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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0450 | Suppose the included angle between non-zero vectors $\vec{a}$ and $\vec{b}$ in the plane is $\frac{\pi}{3}$. If $|\vec{a}|, |\vec{b}|, |\vec{a} + \vec{b}|$ form arithmetic sequence in order, find the value of $|\vec{a}| : |\vec{b}| : |\vec{a} + \vec{b}|$. | [
"Denote $s = |\\vec{a}|$, $t = |\\vec{b}|$, and then $s, t > 0$. Note that the included angle between $\\vec{a}$ and $\\vec{b}$ is $\\frac{\\pi}{3}$, and we have\n$$\n\\begin{aligned}\n|\\vec{a} + \\vec{b}|^2 &= \\vec{a}^2 + (\\vec{b})^2 + 2\\vec{a} \\cdot \\vec{b} \\\\\n&= s^2 + t^2 + 2st \\cos \\frac{\\pi}{3} \\\... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 3:5:7 | |
0dg2 | Find all values of $a$ for which the equation $x^3 + a x^2 + 56x - 4 = 0$ has 3 roots forming consecutive terms of a geometric progression. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series"
] | English | proof and answer | a = -28 * 2^(1/3) | |
03n4 | Let $f$ be a function from the set of positive integers to itself such that, for every $n$, the number of positive integer divisors of $n$ is equal to $f(f(n))$. For example, $f(f(6)) = 4$ and $f(f(25)) = 3$. Prove that if $p$ is prime then $f(p)$ is also prime. | [
"Let $d(n) = f(f(n))$ denote the number of divisors of $n$ and observe that $f(d(n)) = f(f(f(n))) = d(f(n))$ for all $n$. Also note that because all divisors of $n$ are distinct positive integers between $1$ and $n$, including $1$ and $n$, and excluding $n-1$ if $n > 2$, it follows that $2 \\le d(n) < n$ for all $n... | Canada | CMO 2017 | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
04r1 | Find all pairs of primes $p$, $q$ for which there exists a positive integer $a$ such that
$$
\frac{pq}{p+q} = \frac{a^2+1}{a+1}.
$$ | [
"First, we will deal with the case when the wanted primes $p$ and $q$ are distinct. Then, the numbers $pq$ and $p+q$ are relatively prime: the product $pq$ is divisible by two primes only (namely $p$ and $q$), while the sum $p+q$ is divisible by neither of these primes.\nWe will look for a positive integer $r$ whic... | Czech Republic | 62nd Czech and Slovak Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (2, 2) and (5, 5) | |
02gk | The squares of an $m \times n$ board are labeled from $1$ to $mn$ so that the squares labeled $i$ and $i+1$ always have a side in common. Show that for some $k$ the squares $k$ and $k+3$ have a side in common. | [
"Consider the center of the $mn$ unit squares and connect two centers if the numbers assigned to the correspondent squares are consecutive. Then we obtain a path with $mn-1$ unit segments. The $mn$ centers determine a lattice with $(m-1)(n-1)$ unit squares. Squares with numbers $k$ and $k+3$ correspond to a lattice... | Brazil | XXIV OBM | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
02n5 | Problem:
Quais são os dois últimos algarismos do número
$$
8 + 88 + 888 + \cdots + \overbrace{88 \cdots 88}^{2008} ?
$$ | [] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 24 | |
0c65 | Determine all positive integers $n$ for which $n^{n+1} + n - 1$ is the sixth power of an integer. | [
"Clearly, $n = 1$ satisfies the required condition. We now proceed to rule out all integers $n > 1$.\n\nIf $n$ is odd, $n \\geq 3$, then $n^{n+1} + n - 1$ falls strictly between the squares of two consecutive integers,\n$$\n\\left( n^{\\left( \\frac{n+1}{2} \\right)} \\right)^2 < n^{n+1} + n - 1 < \\left( n^{\\left... | Romania | SELECTION TESTS FOR THE 2019 BMO AND IMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | English | proof and answer | 1 | |
00gp | Students in a class form groups each of which contains exactly three members such that any two distinct groups have at most one member in common. Prove that, when the class size is $46$, there is a set of $10$ students in which no group is properly contained. | [
"We let $C$ be the set of all $46$ students in the class and let\n$$\ns := \\max \\{ |S| : S \\subseteq C \\text{ such that } S \\text{ contains no group properly } \\}.\n$$\nThen it suffices to prove that $s \\geq 10$. (If $|S| = s > 10$, we may choose a subset of $S$ consisting of $10$ students.)\n\nSuppose that ... | Asia Pacific Mathematics Olympiad (APMO) | XX Asian Pacific Mathematics Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j8q | Problem:
Let $a$, $b$, $c$ be not necessarily distinct integers between $1$ and $2011$, inclusive. Find the smallest possible value of $\frac{a b + c}{a + b + c}$. | [
"Solution:\nAnswer: $\\frac{2}{3}$\n\nWe have\n$$\n\\frac{a b + c}{a + b + c} = \\frac{a b - a - b}{a + b + c} + 1\n$$\nWe note that $\\frac{a b - a - b}{a + b + c} < 0 \\Leftrightarrow (a - 1)(b - 1) < 1$, which only occurs when either $a = 1$ or $b = 1$. Without loss of generality, let $a = 1$. Then, we have a va... | United States | Harvard-MIT November Tournament | [
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2/3 | |
03wg | A right triangle $ABC$, with $\angle BAC = 90^\circ$, is inscribed in the circle $\Gamma$. The point $E$ lies in the interior of the arc $\widearc{BC}$ (not containing $A$), with $EA > EC$. The point $F$ lies on the ray $EC$ with $\angle EAC = \angle CAF$. The segment $BF$ meets $\Gamma$ again at $D$ (other than $B$). ... | [
"Let $M$ and $N$ be the feet of the perpendiculars from $O$ to the lines $DF$ and $DE$, respectively. Because $O$ is the circumcenter of the triangle $DEF$, the triangles $EOD$ and $ODF$ are both isosceles with $EO = DO = FO$. It follows that\n$$\n\\angle EOF = \\angle EOD + \\angle DOF = 2\\angle NOD + 2\\angle DO... | China | China Girls' 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 > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordin... | English | proof only | null | |
0h83 | A square $n \times n$ is divided into $n^2$ cells. In total $n^2$ tokens are placed at some cell. During every round, a player can move one token from cell $A$ to cell $B$, and one token from cell $A$ to cell $C$, provided that cell $A$ contained at least two tokens, $B$ and $C$ are symmetric with respect to $A$, and f... | [
"**Answer:** a) impossible; b) possible.\n\na) Consider square $n \\times n$, for positive integer $n$. Let us denote the most left column by $1$, adjacent column by $2$, and so on, the most right column by $n$. For any cell $c$ we denote function $w(c) = b$, provided that cell $c$ is located in the column $b$. For... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | a) impossible; b) possible | |
06oo | A holey triangle is an upward equilateral triangle of side length $n$ with $n$ upward unit triangular holes cut out. A diamond is a $60^{\circ}$-$120^{\circ}$ unit rhombus. Prove that a holey triangle $T$ can be tiled with diamonds if and only if the following condition holds: Every upward equilateral triangle of side ... | [
"Let $T$ be a holey triangle. The unit triangles in it will be called cells. We say simply \"triangle\" instead of \"upward equilateral triangle\" and \"size\" instead of \"side length.\"\n\nThe necessity will be proven first. Assume that a holey triangle $T$ can be tiled with diamonds and consider such a tiling. L... | IMO | IMO 2006 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0air | Solve the equation $x^2 + y^4 + 1 = 6^z$ in the set of integers. | [
"It is obvious that $z \\geq 0$. If $z \\geq 2$, then $x^2 + y^4 + 1 \\equiv 0 \\pmod{4}$, i.e. $x^2 + y^4 \\equiv 3 \\pmod{4}$. This is not possible because the remainders of squares of integers after division by $4$ are $0$ or $1$. According to that, $0 \\leq z < 2$.\n\nIf $z = 0$, then $x = y = 0$.\n\nIf $z=1$, ... | North Macedonia | Macedonian Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | (x, y, z) = (0, 0, 0), (2, 1, 1), (-2, 1, 1), (2, -1, 1), (-2, -1, 1) | |
0eou | $ABCD$ is a rectangle and $P$ is a point on $BC$. If the area of triangle $ABP$ is one third of the area of the rectangle, then the ratio $BP : PC$ is
(A) $5 : 2$
(B) $3 : 2$
(C) $2 : 1$
(D) $3 : 1$
(E) $9 : 4$ | [
"$\\frac{\\text{area } \\triangle ABP}{\\text{area } ABCD} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} \\cdot BP \\cdot AB}{BC \\cdot AB} = \\frac{1}{3}$ so $\\frac{\\frac{1}{2} BP}{BC} = \\frac{1}{3}$ and therefore $\\frac{BP}{BC} = \\frac{2}{3}$\nand then $BP : PC$ is $\\frac{2}{3} : \\frac{1}{3} = 2 : 1$"
] | South Africa | South African Mathematics Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals"
] | English | MCQ | C | |
0g89 | 給定 $6 \times 6$ 的方格，並記第一列六個方格座標為 $(1,1), (1,2), \cdots, (1,6)$，其餘類推。對於任意的 $k = 0,1,\cdots,5$，滿足 $i-j \equiv k \pmod 6$ 的六個格子 $(i,j)$ 稱為在同一條對角線上（故共有六條對角線）。試問：能否將 $1,2,\cdots,36$ 填入 $6 \times 6$ 的方格中，同時滿足
(1) 每一列的和都相等。
(2) 每一行的和都相等。
(3) 每一條對角線的和都相等。
There's a $6 \times 6$ chess board, which we label the squares in the f... | [
"不可能。歸謬證法，假設可以填成功，則這個和必為\n$$\nS = \\frac{1}{6}(1 + 2 + \\cdots + 36) = 111.\n$$\n將 $6 \\times 6$ 的方格分成四類：\n1. A: 座標為 (奇, 奇) 的格子。\n2. B: 座標為 (奇, 偶) 的格子。\n3. C: 座標為 (偶, 奇) 的格子。\n4. D: 座標為 (偶, 偶) 的格子。\n則依題目假設, $A + B = 3S$, $B + D = 3S$, $A + D = 3S$, 全部加起來得\n$$\n2(A + B + D) = 9S,\n$$\n但左邊為偶數而右邊為奇數, 矛盾!"
