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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0cbi | The positive real numbers $a$, $b$, $c$ are such that $a + b + c = 3$. Prove that the following inequality holds: $a^2 + b^2 + c^2 + a^2b + b^2c + c^2a \ge 6$. | [
"By adding $2ab + 2bc + 2ca$ to both sides, the inequality becomes:\n$$\n(a + b + c)^2 + a^2b + b^2c + c^2a \\ge 6 + 2ab + 2bc + 2ca.\n$$\nThus, we have to prove that $a^2b + b^2c + c^2a + 3 \\ge 2ab + 2bc + 2ca$. Since $a + b + c = 3$, the previous inequality is equivalent to:\n$$\n(b + a^2b) + (c + b^2c) + (a + c... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0h8t | There is isosceles obtuse triangle $ABC$ with vertex in point $B$ given. Perpendicular bisector to side $BC$ intersects lines $AC$ and $AB$ in points $K$ and $M$ respectively. Prove, that point, symmetric to point $A$ with respect to line $BK$, is on line $CM$.
(Anton Trigub) | [
"Let $A_1$ be point, symmetric to $A$ with respect to $BK$ (Fig. 37). At first, $\\angle BA_1K = \\angle BAK = \\angle BCK$, thus quadrilateral $BA_1CK$ is cyclic. Then,\n$$\n\\angle A_1CB = \\angle A_1KB = \\angle AKB = 2\\angle BCA = \\angle MBC = \\angle MCB,\n$$\nSo we get, that points $C$, $A_1$, $M$ are on on... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0iu5 | Problem:
Admiral Ackbar needs to send a 5-character message through hyperspace to the Rebels. Each character is a lowercase letter, and the same letter may appear more than once in a message. When the message is beamed through hyperspace, the characters come out in a random order. Ackbar chooses his message so that th... | [
"Solution:\n\nIf there is more than one distinct letter sent in the message, then there will be at most a $1/5$ chance of transmitting the right message. So the message must consist of one letter repeated five times, so there are 26 possible messages."
] | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 26 | |
0cyh | Prove that for any positive integer $n$ there is an equiangular hexagon whose side-lengths are $n+1, n+2, \ldots, n+6$ in some order. | [
"Assume that the equiangular hexagon has the side-lengths $a_{1}, a_{2}, \\ldots, a_{6}$. Since all angles of the hexagon are $120^{\\circ}$, extending its sides we get an equilateral triangle.\nIt is clear that\n$$\na_{1}+a_{2}+a_{6}=a_{2}+a_{3}+a_{4}=a_{4}+a_{5}+a_{6}\n$$\nthat is\n$$\na_{1}+a_{6}=a_{3}+a_{4} \\q... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0bqc | Problem:
Fie mulțimea $G = (-k, k)$, $k > 0$ și $x * y = \frac{k^{2}(x + y)}{k^{2} + x y}$, $x, y \in G$. Arătați că:
a) $(G, *)$ este grup abelian.
b) $\frac{k}{3} * \frac{k}{5} * \ldots * \frac{k}{2 n + 1} < \frac{k}{2} * \frac{k}{4} * \ldots * \frac{k}{2 n}$, $n \in \square^{*}$.
(Supliment G.M. nr. 10/2015) | [] | Romania | Olimpiada Natională de Matematică | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
09d3 | $n \times n$ хүснэгтийн нүд бүрийг өгөгдсөн 3 өнгийн аль нэгээр нь дурын аргаар будахад дор хаяж 3 нүд нь ижил өнгөөр будагдсан мөр эсвэл багана ямагт олддог байх $n$-ийн хамгийн бага утгыг ол. | [
"▶ Хариу: $n = 7$.\n\n$$\nn = 7 \\text{ үед } 7^2 = 49 = 3 \\cdot 16 + 1 \\Rightarrow \\text{дор хаяж } 17 \\text{ квадрат ижил} \\\\\n\\text{өнгөөр будагдана. } 17 = 7 \\cdot 2 + 3 \\Rightarrow 7 \\text{ мөрөөс нэг мөр нь}\n$$\n\nдөр хачж 3 ижил өнгийн нүд агуулна. $n = 6$ үед эсрэг жишээ.\n\n$$\nA \\begin{cases} ... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Mongolian | proof and answer | 7 | |
0b6k | All sides and diagonals of a convex $n$-gon, $n \ge 3$, are coloured one of two colours. Show that there exist $\lfloor (n+1)/3 \rfloor$ pairwise disjoint monochromatic segments. (Two segments are disjoint if they do not share an endpoint or an interior point.) | [
"If all sides are monochromatic, then the assertion is clearly true. Otherwise, delete a vertex incident with two sides of different colours together with its neighbours, delete all sides and diagonals incident with these three vertices and apply induction."
