = P(0)$. Selecting $x = 1$ in the given, we have that $P(0) + P(1) + P(2) = P(1)$, so $P(2) = -c$. Selecting $x = 2$, we find that\n$$\n-c = c + 2 P(1) + 4(-c) \\Longrightarrow P(1) = c .\n$$\nTherefore, $P(x) = c + c x - c x^{2}$. As $c \\neq 0$, we solve $1 + x - x^{2} = 0$ to get the answers... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Algebra > Algebraic Expressions > Polynomials",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (1 + sqrt(5))/2 and (1 - sqrt(5))/2 | |
05wi | Problem:
Trouver tous les quadruplets de réels $(a, b, c, d)$ tels que
$$
a = b c + c d, \quad b = c d + d a, \quad c = d a + a b \text{ et } d = a b + b c
$$ | [
"Solution:\nSoit $(a, b, c, d)$ une solution éventuelle. Une première étape consiste à factoriser les membres de droite de nos égalités :\n$$\na = c \\times (b + d), \\quad b = d \\times (a + c), \\quad c = a \\times (b + d) \\text{ et } d = b \\times (a + c).\n$$\nMaintenant, si $a = 0$, l'énoncé indique que $c = ... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (0, 0, 0, 0) and (1/2, 1/2, 1/2, 1/2) | |
0l63 | Problem:
Let $ABCD$ be a parallelogram, and let $O$ be a point inside $ABCD$. Suppose the circumcircles of triangles $OAB$ and $OCD$ intersect at $P \neq O$, and the circumcircles of triangles $OBC$ and $OAD$ intersect at $Q \neq O$. Prove $\angle POQ$ equals one of the angles of quadrilateral $ABCD$.
![](attached_ima... | [
"Solution:\nIn what follows, all angles are directed.\n\nClaim 1. The points $P$ and $Q$ are symmetric over the center of $ABCD$.\n\nProof. Note that\n$$\n\\angle APB = \\angle AOB = \\angle OAD + \\angle CBO = \\angle OQD + \\angle CQO = \\angle CQD.\n$$\nSimilar equalities hold for each pair of opposite sides, so... | United States | HMIC | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0be4 | A die is an unitary cube with numbers from $1$ to $6$ written on its faces, so that each number appears once and the sum of the numbers on any two opposite faces is $7$. We construct a large $3 \times 3 \times 3$ cube using $27$ dice. Find all possible values of the sum of numbers which can be seen on the faces of the ... | [
"We will say that a die of the large cube is of type I, type II or type III according to the number of its faces which are visible (e.g. a die sharing a vertex with the large die is a type III die). Every large cube contains $6$ type I, $12$ type II and $8$ type III dice.\n\nThe minimum sum is obtained when each ty... | Romania | 64th Romanian Mathematical Olympiad - Final Round | [
"Geometry > Solid Geometry > 3D Shapes",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | All integers from 90 to 288 inclusive | |
0iok | Problem:
$ABCD$ is a convex quadrilateral such that $AB < AD$. The diagonal $\overline{AC}$ bisects $\angle BAD$, and $m \angle ABD = 130^{\circ}$. Let $E$ be a point on the interior of $\overline{AD}$, and $m \angle BAD = 40^{\circ}$. Given that $BC = CD = DE$, determine $m \angle ACE$ in degrees. | [
"Solution:\n\nFirst, we check that $ABCD$ is cyclic. Reflect $B$ over $\\overline{AC}$ to $B'$ on $\\overline{AD}$, and note that $B'C = CD$. Therefore, $m \\angle ADC = m \\angle B'DC = m \\angle CB'D = 180^{\\circ} - m \\angle AB'C = 180^{\\circ} - m \\angle CBA$.\n\nNow $m \\angle CBD = m \\angle CAD = 20^{\\cir... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 55 | |
06ly | Let $n$ be a positive integer. Is it possible to express $n^2 + 3n + 3$ into the form $ab$ with $a$ and $b$ being positive integers, and such that the difference between $a$ and $b$ is smaller than $2\sqrt{n} + 1$? | [
"No, it is not possible. Indeed, if $n^2 + 3n + 3 = ab$, then\n$$\n(a+b)^2 = (a-b)^2 + 4ab \\ge 4ab = 4n^2 + 12n + 12 > 4n^2 + 12n + 9 = (2n+3)^2.\n$$\nSince both sides are squares, this yields $(a+b)^2 \\ge (2n+4)^2$. It follows that\n$$\n(a-b)^2 = (a+b)^2 - 4ab \\ge (2n+4)^2 - (4n^2 + 12n + 12) = 4n + 4 = 4(n+1).... | Hong Kong | Year 2021 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
018p | Two persons play the following game with positive integers. The initial number is $2011^{2011}$. Each move consists of subtraction by an integer between $1$ and $2010$ inclusive, or division by $2011$, rounding down when necessary. The player who obtains a non-positive integer wins. Who will win this game: the first pl... | [
"Though the problem is taken from the recent article (A. Guo. Winning strategies for aperiodic subtraction games // arXiv: 1108.1239v2), it could be known for the smaller numbers, say, for $2$ instead of $2011$.\n\nThe initial numbers $N$ for which the second player has a winning strategy are those ones that have o... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | the second player | |
0g6f | 給定一大於 $5$ 的正整數 $n$. 試求出所有的實數 $a$, 使得存在非負實數 $x_1, x_2, \dots, x_n$ 滿足
$$
\sum_{k=1}^{n} kx_{k} = a, \quad \sum_{k=1}^{n} k^{3}x_{k} = a^{2}, \quad \sum_{k=1}^{n} k^{5}x_{k} = a^{3}.
$$ | [
"假設 $\\{x_i\\}$ 滿足題設，則\n$$\n\\sum_{k=1}^{n} kx_{k} = a, \\quad \\sum_{k=1}^{n} k^{3}x_{k} = a^{2}, \\quad \\sum_{k=1}^{n} k^{5}x_{k} = a^{3}.\n$$\n由歌西不等式得\n$$\n\\begin{aligned}\naa^3 &= \\left(\\sum_{k=1}^{n} kx_k\\right) \\left(\\sum_{k=1}^{n} k^5 x_k\\right) \\\\ &\\geq \\left(\\sum_{k=1}^{n} k^3 x_k\\right)^2 = ... | Taiwan | 二〇一二數學奧林匹亞競賽第二階段選訓營 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | a = 0 or a = i^2 for some integer i with 1 ≤ i ≤ n | |
0684 | Let $A B \Gamma$ be an isosceles acute angled triangle with $AB = A\Gamma$. Let $\Gamma\Delta$ be an altitude of the triangle. The circle $c_2(\Gamma, \Gamma\Delta)$ intersects $A\Gamma$ at point $K$, the extension of $A\Gamma$ at point $Z$ and the circle $c_1(B, B\Delta)$ at point $E$. Finally, $\Delta Z$ intersects t... | [
"a.\nFrom the right angled triangle $B\\Gamma\\Delta$ we have: $\\hat{\\Gamma}_1 = 90^\\circ - \\hat{B}$.\nThe line joining the centers of the circles $c_1$ and $c_2$ is the perpendicular bisector of their common chord $\\Delta E$.\nIf $T$ is the point of intersection of the lines $B\\Gamma$ and $\\Delta E$, then:\... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null |
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