] | Taiwan | 二〇一四年國際數學奧林匹亞競賽第三階段選訓營 獨立研究(一) | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Impossible | |
0hsa | Problem:
If $f(1) = 1$ and $f(1) + f(2) + \cdots + f(n) = n^{2} f(n)$ for every integer $n \geq 2$, evaluate $f(2008)$. | [
"Solution:\n$n^{2} f(n) - f(n) = f(1) + f(2) + \\cdots + f(n-1) = (n-1)^{2} f(n-1)$\nhence $f(n) = \\frac{(n-1)^{2}}{n^{2} - 1} f(n-1) = \\frac{n-1}{n+1} f(n-1)$.\n\nThus\n$$\n\\begin{aligned}\nf(2008) &= \\frac{2007}{2009} f(2007) \\\\\n&= \\frac{2007}{2009} \\cdot \\frac{2006}{2008} f(2006) \\\\\n&= \\frac{2007}{... | United States | Berkeley Math Circle Monthly Contest 7 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2/(2009*2008) | |
0h2j | $$
\max\{|x|-1, x^2-1\} \leq \min\{x^2-1, 1-x\},
$$
where $\max\{a,b\} = \begin{cases} a, & \text{if } a \ge b \\ b, & \text{if } a < b \end{cases}$, and $\min\{a,b\} = \begin{cases} b, & \text{if } a \ge b \\ a, & \text{if } a < b \end{cases}$. | [
"It is clear that\n$$\n\\max\\{|x|-1, x^2-1\\} \\geq x^2-1 \\geq \\min\\{x^2-1, 1-x\\},\n$$\nso the inequality from the problem condition can be satisfied only if\n$$\n\\max\\{|x|-1, x^2-1\\} = x^2-1 = \\min\\{x^2-1, 1-x\\},\n$$\nand this, in turn, implies that\n$$\n|x|-1 \\leq x^2-1 \\leq 1-|x|.\n$$\nTherefore, $1... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | x ∈ {-1, 0, 1} | |
0l26 | Problem:
Let $p$ denote the proportion of teams, out of all participating teams, who submitted a negative response to problem 5 of the Team round (e.g. "there are no such integers"). Estimate $P=\lfloor 10000 p\rfloor$. An estimate of $E$ earns $\max (0,\lfloor 20-|P-E| / 20\rfloor)$ points.
If you have forgotten, pr... | [
"Solution:\n\nOf the 88 teams competing in this year's Team round, 49 of them answered negatively, 9 (correctly) provided a construction, 16 answered ambiguously or did not provide a construction, and the remaining 14 teams did not submit to problem 5. Thus $p=\\frac{49}{88} \\approx 0.5568$."
] | United States | HMMT February 2024 Guts Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | final answer only | 5568 | |
0642 | Problem:
Die Menge aller positiver rationaler Zahlen sei mit $\mathbb{Q}^{+}$ bezeichnet. Man bestimme alle Funktionen $f: \mathbb{Q}^{+} \rightarrow \mathbb{Q}^{+}$ mit der Eigenschaft
$$
f\left(x^{2} f(y)^{2}\right) = f\left(x^{2}\right) f(y) \tag{*}
$$
für alle $x, y \in \mathbb{Q}^{+}$. | [
"Solution:\n\nWir beginnen mit einigen für die Lösungsmenge wichtigen Eigenschaften der Grundmenge $\\mathbb{Q}^{+}$. Bekanntlich liegt das Produkt $a b$ zweier Zahlen $a, b \\in \\mathbb{Q}^{+}$ wieder in $\\mathbb{Q}^{+}$, ebenso wie das Inverse $\\frac{1}{a}$ zu jeder Zahl $a \\in \\mathbb{Q}^{+}$ (was für $\\sq... | Germany | 1. Auswahlklausur | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof and answer | All solutions are obtained as follows. Let U be any subgroup of the positive rationals under multiplication. Fix representatives a_i of the cosets of the quotient of the positive rationals by U, and choose elements u_i in U, one for each representative. Define f on the positive rationals by:
- For any square argument x... | |
0a80 | Problem:
Let $ABCD$ be a convex quadrilateral. We assume that there exists a point $P$ inside the quadrilateral such that the areas of the triangles $ABP$, $BCP$, $CDP$, and $DAP$ are equal. Show that at least one of the diagonals of the quadrilateral bisects the other diagonal. | [
"Solution:\n(See Figure 9.) We first assume that $P$ does not lie on the diagonal $AC$ and the line $BP$ meets the diagonal $AC$ at $M$. Let $S$ and $T$ be the feet of the perpendiculars from $A$ and $C$ on the line $BP$. The triangles $APB$ and $CBP$ have equal area. Thus $AS = CT$. If $S \\neq T$, then the right ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 11 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
018q | A deck consists of $3n$ cards, $n$ each colored red, green and blue in denominations $1$ through $n$. We choose a subset $S$ of the denominations and deal all cards of the chosen denominations into three equal size hands to players designated red, green and blue in such a way that no player receives a card of her own c... | [
"Partition the set of denominations $D = \\{1, 2, \\dots, k\\}$ occurring in red's hand into three blocks: $A$, those appearing on both blue and green cards (in red's hand); $B$, those appearing on blue cards only; $C$, those appearing on green cards only. Set $|A| = a$, $|B| = b$, $|C| = c$. Thus $a + b + c = k$ a... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof and answer | \binom{n}{k}\binom{2k}{k} | |
0bux | Problem:
Fie $(a_{n})_{n \geq 0}$ un șir de numere reale astfel încât $a_{0} \in \mathbb{R}$ și $a_{n+1}=a_{n}^{2}-8 a_{n}+18, \forall n \geq 0$.
a) Să se arate că dacă șirul $(a_{n})_{n \geq 0}$ este convergent atunci $\exists p \in \mathbb{N}$ astfel încât $a_{n+1}=a_{n}, \forall n \geq p$.
b) Determinați mulțimea... | [] | Romania | Olimpiada Nationala de Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | {2, 3, 4, 5, 6} | |
00z0 | Problem:
Let $a < b < c$ be three positive integers. Prove that among any $2c$ consecutive positive integers there exist three different numbers $x, y, z$ such that $abc$ divides $xyz$. | [
"Solution:\nFirst we show that among any $b$ consecutive numbers there are two different numbers $x$ and $y$ such that $ab$ divides $xy$. Among the $b$ consecutive numbers there is clearly a number $x'$ divisible by $b$, and a number $y'$ divisible by $a$. If $x' \\neq y'$, we can take $x = x'$ and $y = y'$, and we... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | null | proof only | null | |
0hv4 | Problem:
Let $x = 2001^{1002} - 2001^{-1002}$ and $y = 2001^{1002} + 2001^{-1002}$. Find $x^2 - y^2$. | [
"Solution:\n\nWe are given $x = 2001^{1002} - 2001^{-1002}$ and $y = 2001^{1002} + 2001^{-1002}$. We are to find $x^2 - y^2$.\n\nRecall that $x^2 - y^2 = (x - y)(x + y)$.\n\nCompute $x + y$ and $x - y$:\n\n$x + y = (2001^{1002} - 2001^{-1002}) + (2001^{1002} + 2001^{-1002}) = 2 \\times 2001^{1002}$\n\n$x - y = (200... | United States | null | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | final answer only | -4 | |
0536 | The numbers $1, 2, \ldots, 2012$ are written on the blackboard in some order, each of them exactly once. Between each two neighboring numbers the absolute value of their difference is written and the original numbers are erased. This is repeated until only one number is left on the blackboard. What is the largest possi... | [
"The largest number on the blackboard cannot increase on any step, because the absolute value of the difference of two nonnegative numbers cannot be greater than the maximum of these two numbers. Since in the beginning all the numbers are different and positive, after the first step the largest possible number is $... | Estonia | Open Contests | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2010 | |
02e1 | $a$, $b$, $c$, $d$ are integers. Show that $x^2 + a x + b = y^2 + c y + d$ has infinitely many integer solutions iff $a^2 - 4b = c^2 - 4d$. | [
"If $a^2 - 4b = c^2 - 4d$, then $a^2 \\equiv c^2 \\pmod{4}$, so $a$ and $c$ have the same parity. So if we take any integer $x$ and then $y$ to be the integer $x - \\frac{a-c}{2}$ we have $x - \\frac{a}{2} = y - \\frac{c}{2}$ and hence $(x - \\frac{a}{2})^2 - \\frac{a^2-4b}{4} = (y - \\frac{c}{2})^2 - \\frac{c^2-4d... | Brazil | VII OBM | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0663 | If $a, b, c$ are positive real numbers with sum $6$, determine the maximal value of the expression:
$$
S = \sqrt[3]{a^2 + 2bc} + \sqrt[3]{b^2 + 2ca} + \sqrt[3]{c^2 + 2ab}.