] | Romania | 2010 DANUBE MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
05iv | Problem:
Soient $x$, $y$ et $z$ des réels strictement positifs tels que
$$
x + y + z \geqslant \frac{1}{x} + \frac{1}{y} + \frac{1}{z}
$$
Montrer que
$$
\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \geqslant \frac{1}{x y} + \frac{1}{y z} + \frac{1}{z x}
$$ | [
"Solution:\nEn mettant au même dénominateur, l'inégalité recherchée se réécrit\n$$\nx^{2} z + y^{2} x + z^{2} y \\geqslant x + y + z\n$$\nMontrons celle-ci en utilisant successivement l'hypothèse et l'inégalité de Cauchy-Schwarz :\n$$\n\\begin{aligned}\n& x + y + z \\leqslant \\frac{(x + y + z)^{2}}{\\frac{1}{x} + ... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0hhn | Let's consider on the Cartesian plane all pairs of distinct points $(A, B)$, each of which has both integer coordinates. Among these pairs of points, find all those for which there exist two distinct points $(X, Y)$ with both integer coordinates, such that quadrilateral $AXBY$ is convex and inscribed.
A quadrilateral i... | [
"First, we will show that for points that are at a distance of $1$ from each other, there are no points $(X, Y)$ that satisfy the condition. Indeed, let us assume that such points exist. Then, $\\angle AXB + \\angle AYB = 180^\\circ$, which means that at least one of these angles is not less than $90^\\circ$. There... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterrals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Ge... | English | proof and answer | All pairs of distinct lattice points except those at unit distance apart; for distance 1 no such points exist, and for any other distance such points can be found. | |
0ix2 | Problem:
A cube has side length $1$. Find the product of the lengths of the diagonals of this cube (a diagonal is a line between two vertices that is not an edge). | [
"Solution:\n\nThere are $12$ diagonals that go along a face and $4$ that go through the center of the cube, so the answer is $\\sqrt{2}^{12} \\cdot \\sqrt{3}^{4} = 576$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Solid Geometry > 3D Shapes"
] | null | final answer only | 576 | |
015x | $M$ is the midpoint of the side $AC$ of triangle $ABC$, $L$ is a point on the segment $BC$. The line $LM$ intersects the ray $BA$ in the point $K$. $P$ is the point on the segment $BM$ such that $PM$ is a bisector of angle $LPK$. The line $\ell$ passes through $A$ and is parallel to $BM$. Prove that the projection of t... | [
"Let $X$ and $S$ be the intersection points of the line $\\ell$ and segments $PK$ and $MK$. Draw the line that passes through the point $C$ and is parallel to $BM$. Let $Y$ and $T$ be the intersection points of this line with rays $PL$ and $ML$.\n\nTriangles $AMS$ and $CMT$ are equal and symmetrical with respect to... | Baltic Way | Baltic Way SHL | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0a2k | Let $ABCD$ be a parallelogram with the property that $|AD| = |BD|$. Now let $P$ and $Q$ be points such that $\triangle ADP$ and $\triangle CDQ$ are equilateral and do not overlap with the parallelogram. Prove that $\angle PQD = 30^\circ$. | [
"There are a lot of line segments of the same length. For example, $|AD| = |DP| = |PA|$ because triangle $\\triangle ADP$ is equilateral. It is given that $|AD| = |BD|$, and finally $|AD| = |BC|$ because $ABCD$ is a parallelogram. Similarly, $|CD| = |DQ| = |QC| = |AB|$. The opposite angles in the parallelogram are ... | Netherlands | Dutch Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07rr | Find all pairs of functions $f, g : \mathbb{Q} \to \mathbb{Q}$ that satisfy the identity
$$
f(x+y) = f(x)g(1) + g(y), \quad \forall x, y \in \mathbb{Q}
$$
and the equation $f(1) = 3$. | [
"We will show that there are exactly two solutions to this functional equation. First, take $x = 0$ to get\n$$\nf(y) = f(0)g(1) + g(y),\n$$\nso $g = f + C$ for some constant $C$. Since $f(1) = 3$, we deduce that\n$$\nf(x+y) = (3+C)f(x) + f(y) + C, \\quad x,y \\in \\mathbb{Q}. \\qquad (15)\n$$\nThe equations $x + y ... | Ireland | Irish | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | Two solutions: (1) f is identically three and g is identically three quarters; (2) f equals the input plus two and g equals the input, for all rational inputs. | |
03dx | There are $n$ towns in a country. Some of the towns are connected with one way roads and between any two towns it is possible to have several roads in both directions. It is known that for any two towns $A$ and $B$ one can travel from $A$ to $B$, or from $B$ to $A$ or both. Find the minimum number of roads that have to... | [
"We prove that the answer is $1$. Consider towns numbered from $1$ to $n$ and let from town $i$ there is a road to town $i+1$, $\\forall i < n$. Obviously the condition is fulfilled and we need at least one new road, thus the answer is greater than $0$.\n\nWe show that there is a town $F$ from which one can travel ... | Bulgaria | Bulgaria 2022 | [
"Discrete Mathematics > Graph Theory"
] | null | proof and answer | 1 | |
06as | Find the number of rectangles satisfying the following properties:
(α) Their vertices are points $(x, y)$ of the plane $Oxy$, with $x, y$ non-negative integers and $x \le 8$, $y \le 8$.
(β) Their sides are parallel to axis
(γ) Their area $E$ satisfies: $30 < E \le 40$. | [
"First we examine which values of the area of rectangles are acceptable:\nSince, $0 < x, y \\le 8$, the integer $40$ is written only as $40 = 5 \\cdot 8$. Since a $5 \\times 8$ rectangle can be put in the $8 \\times 8$ rectangle with $4$ ways horizontally and with $4$ ways vertically we have totally $8$ such rectan... | Greece | 40th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Other",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | final answer only | 43 | |
0fmp | Determine the smallest integer $k$ for which the following story could hold true: In a chess tournament with 24 players, every pair of players plays at least two and at most $k$ games against each other. In the end of the tournament, it turns out that every player has played a different number of games. | [
"The answer is $k = 4$. If $k = 3$ was possible, then every player plays either 2 or 3 games against each of the other 23 players. Hence he plays at least $2 \\cdot 23 = 46$ and at most $3 \\cdot 23 = 69$ games. It is impossible that there is a player $A$ who has played 46 games (and hence 2 games against every oth... | Spain | Mediterranean Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | Spanish | proof and answer | 4 | |
06vt | Let $a$ be a positive integer. We say that a positive integer $b$ is $a$-good if $\binom{a n}{b}-1$ is divisible by $a n+1$ for all positive integers $n$ with $a n \geqslant b$. Suppose $b$ is a positive integer such that $b$ is $a$-good, but $b+2$ is not $a$-good. Prove that $b+1$ is prime. | [
"We first show that $b$ is $a$-good if and only if $b$ is even, and $p \\mid a$ for all primes $p \\leqslant b$.\nTo start with, the condition that $a n+1 \\left\\lvert\\,\\binom{ a n}{b}-1\\right.$ can be rewritten as saying that\n$$\n\\frac{a n(a n-1) \\cdots(a n-b+1)}{b!} \\equiv 1 \\quad(\\bmod a n+1) .\n$$\nSu... | IMO | IMO 2019 Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0ikg | Problem:
For how many ordered triplets $(a, b, c)$ of positive integers less than $10$ is the product $a \times b \times c$ divisible by $20$? | [
"Solution:\n\nOne number must be $5$. The other two must have a product divisible by $4$. Either both are even, or one is divisible by $4$ and the other is odd. In the former case, there are $48 = 3 \\times 4 \\times 4$ possibilities: $3$ positions for the $5$, and any of $4$ even numbers to fill the other two. In ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | 102 | |
05lr | Problem:
Trouver toutes les fonctions $f: \mathbb{N} \rightarrow \mathbb{N}$ telles que $2 m n + m f(m) + n f(n)$ est un carré parfait pour tous entiers positifs $m$ et $n$. | [
"Solution:\n\nPour simplifier, posons $F(m, n) = 2 m n + m f(m) + n f(n)$. Alors $F(m, 0) = m f(m)$ est un carré pour tout $m \\geqslant 1$.\n\nOn peut donc écrire $f(p) = p a^{2}$ pour un nombre premier $p$. Supposons que $a \\geqslant 2$. Alors $(a p)^{2} + 2 p + f(1) = F(p, 1) > (a p^{2})$ est un carré, donc $(a... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | f(n) = n for all n | |
0jra | Problem:
Let $S$ be the set of discs $D$ contained completely in the set $\{(x, y): y<0\}$ (the region below the $x$-axis) and centered (at some point) on the curve $y=x^{2}-\frac{3}{4}$. What is the area of the union of the elements of $S$? | [
"Solution:\n\nAnswer: $\\frac{2 \\pi}{3}+\\frac{\\sqrt{3}}{4}$\n\nSolution 1. An arbitrary point $\\left(x_{0}, y_{0}\\right)$ is contained in $S$ if and only if there exists some $(x, y)$ on the curve $\\left(x, x^{2}-\\frac{3}{4}\\right)$ such that $\\left(x-x_{0}\\right)^{2}+\\left(y-y_{0}\\right)^{2}<y^{2}$, si... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 2π/3 + √3/4 | |
0bl9 | Let $n \ge 2$ be a positive integer. Find all real $x > 0$ such that
$$
\left[\frac{1}{x}\right] + \left[\frac{1}{x^n}\right] = n.
$$ | [] | Romania | SHORTLISTED PROBLEMS FOR THE 66th NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | x ∈ (n^{-1/n}, (n-1)^{-1/n}] | |
0hbv | Consider the $m \times n$ table, $m, n \ge 2$ ($m$ rows are enumerated $1, 2, \ldots, m$ and $n$ columns are enumerated $1, 2, \ldots, n$), which is filled with positive integers. Let $b_i$ be the $lcm$ (least common multiple) of all numbers in the $i^{th}$ row, $1 \le i \le m$, and let $B$ be the $gcd$ (greatest commo... | [
"**Answer:** $B$ is divisible by $C$.\n\nConsider any prime number $p$, and its power for each number in the table. Replace all the numbers in the table with a power of the chosen prime number. Let the $m \\times n$ table is filled with $\\alpha_{i,j}$, $i=1, \\overline{m}$, $j=1, \\overline{n}$. Now $\\beta_i$ is ... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | B is divisible by C | |
0bsm | Let $a \in \mathbb{R}$ and $f : (0, \infty) \to (0, \infty)$. Prove that the following two statements are equivalent:
(i) $\lim_{x \to \infty} \frac{f(x)}{x^{a+\varepsilon}} = 0$ and $\lim_{x \to \infty} \frac{f(x)}{x^{a-\varepsilon}} = \infty$, for all $\varepsilon > 0$;
(ii) $\lim_{x \to \infty} \frac{\ln f(x)}{\ln x... | [
"(i)⇒(ii). Let $\\varepsilon > 0$; according to (i), there exists $m_1 > 0$ such that $\\frac{f(x)}{x^{a+\\varepsilon}} < 1, \\forall x > m_1$ and $m_2 > 0$ such that $\\frac{f(x)}{x^{a-\\varepsilon}} > 1, \\forall x > m_2$. If we denote $m = \\max\\{m_1, m_2, 1\\}$, then $x^{a-\\varepsilon} < f(x) < x^{a+\\varepsi... | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions"
] | English | proof only | null | |
0dso | Suppose $f: N \to N$ is a function such that
$$
f^n(n) = 2n
$$
for all $n \in N$. Must $f(n) = n + 1$ for all $n$? | [] | Singapore | Singapore International Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | Yes | |