$$ | [
"We use the inequality of arithmetic–geometric mean as follows:\n\n$$\n\\sqrt[3]{a^2+2bc} = \\frac{1}{\\sqrt[3]{12^2}} \\sqrt[3]{(a^2+2bc) \\cdot 12 \\cdot 12} \\le \\frac{1}{\\sqrt[3]{12^2}} \\cdot \\frac{a^2+2bc+12+12}{3} = \\frac{1}{3\\sqrt[3]{12^2}} (a^2+2bc+24),\n$$\n$$\n\\sqrt[3]{b^2+2ca} = \\frac{1}{\\sqrt[3... | Greece | 28th Hellenic Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof and answer | 3\sqrt[3]{12} | |
00z7 | Problem:
The real numbers $x_{1}, x_{2}, \ldots, x_{1996}$ have the following property: for any polynomial $W$ of degree $2$ at least three of the numbers $W\left(x_{1}\right), W\left(x_{2}\right), \ldots, W\left(x_{1996}\right)$ are equal. Prove that at least three of the numbers $x_{1}, x_{2}, \ldots, x_{1996}$ are ... | [
"Solution:\n\nLet $m = \\min \\{ x_{1}, \\ldots, x_{1996} \\}$. Then the polynomial $W(x) = (x - m)^{2}$ is strictly increasing for $x \\geq m$. Hence if $W\\left(x_{i}\\right) = W\\left(x_{j}\\right)$ we must have $x_{i} = x_{j}$, and the conclusion follows."
] | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ggm | 令 $I, O, H, \Omega$ 分別為三角形 $ABC$ 的內心、外心、垂心與外接圓。設 $AI$ 與 $\Omega$ 交於 $M \neq A$, $IH$ 與 $BC$ 交於 $D$, $MD$ 與 $\Omega$ 交於 $E \neq M$。
證明：直線 $OI$ 與 $\triangle IHE$ 的外接圓相切。 | [
"取一點 $X$ 使得 $\\triangle XHI \\sim \\triangle XIO$。我們先證明 $X$ 位於 $\\Omega$ 上：取 $Y, Z$ 使得 $\\triangle XAY \\sim \\triangle XBZ \\sim \\triangle XHI$，則由旋似，$\\triangle IAH \\sim \\triangle OYI$。這告訴我們 $\\angle OYI = \\angle IAH = \\angle OAI$，即 $A, O, I, Y$ 共圓。同理有 $B, O, I, Z$ 共圓。設 $P$ 為 $AY$ 與 $BZ$ 的交點，則一樣由旋似，$\\triangl... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究(一) | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-... | Chinese; English | proof only | null | |
0hul | Problem:
Prove that
$$
-\frac{1}{2} \leq \frac{(x+y)(1-x y)}{\left(1+x^{2}\right)\left(1+y^{2}\right)} \leq \frac{1}{2}
$$
for all real numbers $x$ and $y$. | [
"Solution:\nSince $x$ and $y$ can be replaced by their negatives, it suffices to prove the right-hand inequality. On cross-multiplication, this is\n$$\n\\begin{aligned}\n0 & \\stackrel{?}{\\leq} \\left(x^{2}+1\\right)\\left(y^{2}+1\\right)-2(x+y)(1-x y) \\\\\n& = x^{2} y^{2}-2 x^{2} y + x^{2} - 2 x y^{2} + 2 x + y^... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
05p5 | Problem:
Soit $ABC$ un triangle tel que $AB < AC$, $H$ son orthocentre, $\Gamma$ son cercle circonscrit, $d$ la tangente à $\Gamma$ en $A$. On considère le cercle de centre $B$ passant par $A$. Il coupe $d$ en $D$ et $(AC)$ en $E$.
Montrer que $D$, $E$, $H$ sont alignés. | [
"Solution:\n\n\n\nOn remarque tout d'abord que la hauteur issue de $B$ dans $ABC$ est la médiatrice de $[AE]$. Soit en effet $L$ son point d'intersection avec $(AC)$. Par la puissance d'un point, $CE \\cdot CA = CB^2 - AB^2 = CL^2 - AL^2 = CA \\cdot (CL - AL)$ donc $LA = LE$.\n\nDonc $HAE$ ... | France | Préparation Olympique Française de Mathématiques | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | null | proof only | null | |
0jwl | Problem:
Denote $\phi=\frac{1+\sqrt{5}}{2}$ and consider the set of all finite binary strings without leading zeroes. Each string $S$ has a "base-$\phi$" value $p(S)$. For example, $p(1101)=\phi^{3}+\phi^{2}+1$. For any positive integer $n$, let $f(n)$ be the number of such strings $S$ that satisfy $p(S)=\frac{\phi^{4... | [
"Solution:\n\nWe write everything in base $\\phi$. Notice that\n$$\n\\frac{\\phi^{48 n}-1}{\\phi^{48}-1}=10\\ldots 010\\ldots 01\\ldots 10\\ldots 01\n$$\nwhere there are $n-1$ blocks of 47 zeros each. We can prove that every valid base-$\\phi$ representation comes from replacing a consecutive string $100$ with a $0... | United States | HMMT November 2017 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | (25 + 3*sqrt(69)) / 2 | |
09z4 | Find all pairs $(p, q)$ of prime numbers such that
$$
p(p^2 - p - 1) = q(2q + 3).
$$ | [
"We show that the only solution is $(p, q) = (13, 31)$.\nFirst suppose that $p = q$. Then $p^2 - p - 1 = 2q + 3 = 2p + 3$, therefore $(p-4)(p+1) = 0$. As neither $4$ and $-1$ are prime numbers, there are no solutions $(p, q)$ with $p = q$.\n\nHence $p \\neq q$, so from the equation, it follows that $p \\mid 2q+3$ a... | Netherlands | BxMO Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (13, 31) | |
004g | Determine los valores de $n$ ($n \in \mathbb{N}$) tales que un cuadrado de lado $n$ se pueda partir en un cuadrado de lado $1$ y cinco rectángulos cuyas medidas de los lados sean $10$ números naturales distintos dos a dos y todos mayores que $1$. | [] | Argentina | XVI Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Spanish | proof and answer | all n ≥ 14 | |
05tb | Problem:
On choisit 5 diviseurs positifs de $10^{2020}$. Montrer qu'il y en a deux dont le produit est un carré. | [
"Solution:\n\nOn commence par regarder la forme que peut prendre un éventuel diviseur de $10^{2020}$. Les diviseurs de $10^{2020}$ sont de la forme $2^{a} 5^{b}$. Le produit de deux diviseurs est donc également un nombre de la forme $2^{a} 5^{b}$ et c'est un carré parfait si et seulement si $a$ et $b$ sont tous les... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
062c | Problem:
Für jede reelle Zahl $x$ mit $0 < x < 1$ sei $y \in (0, 1)$ diejenige Zahl, deren $n$-te Nachkommastelle die $(2n)$-te Nachkommastelle von $x$ ist. Man beweise: Wenn $x$ rational ist, dann ist auch $y$ rational. | [
"Solution:\n\nEine rationale Zahl besitzt eine abbrechende oder eine periodische Dezimalbruchentwicklung. Eine abbrechende Dezimalbruchentwicklung lässt sich durch Anhängen von Nullen ins Unendliche fortsetzen und entspricht daher einer periodischen Dezimalbruchentwicklung mit der Periodenlänge $1$.\n\nIm Folgenden... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Number Theory > Modular Arithmetic",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof only | null | |
0h32 | Is it possible to write numbers $1, 2, 3, 4, 5, 6, 7, 8$ at the vertices of an octagon (one number at each vertex) so that the sum of numbers in each three consecutive vertices is greater than:
a) $13$;
b) $11$;
c) $12$? | [
"Нехай біля вершин правильного восьмикутника послідовно записані числа $a_1, a_2, a_3, a_4, a_5, a_6, a_7, a_8$.\n\na) Припустимо, що числа можна розташувати потрібним чином. Тоді\n$$\n\\begin{cases} a_1 + a_2 + a_3 \\geq 14, \\\\ a_2 + a_3 + a_4 \\geq 14, \\\\ \\vdots \\\\ a_8 + a_1 + a_2 \\geq 14. \\end{cases}\n$... | Ukraine | Ukrainian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | a) No; b) Yes; c) No | |
09pe | Prove that the polynomial $P(X) = X^8 + 61X + 2025$ is not the product of two non-constant polynomials with integer coefficients.
(Otgonbayar Uuye) | [] | Mongolia | MMO2025 Round 2 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0f83 | Problem:
In the triangle $ABC$, the angle $C$ is obtuse and $D$ is a fixed point on the side $BC$, different from $B$ and $C$. For any point $M$ on the side $BC$, different from $D$, the ray $AM$ intersects the circumcircle $S$ of $ABC$ at $N$. The circle through $M$, $D$ and $N$ meets $S$ again at $P$, different from... | [
"Solution:\n\nTake $A'$ on the circle $S$ such that $AA'$ is parallel to $BC$. Let the ray $AD$ meet $S$ again at $P'$. Then $\\angle MNP' = \\angle ANP'$ (same angle) $= \\angle AA'P'$ ($A'AP'N$ cyclic) $= \\angle A'DB$ ($BC$ parallel to $AA'$) $= \\angle MDP$ (opposite angles). So $MDNP'$ is cyclic, so $P$ must b... | Soviet Union | 22nd ASU | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | Let P be the second intersection of the ray AD with the circumcircle of triangle ABC; the minimizing position of M on BC is the foot of the perpendicular from P to BC. | |
0cnm | Find the least possible denominator in the irreducible fraction which is the sum of two irreducible fractions with denominators $600$ and $700$. (I. Bogdanov) | [
"Ответ. $2^3 \\cdot 3 \\cdot 7 = 168$.\n\nПусть наши дроби $\\frac{a}{600}$ и $\\frac{b}{700}$. Тогда $a$ взаимно просто с $6$, а $b$ — с $7$. Поэтому числитель их суммы $\\frac{7a+6b}{4200}$ взаимно прост как с $6 = 2 \\cdot 3$, так и с $7$. Поскольку $4200 = 2^3 \\cdot 3 \\cdot 7 \\cdot 5^2$, это означает, что зн... | Russia | Russian mathematical olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English; Russian | proof and answer | 168 | |
030f | Problem:
Seja $ABCD$ um quadrado de lado $28~\mathrm{cm}$. Seja $P$ um ponto interior ao quadrado e $E$ um ponto no lado $CD$ tal que $PE$ é perpendicular a $CD$. Além disso, $AP = BP = PE$. Encontre o comprimento de $AP$.
 | [
"Solution:\n\nSeja $M$ o ponto de interseção da reta $EP$ com o lado $AB$. Como $EP$ é perpendicular a $CD$, então $EM$ é perpendicular a $AB$. Além disso, como o triângulo $ABP$ é isósceles, $M$ é o ponto médio de $AB$. Se o comprimento de $AP$ é $x$, segue do Teorema de Pitágoras que\n$$\n\\begin{aligned}\nAP^2 &... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 17.5 cm | |
03lr | Problem:
Consider a round-robin tournament with $2n+1$ teams, where each team plays each other team exactly once. We say that three teams $X$, $Y$ and $Z$ form a cycle triplet if $X$ beats $Y$, $Y$ beats $Z$, and $Z$ beats $X$. There are no ties.
a. Determine the minimum number of cycle triplets possible.
b. Determi... | [
"Solution:\n\na. The minimum is $0$, which is achieved by a tournament in which team $T_{i}$ beats $T_{j}$ if and only if $i > j$.\n\nb. Any set of three teams constitutes either a cycle triplet or a \"dominated triplet\" in which one team beats the other two; let there be $c$ of the former and $d$ of the latter. T... | Canada | 38th Canadian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Minimum: 0; Maximum: n(n+1)(2n+1)/6 | |
0feb | Problem:
Sea $a \neq 1$ un número real positivo y $n$ un entero positivo. Demostrar que $n^{2}<\frac{a^{n}+a^{-n}-2}{a+a^{-1}-2}$. | [
"Solution:\nLa desigualdad dada $n^{2}<\\frac{a^{n}+a^{-n}-2}{a+a^{-1}-2}$ es equivalente a $n^{2}<\\frac{\\left(a^{\\frac{n}{2}}-a^{-\\frac{n}{2}}\\right)^{2}}{\\left(a^{\\frac{1}{2}}-a^{-\\frac{1}{2}}\\right)^{2}}$, que a su vez equivale a que $n<\\frac{\\alpha^{n}-\\alpha^{-n}}{\\alpha-\\alpha^{-1}}$, siendo $\\... | Spain | null | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0bzo | Find the distinct positive integers $a$, $b$, $c$, $d$, such that the following conditions hold:
(1) exactly three of the four numbers are prime numbers;
(2) $a^2 + b^2 + c^2 + d^2 = 2018$. | [] | Romania | 69th Romanian Mathematical Olympiad - Final Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 2, 3, 41, 18 | |
0bxo | Consider the set $A$ of positive 3-digit integers $abc$ consisting of three consecutive, not necessarily ordered, digits.
a) Determine the cardinal of the set $A$.
b) Prove that for any choice of some elements of the set $A$, the sum of the chosen elements cannot equal 2017. | [
"a) $\\{0, 1, 2\\}$ generate 4 numbers.\nEach of the triplets $\\{1, 2, 3\\}$, $\\{2, 3, 4\\}$, ..., $\\{7, 8, 9\\}$ generate 6 numbers.\nThe cardinal of the set $A$ is $4 + 7 \\cdot 6 = 46$.\n\nb) Any number in $A$ is a multiple of 3. 2017 being not a multiple of 3, the sum of any numbers in $A$ cannot be 2017."
] | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization"
] | English | proof and answer | 46; no sum of chosen elements can equal 2017 | |
0kfa | Problem:
Let $ABC$ be a scalene triangle with angle bisectors $AD$, $BE$, and $CF$ so that $D$, $E$, and $F$ lie on segments $BC$, $CA$, and $AB$ respectively. Let $M$ and $N$ be the midpoints of $BC$ and $EF$ respectively. Prove that line $AN$ and the line through $M$ parallel to $AD$ intersect on the circumcircle of... | [
"Solution:\n\n\n\nLet $X, Y$ be on $AB, AC$ such that $CX \\parallel BE$ and $BY \\parallel CF$. Then $BX = BC = CY$. Let $Z$ be the midpoint of $XY$. Then $\\overrightarrow{MZ} = \\frac{1}{2}(\\overrightarrow{BX} + \\overrightarrow{CY})$, which bisects the angle between $BX$ and $CY$ becau... | United States | HMMT February 2020 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"... | null | proof only | null | |
063g | Problem:
Es sei $n$ eine positive natürliche Zahl. Im Folgenden betrachten wir Paare von Elementen der Menge $\{1,2, \ldots, n\}$, die jeweils kein gemeinsames Element haben.
Man bestimme mit Beweis die größtmögliche Anzahl solcher Paare, für welche die Summen ihrer Elemente paarweise verschieden und nicht größer als ... | [
"Solution:\n\nEs sei $k$ die Anzahl möglicher Paare. Dann lässt sich die Summe $S$ der in ihnen vorkommenden $2 k$ Zahlen in zwei Richtungen abschätzen:\n\n$S \\geq 1+2+\\ldots+2 k = k(2 k+1)$ (alle Zahlen sind verschieden) und\n\n$S \\leq n+(n-1)+\\ldots+(n-k+1) = n k - \\frac{1}{2} k(k-1)$ (alle Summen sind versc... | Germany | 1. Auswahlklausur | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | floor((2n-1)/5) | |
0gkd | In a square of side length $5$, a figure is placed such that the distance between any two of its points is not equal to $0.001$, $0.002 \times \sqrt{3}$, $0.002$. Prove that the area of this figure does not exceed $3.575$. | [
"Let $F$ denote this figure. Consider the translations of $F$ under the following vectors in the plane: $\\vec{u}_1 = 0.001\\vec{i}$, and\n$$\n\\vec{u}_{k+1} = R_{\\pi/3}(\\vec{u}_k), \\quad k = 1, 2, 3, 4, 5,\n$$\nwhere $R_{\\pi/3}$ denotes the rotation by $\\pi/3$ counterclockwise. Now let $F_0 = F$ and, for $k =... | Thailand | Thai Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof only | null | |
0a6d | Problem:
Determine all positive integers $n$ less than $2024$ such that for all positive integers $x$, the greatest common divisor of $9x + 1$ and $nx + 1$ is $1$. | [
"Solution:\nLet $m = n - 9$ and for the sake of contradiction assume $m$ has a prime factor $p$ with $p \\neq 3$. Express $p = 9q + r$ where $0 \\leq r \\leq 8$ ($r$ is the remainder when $p$ is divided by $9$ and $q$ may be zero). Since $\\gcd(9, p) = 1$ we must have $r = 1, 2, 4, 5, 7$ or $8$. For each of these p... | New Zealand | NZMO Round Two | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n"
] | null | proof and answer | 6, 8, 10, 12, 18, 36, 90, 252, 738 | |
04an | Find all sequences $a : \mathbb{N} \rightarrow \mathbb{N}$ such that $a_n + a_{n+1} = a_{n+2}a_{n+3} - 200$. | [
"By subtracting equalities\n$$\n\\begin{aligned}\na_n + a_{n+1} &= a_{n+2}a_{n+3} - 200 \\\\\na_{n+1} + a_{n+2} &= a_{n+3}a_{n+4} - 200\n\\end{aligned}\n$$\nwe get\n$$\na_n - a_{n+2} = a_{n+3} (a_{n+2} - a_{n+4}). \\qquad (4.3)\n$$\nSince $a_{n+3} > 0$, we have\n$$\n\\begin{array}{lcll}\na_n > a_{n+2} & \\text{if a... | Croatia | CroatianCompetitions2011 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | All sequences a: N → N satisfying a_n + a_{n+1} = a_{n+2} a_{n+3} − 200 are exactly those of the following forms:
1) Two-periodic sequences with constants a on odd indices and b on even indices where (a − 1)(b − 1) = 201. Explicitly, (a, b) ∈ {(2, 202), (202, 2), (4, 68), (68, 4)}.