0jpm | Problem:
Let $n$ be the smallest positive integer such that any positive integer can be expressed as the sum of $n$ integer $2015$th powers. Find $n$. If your answer is $a$, your score will be $\max \left(20-\frac{1}{5}\left|\log _{10} \frac{a}{n}\right|, 0\right)$, rounded up. | [
"Solution:\n$2^{2015}+\\left\\lfloor\\left(\\frac{3}{2}\\right)^{2015}\\right\\rfloor-2$\nIn general, if $k \\leq 471600000$, then any integer can be expressed as the sum of $2^{k}+\\left\\lfloor\\left(\\frac{3}{2}\\right)^{k}\\right\\rfloor-2$ integer $k$th powers. This bound is optimal.\nThe problem asking for th... | United States | HMMT November 2015 | [
"Number Theory > Other"
] | null | final answer only | 2^{2015} + floor((3/2)^{2015}) - 2 | |
03nc | Problem:
Let $k$ be a given even positive integer. Sarah first picks a positive integer $N$ greater than $1$ and proceeds to alter it as follows: every minute, she chooses a prime divisor $p$ of the current value of $N$, and multiplies the current $N$ by $p^{k}-p^{-1}$ to produce the next value of $N$. Prove that there... | [
"Solution:\nNote that $1009$ is prime. We will show that if $k=1009^{m}-1$ for some positive integer $m$, then Sarah's number must at some point be divisible by $2018$. Let $P$ be the largest divisor of $N$ not divisible by a prime congruent to $1$ modulo $1009$. Assume for contradiction that $N$ is never divisible... | Canada | Canadian Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0j5p | Problem:
Collinear points $A$, $B$, and $C$ are given in the Cartesian plane such that $A=(a, 0)$ lies along the $x$-axis, $B$ lies along the line $y=x$, $C$ lies along the line $y=2x$, and $AB / BC = 2$. If $D=(a, a)$, the circumcircle of triangle $ADC$ intersects $y=x$ again at $E$, and ray $AE$ intersects $y=2x$ at... | [
"Solution:\n\n\n\nLet points $O$, $P$, and $Q$ be located at $(0,0)$, $(a, 2a)$, and $(0,2a)$, respectively. Note that $BC / AB = 1/2$ implies $[OCD]/[OAD] = 1/2$, so since $[OPD] = [OAD]$, $[OCD]/[OPD] = 1/2$. It follows that $[OCD] = [OPD]$. Hence $OC = CP$. We may conclude that triangles... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem"
] | null | proof and answer | 7 | |
0gfu | 設 $ABC$ 是不等邊的銳角三角形，其內心為 $I$，外接圓為 $\Gamma$。直線 $AI$ 與 $\Gamma$ 再交於點 $M$。令 $N$ 為 $BC$ 的中點，而 $T$ 為 $\Gamma$ 上滿足 $IN \perp MT$ 的一點。設 $\ell$ 為通過 $I$ 且與 $AI$ 垂直的直線。令 $\ell$ 分別與直線 $TB, TC$ 交於點 $P, Q$。證明 $PB = CQ$。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 國際競賽實作(一) | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Advanced Config... | Chinese; English | proof only | null | |
0e1a | The diagonal $BD$ of the quadrilateral $ABCD$ divides this quadrilateral into an acute triangle $ABD$ and an equilateral triangle $BCD$. Let $O$ be the orthocentre of the triangle $ABD$. Prove:
a. if the triangles $ABD$ and $OCD$ are congruent, then $AB \perp BC$;
b. if $\angle CBA = 90^\circ$, then the triangles $AB... | [
"a. The points $C$ and $O$ lie on the bisector of the segment $BD$, so $\\angle DCO = 30^\\circ$. Since the triangles $ABD$ and $OCD$ are congruent, we have $\\angle DBA = \\angle DCO = 30^\\circ$. This implies $\\angle CBA = \\angle CBD + \\angle DBA = 90^\\circ$.\n\n\n\nb. If $\\angle ABC... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0k3y | Problem:
Randall proposes a new temperature system called Felsius temperature with the following conversion between Felsius $\{ \}^{\circ} E$, Celsius $\{ \}^{\circ} C$, and Fahrenheit $\{ \}^{\circ} F$:
$$
\{ \}^{\circ} E = \frac{7 \times \{ \}^{\circ} C}{5} + 16 = \frac{7 \times \{ \}^{\circ} F - 80}{9}.