2) Sequences with all odd terms equ... | |
0byo | Let $n$ be an integer greater than $1$ and let $X$ be an $n$-element set. A non-empty collection of subsets $A_1, \dots, A_k$ of $X$ is *tight* if the union $A_1 \cup \dots \cup A_k$ is a proper subset of $X$ and no element of $X$ lies in exactly one of the $A_i$s. Find the largest cardinality of a collection of proper... | [
"*Note.* A subset $A$ of $X$ is proper if $A \\ne X$. The sets in a collection are assumed to be distinct. The whole collection is assumed to be a subcollection.\n\n*First solution.* (Ilya Bogdanov) The required maximum is $2n-2$. To describe a $(2n-2)$-element collection satisfying the required conditions, write $... | Romania | THE Tenth ROMANIAN MASTER OF MATHEMATICS | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 2n - 2 | |
01dc | $ABCD$ is a convex quadrilateral such that $AB = AD$. $T$ is a point on the diagonal $AC$ such that $\angle ABT + \angle ADT = \angle BCD$. Prove that $AT + AC \geq AB + AD$. | [
"Let $T'$ be a point on the ray $AC$ such that $AT' \\cdot AC = AB^2 = AD^2$. Then triangles $ACB$ and $ABT'$ are similar (by two sides and the angle between them), hence $\\triangle ACB = \\triangle ABT'$. The triangles $ACD$ and $ADT'$ are similar by analogous reasons, therefore $\\triangle ACD = \\triangle ADT'$... | Baltic Way | Baltic Way 2016 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-H... | null | proof only | null | |
0ilc | Problem:
Compute $(x-a)(x-b)\cdots(x-z)$ | [] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 0 | |
01px | Three of six segments (four sides and two diagonals of a parallelogram) are painted red, and three others are painted green.
Prove that one can construct a triangle using the segments of the same color as its sides. | [
"Let $BC = AD = a$, $AB = CD = b$, $AC = d_1$, $BD = d_2$. Let for the definiteness $\\angle ADC \\ge 90^\\circ$. Then $d_1 > a$, $d_1 > b$. Without loss of generality we suppose that $a \\ge b$. Then $a+a \\ge a+b > d_1 \\ge d_2$. Exactly two cases are possible.\n\n1. The diagonals have the different colors. If tw... | Belarus | BelarusMO 2013_s | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0b6e | Find all continuous functions $f : \mathbb{R} \to \mathbb{R}$ with the property that for every $a \in \mathbb{R}$ there exists $b \in \mathbb{R}$ such that, for every $x \in \mathbb{R}$,
$$
f(f(x) + a) = f(x) + b.
$$ | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | All continuous solutions are either constant functions f(x) = c for some real c, or affine functions of the form f(x) = x + c for some real c. | |
0k9h | Problem:
In an election for the Peer Pressure High School student council president, there are $2019$ voters and two candidates Alice and Celia (who are voters themselves). At the beginning, Alice and Celia both vote for themselves, and Alice's boyfriend Bob votes for Alice as well. Then one by one, each of the remain... | [
"Solution:\n\nLet $P_{n}(m)$ be the probability that after $n$ voters have voted, Alice gets $m$ votes. We show by induction that for $n \\geq 3$, the ratio $P_{n}(2): P_{n}(3): \\cdots: P_{n}(n-1)$ is equal to $1: 2: \\cdots:(n-2)$. We take a base case of $n=3$, for which the claim is obvious. Then suppose the cla... | United States | HMMT February 2019 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 1513/2017 | |
0evq | Let $n (\ge 2)$ be a positive integer. Alice distributes $2n$ candies into $4n$ boxes $B_1, B_2, \dots, B_{4n}$. After checking the number of candies Alice puts in each box, Bob chooses $2n$ boxes $B_{k_1}, B_{k_2}, \dots, B_{k_{2n}}$ out of the $4n$ boxes satisfying the following, and then takes all candies in the cho... | [
"The answer is $n$.\nIf Alice puts one candy in each of boxes $B_1, B_2, \\dots, B_{4n-1}$, then Bob can choose at most $n$ out of the $2n$ boxes, so Alice can get exactly $n$ candies.\n\nNow we prove that Bob can take at least $n$ candies. Let $b_i$ be the number of candies Alice puts in $B_i$ for $i = 1, 2, \\dot... | South Korea | The 37th Korean Mathematical Olympiad Final Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n | |
0eje | Problem:
Naj bo $x \in \mathbb{R}$ in $x>0$, $x \neq 4$, $x \neq 9$. Kateri izraz je ekvivalenten izrazu $\left((2+\sqrt{x})^{-1}+1\right)^{-1}$?
(A) $\frac{x+\sqrt{x}-6}{x-9}$
(B) $\frac{x-\sqrt{x}-6}{x-9}$
(C) $\frac{x-\sqrt{x}+6}{x-9}$
(D) $\frac{x-\sqrt{x}-6}{x-4}$
(E) $\frac{x-\sqrt{x}+6}{x-4}$ | [
"Solution:\n\n$\\left((2+\\sqrt{x})^{-1}+1\\right)^{-1}$ preoblikujemo v $\\left(\\frac{1}{2+\\sqrt{x}}+1\\right)^{-1}=\\left(\\frac{3+\\sqrt{x}}{2+\\sqrt{x}}\\right)^{-1}=\\frac{2+\\sqrt{x}}{3+\\sqrt{x}}$, $x \\neq 9$, $x \\neq 4$. Izraz še racionaliziramo in dobimo $\\frac{x-\\sqrt{x}-6}{x-9}$. Pravilen je odgovo... | Slovenia | 21. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | B | |
0k61 | Problem:
Prove that for all positive integers $n$, all complex roots $r$ of the polynomial
$$
P(x) = (2 n) x^{2 n} + (2 n-1) x^{2 n-1} + \cdots + (n+1) x^{n+1} + n x^{n} + (n+1) x^{n-1} + \cdots + (2 n-1) x + 2 n
$$
lie on the unit circle (i.e. $|r|=1$). | [
"Solution:\nNote that neither $0$ nor $1$ are roots of the polynomial. Consider the function\n$$\nQ(x) = \\frac{P(x)}{x^{n}} = (2 n) x^{n} + (2 n) x^{-n} + (2 n-1) x^{n-1} + (2 n-1) x^{-n+1} + \\cdots + (n+1) x^{1} + (n+1) x^{-1} + n.\n$$\nAll $2 n$ of the complex roots of $P(x)$ will be roots of $Q(x)$.\nIf $|x|=1... | United States | HMMT February 2019 Team Round | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof only | null | |
0b6s | Given an integer number $n \ge 3$, determine the real numbers $x_1, x_2, \dots, x_n$ minimising $(n-1)(x_1^2 + x_2^2 + \dots + x_n^2) + nx_1x_2 \dots x_n$, subject to $x_k \ge 0$, $k = 1, 2, \dots, n$, and $x_1 + x_2 + \dots + x_n = n$. | [
"Let $f(x_1, x_2, \\dots, x_n) = (n-1)(x_1^2 + x_2^2 + \\dots + x_n^2) + nx_1x_2 \\dots x_n$. Assuming $0 \\le x_1 \\le x_2 \\le \\dots \\le x_n$, we first show that\n$$\nf(x_1, x_2, \\dots, x_n) \\ge f(x_1, (x_2+x_n)/2, x_3, \\dots, x_{n-1}, (x_2+x_n)/2).\n$$\n\nTo begin with, notice that $x_n \\ge 1$, to get\n$$\... | Romania | 2010 DANUBE MATHEMATICAL COMPETITION | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | Minimum value: n^2. It is attained either when all variables are equal to 1, or when exactly one variable is 0 and the remaining n−1 variables are all equal to n/(n−1). | |
0l81 | Let $n \ge 2$ be a positive integer. Let $a_1, a_2, \dots, a_n$ be a sequence of positive integers such that
$$
\gcd(a_1, a_2), \gcd(a_2, a_3), \dots, \gcd(a_{n-1}, a_n)
$$
is a strictly increasing sequence. Find, in terms of $n$, the maximum possible value of
$$
\frac{1}{a_1} + \frac{1}{a_2} + \dots + \frac{1}{a_n}.
$... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 2 | |
037k | Problem:
A plane intersects a tetrahedron $ABCD$ and divides the medians of the triangles $DAB$, $DBC$ and $DCA$ through $D$ in ratios $1:2$, $1:3$ and $1:4$ from $D$, respectively. Find the ratio of the volumes of the two parts of the tetrahedron cut by the plane.
Oleg Mushkarov | [
"Solution:\nLet the plane meet the edges $DA$, $DB$ and $DC$ at points $P$, $Q$ and $R$, respectively. Set\n$$\n\\frac{DP}{DA} = x, \\quad \\frac{DQ}{DB} = y, \\text{ and } \\frac{DR}{DC} = z\n$$\nLet $M$ be the midpoint of $AB$ and $L = DM \\cap PQ$. It follows from the condition of the\n\... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 1:47 | |
0f0x | Problem:
Find all three-digit decimal numbers $a_1a_2a_3$ which equal the mean of the six numbers $a_1a_2a_3$, $a_1a_3a_2$, $a_2a_1a_3$, $a_2a_3a_1$, $a_3a_1a_2$, $a_3a_2a_1$. | [] | Soviet Union | ASU | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 111, 222, 333, 444, 555, 666, 777, 888, 999, 370, 407, 481, 518, 592, 629 | |
0hgp | For positive real numbers $a$, $b$, $c$, which satisfy the condition $ab + bc + ca = 1$, prove the following inequality:
$$
\left(\sqrt{bc} + \frac{1}{2a + \sqrt{bc}}\right) \cdot \left(\sqrt{ca} + \frac{1}{2b + \sqrt{ca}}\right) \cdot \left(\sqrt{ab} + \frac{1}{2c + \sqrt{ab}}\right) \ge 8abc.
$$ | [
"From the statement it follows, that\n$$\n\\frac{1}{2a + \\sqrt{bc}} = \\frac{ab + bc + ca}{2a + \\sqrt{bc}} = \\frac{bc + a(b + c)}{2a + \\sqrt{bc}} \\ge \\frac{bc + 2a\\sqrt{bc}}{2a + \\sqrt{bc}} = \\sqrt{bc},\n$$\nimplying\n$$\n\\sqrt{bc} + \\frac{1}{2a + \\sqrt{bc}} \\ge 2\\sqrt{bc}.\n$$\nIf we write two simila... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
02nc | Problem:
Digite numa calculadora um número qualquer de três algarismos. Em seguida, digite o mesmo número obtendo, assim, um número de seis algarismos, da forma $abcabc$. Divida esse número por $7$, divida o resultado por $11$ e, finalmente, divida o número obtido por $13$. O que aconteceu? Por que você obteve esse re... | [
"Solution:\n\nSeja o número de três algarismos $n = abc$. Ao digitá-lo duas vezes, obtemos o número de seis algarismos $abcabc$.\n\nEsse número pode ser escrito como:\n$$\nabcabc = 1000 \\times n + n = n \\times 1001\n$$\n\nAgora, dividimos $n \\times 1001$ por $7$, depois por $11$ e, finalmente, por $13$.\n\nNote ... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | The final result is the original three-digit number. | |
09q8 | Problem:
Zij $ABCD$ een koordenvierhoek met de eigenschap dat $\angle ABD = \angle DBC$. Zij $E$ het snijpunt van de diagonalen $AC$ en $BD$. Zij $M$ het midden van $AE$ en $N$ het midden van $DC$. Bewijs dat $MBCN$ een koordenvierhoek is. | [
"Solution:\n\nOplossing I. Omdat $ABCD$ een koordenvierhoek is, weten we dat $\\angle BDC = \\angle BAC = \\angle BAE$. Daarnaast volgt uit het gegeven dat $\\angle CBD = \\angle EBA$. Samen geeft dit $\\triangle DCB \\sim \\triangle AEB$ (hh). We gaan nu bewijzen dat hieruit volgt $\\triangle NCB \\sim \\triangle ... | Netherlands | Dutch TST | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
090j | Let $n$ be a positive integer. Consider triples $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, \dots, $(x_n, y_n, z_n)$ which consist of integers $1, 2, \dots, 100$ and satisfy the following condition:
For all infinite sequences $a_1, a_2, \dots$ which consist of integers $1, 2, \dots, 100$, there exist a positive integer $i$ a... | [
"**Lemma.** Let $x, y, z$ be integers with $1 \\le x, y, z \\le N$. Then at least one of $(x, y, z)$, $(y, z, x)$, or $(z, x, y)$ belongs to $S$.\n\n**proof.** Consider a sequence $a_1 = x, a_2 = y, a_3 = z, a_4 = x, a_5 = y, a_6 = z, \\dots$. Then, for every positive integer $i$, $(a_i, a_{i+1}, a_{i+2})$ is one o... | Japan | The 35th Japanese Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 333400 | |
0iod | Problem:
Suppose that $\omega$ is a primitive $2007^{\text{th}}$ root of unity. Find $\left(2^{2007}-1\right) \sum_{j=1}^{2006} \frac{1}{2-\omega^{j}}$.
For this problem only, you may express your answer in the form $m \cdot n^{k}+p$, where $m, n, k$, and $p$ are positive integers. Note that a number $z$ is a primitiv... | [
"Solution:\nAnswer: $2005 \\cdot 2^{2006}+1$. Note that\n$$\n\\begin{aligned}\n& \\frac{1}{z-\\omega}+\\cdots+\\frac{1}{z-\\omega^{2006}}=\\frac{\\sum_{j=1}^{2006} \\prod_{i \\neq j}\\left(z-\\omega^{i}\\right)}{(z-\\omega) \\cdots\\left(z-\\omega^{2006}\\right)} \\\\\n& \\quad=\\frac{\\frac{\\mathrm{d}}{\\mathrm{~... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 2005 · 2^{2006} + 1 | |
0i61 | Problem:
A permutation of a finite set is a one-to-one function from the set to itself; for instance, one permutation of $\{1,2,3,4\}$ is the function $\pi$ defined such that $\pi(1)=1, \pi(2)=3$, $\pi(3)=4$, and $\pi(4)=2$. How many permutations $\pi$ of the set $\{1,2, \ldots, 10\}$ have the property that $\pi(i) \ne... | [
"Solution:\nFor each such $\\pi$, the elements of $\\{1,2, \\ldots, 10\\}$ can be arranged into pairs $\\{i, j\\}$ such that $\\pi(i)=j ; \\pi(j)=i$. Choosing a permutation $\\pi$ is thus tantamount to choosing a partition of $\\{1,2, \\ldots, 10\\}$ into five disjoint pairs. There are 9 ways to pair off the number... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof and answer | 945 | |
006o | Decimos que un número es capicúa si al invertir el orden de sus cifras se obtiene el mismo número. Hallar todos los números que tienen al menos un múltiplo que es capicúa. | [] | Argentina | XIX Olimpiada de Matemática de Países del Cono Sur | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | Spanish | proof and answer | All positive integers not divisible by 10 | |
0jy1 | Problem:
Consider the $n \times n$ "multiplication table" below. The numbers in the first column multiplied by the numbers in the first row give the remaining numbers in the table.
| 1 | 2 | 3 | $\cdots$ | $n$ |
| :---: | :---: | :---: | :---: | :---: |
| 2 | 4 | 6 | $\cdots$ | $2 n$ |
| 3 | 6 | 9 | $\cdots$ | $3 n$ ... | [
"Solution:\n\nThe minimum is achieved by staying on the perimeter of the grid, and the maximum by staying as close to the main diagonal as possible. To see why this is true, tilt the grid 45 degrees:\n\n\n\nNow every path must include exactly one number from each row. The $n^{\\text{th}}$ r... | United States | 19th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Minimum sum: n(n^2 + 2n − 1)/2. Maximum sum: n(n + 1)(4n − 1)/6. | |
0g91 | 對於一個實數數列 $x_1, x_2, \dots, x_n$, 我們定義它的權重為
$$
\max_{1 \le i \le n} |x_1 + \dots + x_i|.
$$
給定 $n$ 個實數, 大衛和喬治想要把它們排列使權重很小。勤勞的大衛找遍所有可能性找到最小可能的權重是 $D$. 貪心的喬治則選了 $x_1$ 使 $|x_1|$ 最小, 在剩下的數字中, 他選了 $x_2$ 使 $|x_1 + x_2|$ 最小, 依此類推。像這樣, 在第 $i$ 步他從剩下的數字中選了 $x_i$ 使 $|x_1 + \dots + x_i|$ 最小。在每一步中, 如果有好幾個數字給出了相同的值, 則他隨機選一個。最後他得到一個權重... | [
"答案：$c = 2$\n\n解法：如果一開始是 $1, -1, 2, -2$, 則大衛會排成 $1, -2, 2, -1$, 喬治會排成 $1, -1, 2, -2$, 則 $D = 1, G = 2$, 所以 $c \\ge 2$.\n\n再來要證明 $G \\le 2D$, 令一開始有 $x_1, x_2, \\dots, x_n$, 假設大衛和喬治分別將他們排成 $d_1, d_2, \\dots, d_n$ 和 $g_1, g_2, \\dots, g_n$, 令\n$$\nM = \\max_{1 \\le i \\le n} |x_i|, \\quad S = |x_1 + \\dots + x_n|, \\q... | Taiwan | 2015 Math Olympiad Second Stage Training Camp | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 2 | |
0cbc | Let $O$ be the common point of the diagonals of the convex quadrilateral $ABCD$ and $M, N, P$ be the midpoints of the segments $AB, CD$ respectively $DO$. Prove that $AP \parallel MN$ if and only if $AD \parallel BN$. | [] | Romania | SHORTLISTED PROBLEMS FOR THE 73rd NMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof only | null | |
0hgh | The sum of coprime integers $m$ and $n$ equals $90$. What largest possible value can be attained by the product $mn$ of these integers? | [
"Make the following transformations:\n$$\nmn = \\left(\\frac{m+n}{2}\\right)^2 - \\left(\\frac{m-n}{2}\\right)^2 = 45^2 - \\left(\\frac{m-(90-m)}{2}\\right)^2 = 2025 - (m-45)^2 \\rightarrow \\max.\n$$\nSo the max will be achieved, when the value of the expression $(m - 45)^2$ is minimized, but here we have to not f... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, First Tour | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 2021 | |
0e81 | Find all natural numbers $n$ for which there exist $a_1, a_2, \dots, a_n \in \mathbb{N}$ such that
$$
a_1 + 2a_2 + \dots + n a_n = 6n
$$
and
$$
\frac{1}{a_1} + \frac{2}{a_2} + \dots + \frac{n}{a_n} = 2 + \frac{1}{n}.