$$
For exam... | [
"Solution:\n\nNotice that $(5 k)^{\\circ} C = (7 k + 16)^{\\circ} E = (9 k + 32)^{\\circ} F$, so Felsius is an exact average of Celsius and Fahrenheit at the same temperature. Therefore we conclude that $x = y = z$, and it is not difficult to compute that they are all equal to $-40$."
] | United States | HMMT February 2018 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -120 | |
07tu | Suppose $\triangle ABC$ is a triangle inscribed in the unit circle. Prove that its area doesn't exceed $3\sqrt{3}/4$, and its perimeter doesn't exceed $3\sqrt{3}$, with equality in both cases iff the triangle is equilateral. | [
"Let $O$ denote the centre of the circle, and let $\\alpha, \\beta, \\gamma$, respectively, be the radian measures of the vertex angles $\\angle A, \\angle B, \\angle C$. Since the circumradius of $ABC$ is $1$, the formula for the circumradius of any triangle tells us that\n$$\n\\frac{a}{\\sin \\alpha} = \\frac{b}{... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Jensen/smoothing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof only | null | |
0ild | Let $n$ be a positive integer. Define a sequence by setting $a_1 = n$ and, for each $k > 1$, letting $a_k$ be the unique integer in the range $0 \le a_k \le k - 1$ for which $a_1 + a_2 + \cdots + a_k$ is divisible by $k$. For instance, when $n = 9$ the obtained sequence is $9, 1, 2, 0, 3, 3, 3, \dots$. Prove that for a... | [
"For $k \\ge 1$, let\n$$\ns_k = a_1 + a_2 + \\cdots + a_k.\n$$\nWe have\n$$\n\\frac{s_{k+1}}{k+1} < \\frac{s_{k+1}}{k} = \\frac{s_k + a_{k+1}}{k} \\le \\frac{s_k + k}{k} = \\frac{s_k}{k} + 1.\n$$\nOn the other hand, for each $k$, $s_k/k$ is a positive integer. Therefore\n$$\n\\frac{s_{k+1}}{k+1} \\le \\frac{s_k}{k}... | United States | USAMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Numb... | null | proof only | null | |
07a3 | Find the smallest positive integer $n$ such that there exist $n$ real numbers in the interval $(-1,1)$ such that their sum is zero and the sum of their squares equals $20$. | [
"Suppose that $a_1, a_2, \\dots, a_n$ satisfies the conditions. First, we have\n$$\n20 = a_1^2 + a_2^2 + \\dots + a_n^2 < \\underbrace{1 + 1 + \\dots + 1}_{n} = n.\n$$\nSo $21 \\le n$. We want to show that $n = 22$ is the answer. So we prove that there are not $21$ numbers $a_1, a_2, \\dots, a_{21}$ in the interval... | Iran | Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 22 | |
0hyb | Problem:
Evaluate $\sum_{n=1}^{\infty} \frac{1}{n \cdot 2^{n-1}}$. | [
"Solution:\nNote that if we take the integral of $f(x)$ in problem 4, we get the function $F(x) = x + \\frac{x^{2}}{2 \\cdot 2} + \\frac{x^{3}}{3 \\cdot 2^{2}} + \\ldots$. Evaluating this integral in the interval $[0,1]$, we get $1 + \\frac{1}{2 \\cdot 2} + \\frac{1}{3 \\cdot 2^{2}} + \\ldots$, which is the desired... | United States | HMMT | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2 ln 2 | |
0js7 | Problem:
The incircle of a triangle $ABC$ is tangent to $BC$ at $D$. Let $H$ and $\Gamma$ denote the orthocenter and circumcircle of $\triangle ABC$. The $B$-mixtilinear incircle, centered at $O_{B}$, is tangent to lines $BA$ and $BC$ and internally tangent to $\Gamma$. The $C$-mixtilinear incircle, centered at $O_{C}... | [
"Solution:\n\nLet the $B$-mixtilinear incircle $\\omega_{B}$ touch $\\Gamma$ at $T_{B}$, $BA$ at $B_{1}$ and $BC$ at $B_{2}$. Define $T_{C} \\in \\Gamma$, $C_{1} \\in CB$, $C_{2} \\in CA$, and $\\omega_{C}$ similarly. Call $I$ the incenter of triangle $ABC$, and $\\gamma$ the incircle.\n\nWe first identify two poin... | United States | HMMT February 2016 | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle t... | null | proof and answer | sqrt((7 + 2*sqrt(13))/3) | |
0lfy | Problem:
Each point with integral coordinates in the plane is coloured white or blue. Prove that one can choose a colour so that for every positive integer $n$ there exists a triangle of area $n$ having its vertices of the chosen colour. | [
"Solution:\n\nIf there exists some $c$-monochromatic horizontal row $y = k$, then if both rows $y = k - 1$ and $y = k + 1$ are $\\bar\\{c\\}$-monochromatic we can find $\\bar\\{c\\}$-monochromatic triangles of any positive integer area, otherwise they must contain at least a $c$-point, and we can find $c$-monochrom... | Zhautykov Olympiad | XI International Zhautykov Olympiad in Sciences | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fkd | Problem:
Probar que para todo entero positivo $n$, $n^{19}-n^{7}$ es divisible por $30$. | [
"Solution:\n$n^{19}-n^{7}=n^{7}\\left(n^{12}-1\\right)=n^{7}\\left(n^{6}+1\\right)\\left(n^{6}-1\\right)=n^{7}\\left(n^{6}+1\\right)\\left(n^{3}+1\\right)\\left(n^{3}-1\\right)$, con lo que en la descomposición de $n^{19}-n^{7}$ aparecen tres números consecutivos, $n-1$, $n$, $n+1$, de los cuales al menos uno es di... | Spain | XLV Olimpiada Matemática Española | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0g6i | 有65對情侶出去玩,每一位男生都有一輛機車,並且都得要負責載一位女生。假設他們能夠安排出一種載法,使得對於任兩輛機車,下面兩命題恰一成立:
(i) 這兩輛機車上的男生互相認識彼此;
(ii) 這兩輛機車上的女生的男朋友互相認識彼此。
試證明:一定可以找到一對情侶,把他們剔除後,剩下的64對情侶仍能夠安排出一個滿足上述條件的載法。 | [
"假設這65對情侶已經選擇了一種符合題目條件的載法,我們證明:一定有一對情侶在同一輛車上 (事實上必恰只有一對),因此把他們剔除後,剩下的64對情侶可以沿用原本的載法,這樣顯然能夠滿足題目條件。\n\n將所有情侶編號1至65,並且定義函數 $f(a) = b$ 表示第 $a$ 號女生被第 $b$ 號男生載。則 $f(a)$ 是個一對一映成的函數,因此若是對所有 $1 \\le a \\le 65$,將 $a$ 不停代入 $f$ 直到其值變回 $a$,並把過程寫成一個環狀,就能得到 $a \\rightarrow f(a) \\rightarrow f(f(a)) \\cdots \\rightarrow f^{(k)}(a)... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Abstract Algebra > Permutations / basic group theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof only | null | |
0b6x | Let $S$ be a subset with 673 elements of the set $\{1, 2, \dots, 2010\}$. Prove that one can find two distinct elements of $S$, say $a$ and $b$ such that $6$ divides $a + b$. | [
"Consider the following sets, each containing 335 elements\n$$\n\\begin{align*}\nA &= \\{6, 12, \\dots, 2010\\}, & B &= \\{3, 9, 15, \\dots, 2007\\}, \\\\\nC &= \\{1, 7, 13, \\dots, 2005\\}, & D &= \\{2, 8, 14, \\dots, 2006\\}, \\\\\nE &= \\{4, 10, 16, \\dots, 2008\\}, & F &= \\{5, 11, 17, \\dots, 2009\\}.\n\\end{a... | Romania | Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