$$ | [
"Each of the sums $a_1 + 2a_2 + \\dots + n a_n$ and $\\frac{1}{a_1} + \\frac{2}{a_2} + \\dots + \\frac{n}{a_n}$ contains $1 + 2 + \\dots + n = \\frac{n(n+1)}{2}$ numbers. From the given data, the arithmetic mean $A$ and the harmonic mean $H$ of numbers $a_1, a_2, a_3, a_3, a_3, \\dots, a_n, a_n, \\dots, a_n$ can be... | Slovenia | Selection Examinations for the IMO 2013 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 3 | |
0khv | Five balls are arranged around a circle. Chris chooses two adjacent balls at random and interchanges them. Then Silva does the same, with her choice of adjacent balls to interchange being independent of Chris's. What is the expected number of balls that occupy their original positions after these two successive transpo... | [
"Label the balls in order 1 through 5. Assume without loss of generality that the first transposition is $1 \\leftrightarrow 2$, resulting in the order 21345. The following table shows the results of the 5 equally likely second transpositions.\n\n| 2nd transposition | result | balls in original position | count |\... | United States | 2021 AMC 10 B Fall | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | MCQ | D | |
08lq | Problem:
Prove that $2^{n}+3^{n}$ is not a perfect cube for any positive integer $n$. | [
"Solution:\nIf $n=1$ then $2^{1}+3^{1}=5$ is not a perfect cube.\n\nPerfect cubes give residues $-1, 0$ and $1$ modulo $9$. If $2^{n}+3^{n}$ is a perfect cube, then $n$ must be divisible by $3$ (consider the congruence $2^{n}+3^{n}=x^{3}$ modulo $9$).\n\nIf $n=3k$ then $2^{3k}+3^{3k} > (3^{k})^{3}$. Also, $(3^{k}+1... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0880 | Problem:
Sia $x_{1} \ldots x_{n}$ una sequenza finita di numeri reali tali che:
i) la somma di 7 suoi termini consecutivi sia sempre strettamente positiva;
ii) la somma di 13 suoi elementi consecutivi sia sempre strettamente negativa.
Quale delle seguenti affermazioni è vera?
(A) La sequenza ha al più 18 termini
(B) L... | [
"Solution:\n\nLa risposta è (A). Basta osservare che se la sequenza ha più di 18 elementi allora la quantità:\n$$\nX=\\left(x_{1}+\\ldots+x_{7}\\right)+\\left(x_{2}+\\ldots+x_{8}\\right)+\\ldots+\\left(x_{13}+\\ldots+x_{19}\\right)\n$$\nè sia strettamente maggiore di zero che strettamente minore di zero, infatti: è... | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | MCQ | A | |
01t1 | Points $B_1$ and $C_1$ are marked respectively on the sides $AB$ and $AC$ of an acute isosceles triangle $ABC$ ($AB = AC$) such that $BB_1 = AC_1$. The points $B$, $C$, and $S$ lie in the same half-plane with respect to the line $B_1C_1$ so that
$$
\angle SB_1C_1 = \angle SC_1B_1 = \angle BAC.
$$
Prove that $B$, $C$, $... | [
"Necessity. Let $B$, $S$, $C$ be collinear. Let $X$ be the midpoint of the segment $B_1C_1$. Then the bisector of $B_1C_1$ passes through $X$ and intersects the side $BC$ at $S$ since $S$ lies on $BC$ and from the condition it follows that $SB_1 = SC_1$ (see Fig. 1).\nLet $O$ be the circumcenter of the triangle $AB... | Belarus | 66th Belarusian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance ... | English | proof only | null | |
0dkp | On the board is written a (not necessarily positive) integer. In a move, if the current number on the board is $a$, one erases it and replaces it by either $a^2 + 2a + 3$ or $5a^2 + 2$ or $a - 119$. Is there a starting number $s$ such that for any positive integer $f$ one can reach $f$ from $s$ with finitely many moves... | [
"Note that all of numbers of form $119k$ and $119k-1$ cannot be represented as $a^2 + 2a + 3$ or $5a^2 + 2$. We can check this claim by consider modulo 7 and 17. Indeed,\n* If $119k = a^2 + 2a + 3 = (a+1)^2 + 2$ then $-2$ is quadratic residue modulo 7, clearly absurd since quadratic residue modulo 7 are 0, 1, 2, 4.... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | No, such a starting number does not exist. | |
0j27 | Problem:
Suppose that $x$ and $y$ are complex numbers such that $x+y=1$ and that $x^{20}+y^{20}=20$. Find the sum of all possible values of $x^{2}+y^{2}$. | [
"Solution:\nWe have $x^{2}+y^{2}+2 x y=1$. Define $a=2 x y$ and $b=x^{2}+y^{2}$ for convenience. Then $a+b=1$ and $b-a=x^{2}+y^{2}-2 x y=(x-y)^{2}=2 b-1$ so that $x, y=\\frac{\\sqrt{2 b-1} \\pm 1}{2}$. Then\n$$\n\\begin{aligned}\nx^{20}+y^{20} & =\\left(\\frac{\\sqrt{2 b-1}+1}{2}\\right)^{20}+\\left(\\frac{\\sqrt{2... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Intermediate Algebra > Complex numbers",
"Discrete Mathematics > Combinatorics > Alge... | null | proof and answer | -90 | |
0h8w | 30 military ships are approaching to the island: 10 destroyers and 20 small ships. All ships are arranged in a circle and all distances between neighboring ships are equal. Two battle ships are defending the island. Each of them has exactly 10 rockets. The first battle ship can launch all 10 rockets at the same time, h... | [
"Consider all possible tens of neighboring military ships. There exist ten neighboring military ships such that there are at least 4 destroyers among them. The first battle ship can hit these 10 targets. Consider the rest 20 targets and divide them into two groups: with odd position and with even position. One of t... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 3 | |
0htm | Problem:
Let $x_{1}, x_{2}, \ldots, x_{1000}, y_{1}, y_{2}, \ldots, y_{1000}$ be 2000 different real numbers, and form the $1000 \times 1000$ matrix whose $(i, j)$-entry is $x_{i}+y_{j}$. If the product of the numbers in each row is 1, show that the product of the numbers in each column is -1. | [
"Solution:\n\nThe given says that $\\left(x_{i}+y_{1}\\right)\\left(x_{i}+y_{2}\\right) \\cdots\\left(x_{i}+y_{1000}\\right)=1$ for each $i=1,2, \\ldots, 1000$. So, if we let $P(x)$ be the polynomial $\\left(x+y_{1}\\right)\\left(x+y_{2}\\right) \\cdots\\left(x+y_{1000}\\right)-1$, the numbers $x_{i}$ are all roots... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0bsc | In an acute triangle, three of the adjacent angles around the orthocenter, made by the altitudes, have their measures directly proportional with $5$, $5$ and $7$, and the sum of the other three angles is $190\degree$. Find the measure of the angles of the triangle.
Constantin Apostol | [
"The sum of the measures of the angles directly proportional with $5$, $5$ and $7$ is $170\\degree$. Their measures are $5x$, $5x$, $7x$, with $5x + 5x + 7x = 170\\degree$, hence $x = 10\\degree$.\n\nThere are three pairs of opposite angles around the orthocenter, so they must be equal. On one side of an altitude m... | Romania | 67th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 50°, 60°, 70° | |
0inf | Problem:
Two sequences of positive integers, $x_{1}, x_{2}, x_{3}, \ldots$ and $y_{1}, y_{2}, y_{3}, \ldots$, are given, such that
$$
y_{n+1} / x_{n+1}>y_{n} / x_{n}
$$
for each $n \geq 1$. Prove that there are infinitely many values of $n$ such that $y_{n}>\sqrt{n}$. | [
"Solution:\n\nSuppose the statement is false. So there are only finitely many values for which $y_{n}>\\sqrt{n}$; suppose there are $R_{1}$ such values. Let $m$ be the largest integer such that $x_{n} / y_{n} \\geq m$ for all $n$ (this is possible since every $x_{n} / y_{n}>0$; note that it may be the case that $m=... | United States | 9th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
008a | There are only coins of $11$ pesos and $13$ pesos in a country (and no bills). An ice-cream shop is about to open and there is a line of customers waiting. Every customer wants to buy a cone of ice-cream and has exactly $155$ pesos. The cone costs $12$ pesos. The salesperson wants to attend everyone by giving back the ... | [
"Having $108$ pesos is sufficient. There is only one way to express $155$ as $11a + 13b$ with $a$, $b$ nonnegative integers: with $a = 7$ and $b = 6$. So each customer has $7$ coins of $11$ and $6$ coins of $13$. Hence to serve a customer as desired it is enough to have available either $6$ coins of $11$ or $5$ coi... | Argentina | Mathematical Olympiad Rioplatense | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | 108 | |
00pd | Let $ABC$ be a triangle and let $O$ be its circumcentre. The internal and external bisectrices of the angle $BAC$ meet the line $BC$ at points $D$ and $E$, respectively. Let further $M$ and $L$ respectively denote the midpoints of the segments $BC$ and $DE$. The circles $ABC$ and $ALO$ meet again at point $N$. Show tha... | [
"We must show that $AN$ is symmedian in the triangle $ABC$.\nLet $R$ denote the radius of the circle $ABC$ and notice that the cross-ratio $(EDBC) = -1$ to deduce that $LA^2 = LD^2 = LB \\cdot LC = LO^2 - R^2 = LO^2 - AO^2$, so $L$ and $O$ are antipodal in the circle $ALO$. It then follows that $N$ is the reflectio... | Balkan Mathematical Olympiad | shortlistBMO 2011 | [
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Triangles > Triangle c... | English | proof only | null | |
09mv | The water in the reservoir evaporates constantly, and water is added through 3 taps. If there were no evaporation, these 3 taps could fill the empty reservoir in 2 weeks. Considering evaporation, it takes six weeks to fill the empty reservoir using only taps 1 and 2. If tap 3 is turned on alone, $\frac{1}{3}$ of the to... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 5 weeks | |
0di1 | Each integer from $1$ to $n$ was colored either red, or blue, and each color was used at least once. It turned out that
* Every red number is a sum of some two distinct blue numbers;
* Every blue number is a difference of some two red numbers.
Find the smallest number $n$ for which such coloring is possible. | [
"The smallest such number is $n = 9$.\nSuppose that $n$ is a number allowing for coloring with the desired properties. Then $n \\ge 2$ since each color needs to be used at least once. Note that $n$ cannot be blue since it is inexpressible as a difference of two numbers between $1$ and $n$, therefore $n$ is red. On ... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Other"
] | English | proof and answer | 9 | |
0jz4 | Problem:
Define a sequence $\{a_{n}\}$ by $a_{1}=1$ and $a_{n}=(a_{n-1})!+1$ for every $n>1$. Find the least $n$ for which $a_{n}>10^{10}$. | [
"Solution:\n\nWe have $a_{2}=2$, $a_{3}=3$, $a_{4}=7$, $a_{5}=7!+1=5041$, and $a_{6}=5041!+1$. But\n$$\n5041!+1 \\gg 5041 \\cdot 5040 \\cdot 5039 > 10^{10}\n$$\nHence, the answer is 6."
] | United States | HMMT November | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 6 | |
05fx | Problem:
Dans un cercle $\mathcal{C}$ de centre $O$, on trace une corde $[CD]$. Soit $\Gamma$ le cercle de diamètre $[CD]$ et $O'$ son centre. La médiatrice de $[CD]$ coupe le cercle $\mathcal{C}$ en deux points $A$ et $B$ tels que $A$ est extérieur à $\Gamma$. Notons $T$ et $T'$ les points de contact des tangentes à ... | [
"Solution:\n\n\n\nComme les triangles $AFT$ et $ATO'$ sont semblables, on a $\\frac{AF}{AT} = \\frac{AT}{AO'}$. Or, $AT^2 = AO'^2 - O'T^2 = AO'^2 - O'C^2$ donc $AF = \\frac{AO'^2 - O'C^2}{AO'}$.\n\nComme les triangles $ACB$ et $AO'C$ sont semblables, on a $\\frac{AB}{AC} = \\frac{AC}{AO'}$.... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
02tt | Problem:
Sejam $D$ um ponto no lado $AB$ do triângulo $\triangle ABC$ e $F$ a interseção de $CD$ e da mediana $AM$. Se $AF = AD$, encontre a razão entre $BD$ e $FM$.
 | [
"Solution:\n\nSeja $E$ o ponto de interseção da reta paralela ao lado $AB$, que passa pelo ponto $M$, com o segmento $CD$. Como $ME \\parallel AB$, segue que $\\angle MED = \\angle EDA$. Além disso, como $\\angle AFD = \\angle EFM$ e o triângulo $\\triangle AFD$ é isósceles, podemos concluir que $\\triangle EFM$ ta... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2 | |
00bs | Find all natural $a$ such that $4x^2 + a$ is a prime for all $x = 0, 1, \dots, a-1$. | [
"Clearly $a$ must be a prime (set $x=0$). More exactly $a$ is an odd prime because $a=2$ is not a solution. Furthermore $a+1$ must be a power of 2. Indeed suppose that $a+1$ has an odd prime divisor $p$. Note that $p \\le \\frac{a+1}{2}$ as $a+1$ is even. Set $x = \\frac{1}{2}(p-1)$; it is clear that $0 < x < a$. W... | Argentina | Argentina_2018 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | a = 3 and a = 7 | |
03im | Problem:
Given a circle of radius $r$ and a tangent line $\ell$ to the circle through a given point $P$ on the circle. From a variable point $R$ on the circle, a perpendicular $RQ$ is drawn to $\ell$ with $Q$ on $\ell$. Determine the maximum of the area of triangle $PQR$. | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 3√3 r^2 / 8 | |
03ej | Given a natural number $n$. We have $n + 1$ balls numbered $1$, $1$, $2$, $3$, ..., $n$ (only the first two are the same). We need to color these balls in $n$ given colors so that every ball is a single color and every color is used at least once. We denote by $a_n$ the number of possible colorings. Find the smallest $... | [
"If $a > 1$, then we have $(n-1)(n-2)/2$ choices for $a$ and $b$, and $n$ choices for their color. The remaining $n-1$ balls (two of which are the same) must be colored in the remaining $n-1$ colors; for this we have $(n-1)!/2$ variants. Therefore, in this case the number of possible colorings is $\\frac{1}{4}n!(n-... | Bulgaria | 2 Bulgarian Winter Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | 13 | |
0b3q | Problem:
Let $PMO$ be a triangle with $PM = 2$ and $\angle PMO = 120^\circ$. Let $B$ be a point on $PO$ such that $PM$ is perpendicular to $MB$, and suppose that $PM = BO$. The product of the lengths of the sides of the triangle can be expressed in the form $a + b \sqrt[3]{c}$, where $a, b, c$ are positive integers, an... | [
"Solution:\nExtend $PM$ to a point $C$ such that $PC \\perp OC$. Since $\\angle PMO = 120^\\circ$, $\\angle CMO = 60^\\circ$ and $\\angle COM = 30^\\circ$. Let $PB = x$ and $MC = a$. Then $CO = a \\sqrt{3}$ and $OM = 2a$. Moreover, $\\triangle PMB$ and $\\triangle PCO$ are similar triangles. Thus, we have\n$$\n\\fr... | Philippines | 24th Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 28 | |
043b | Given integer $n \ge 2$. Find the least positive integer $m$, such that there are $n^2$ distinct positive real numbers $x_{i,j}$ ($1 \le i, j \le n$) satisfying the following conditions:
(1) For every $i, j$,
$$
x_{i,j} = \max\{x_{i,1}, x_{i,2}, \dots, x_{i,j}\} \text{ or}
$$
$$
x_{i,j} = \max\{x_{1,j}, x_{2,j}, \dots,... | [
"$m = \\lfloor \\frac{n+3}{2} \\rfloor$.\n\nPut these $n^2$ numbers $x_{i,j}$ in an $n \\times n$ table: call $x_{i,j}$ a \"row pivot\" if $x_{i,j} = \\max\\{x_{i,1}, x_{i,2}, \\dots, x_{i,j}\\}$; call $x_{i,j}$ a \"column pivot\", if $x_{i,j} = \\max\\{x_{1,j}, x_{2,j}, \\dots, x_{i,j}\\}$.\n\nFirst, prove $m \\ge... | China | China National Team Selection Test | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | floor((n+3)/2) | |
09rm | Problem:
Laat $a$, $b$ en $c$ positieve reële getallen zijn met $a b c=1$. Bewijs dat
$$
a+b+c \geq \sqrt{\frac{(a+2)(b+2)(c+2)}{3}}
$$ | [
"Solution:\n\nLaat $x$, $y$ en $z$ positieve reële getallen zijn zodat $x^{3}=a$, $y^{3}=b$ en $z^{3}=c$. Dan geldt $x^{3} y^{3} z^{3}=1$, dus ook $x y z=1$. We kunnen nu $a+2$ schrijven als $x^{3}+2 x y z = x(x^{2}+2 y z)$, en analoog kunnen we $b+2$ en $c+2$ herschrijven, waarmee we krijgen\n$$(a+2)(b+2)(c+2) = x... | Netherlands | MO-selectietoets | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0b3s | Problem:
Find the sum of all positive integers $n$, $1 \leq n \leq 5000$, for which
$$
n^{2}+2475 n+2454+(-1)^{n}
$$
is divisible by $2477$. (Note that $2477$ is a prime number.) | [
"Solution:\nThe problem is equivalent to finding all values of $n$ that satisfy the following congruence:\n\n$$\nn^{2}-2 n-23+(-1)^{n} \\equiv 0 \\pmod{2477}\n$$\n\nIf $n$ is odd, then $(-1)^{n} = -1$. Then, we have:\n\n$$\n\\begin{gathered}\nn^{2}-2 n-24 \\equiv 0 \\pmod{2477} \\\\\n(n-6)(n+4) \\equiv 0 \\pmod{247... | Philippines | 24th Philippine Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | final answer only | 9912 | |
0d6v | A circle with center $O$ passes through points $A$ and $C$ and intersects the sides $AB$ and $BC$ of triangle $ABC$ at points $K$ and $N$, respectively. The circumcircles of triangles $ABC$ and $KBN$ meet at distinct points $B$ and $M$. Prove that $\angle OMB = 90^{\circ}$. | [
"\nConsider an inversion with center $B$ and radius $r > 0$. The points $A'$, $B'$, $C'$ are collinear and the points $K'$, $N'$, $M'$ are, too.\nThe circumcircle of quadrilateral $ACNK$ becomes the circumcircle $\\omega$ of quadrilateral $A' C' N' K'$. In case $B$ is outside $\\omega$, let... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced ... | English | proof only | null |
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