057k | Find all prime numbers $p$ such that $\frac{p-1}{2}$ and $\frac{p+1}{4}$ are prime numbers, too. | [
"Let $q = \\frac{p-1}{2}$ and $r = \\frac{p+1}{4}$; then $p = 4r-1$ and $q = \\frac{4r-2}{2} = 2r-1$. Consider all remainders that can be left when $r$ is divided by $3$:\n* If $r \\equiv 1 \\pmod{3}$ then $4r-1 \\equiv 0 \\pmod{3}$, i.e., $4r-1$ is divisible by $3$. Thus $p=3$. But then $r=1$ which is not a prime.... | Estonia | Open Contests | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 7 and 11 | |
0287 | Problem:
a) Num triângulo de lados $a$, $b$ e $c$, vale sempre que a soma de dois lados é maior do que o terceiro lado. Por exemplo, no triângulo a seguir, de lados $a$, $b$ e $c$,
vale a desigualdade $a < b + c$. Além disso, valem outras duas desigualdades. Quais são?
b) Na figura abaixo pod... | [
"Solution:\na) As outras desigualdades são $b < a + c$ e $c < a + b$.\n\nb) Conforme a sugestão, desenhamos um retângulo idêntico em cima do retângulo original, com o segmento de comprimento $b$ refletido:\n\nEm seguida, traçamos o segmento tracejado abaixo:\n\nApli... | Brazil | null | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometr... | null | proof and answer | b < a + c and c < a + b; a + b > 10; and the quadrilateral’s perimeter is at least twice the square’s diagonal. | |
07n9 | In the mathematical talent show called "The $X^2$-factor", contestants are scored by a panel of 8 judges. Each judge awards a score of 0 ('fail'), $X$ ('pass'), or $X^2$ ('pass with distinction'). Three of the contestants were Ann, Barbara and David. Ann was awarded the same score as Barbara by exactly 4 judges. David ... | [
"**First Solution:** Represent each \"score sheet\" by a 8-digit ternary string with digits from $\\{0, 1, 2\\}$. Without loss of generality we may assume that Ann's score sheet reads 00001111, and that Barbara's score sheet reads 00002222. The total number of possible score sheets is $3^8$. We will count the numbe... | Ireland | Ireland | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | English | proof and answer | 5505 | |
0735 | Let $w_a, w_b, w_c$ be the lengths of the internal angle bisectors of a triangle $ABC$ with sides $a, b, c$.
Let $R$ be its circum-radius. Prove that
$$
\frac{b^2 + c^2}{w_a} + \frac{c^2 + a^2}{w_b} + \frac{a^2 + b^2}{w_c} > 4R.
$$ | [
"We use the standard\n$$\nw_a = \\frac{2bc \\cos(A/2)}{(b+c)}, \\text{ etc.}\n$$\n\nThe inequality takes the form\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c)}{4Rbc \\cos(A/2)} > 2.\n$$\nThis may be put in the form\n$$\n\\sum_{\\text{cyclic}} \\frac{(b^2 + c^2)(b+c) \\sin(A/2)}{2abc} > 1.\n$$\nBut note that ... | India | Indija TS 2007 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > ... | null | proof only | null | |
07fv | Given an inscribed pentagon $ABCDE$ with circumcircle $\Gamma$. Line $\ell$ passes through vertex $A$ and is tangent to $\Gamma$. Points $X, Y$ lie on $\ell$ so that $A$ lies between $X$ and $Y$. Circumcircle of triangle $\triangle XED$ intersects segment $AD$ at $Q$ and circumcircle of triangle $\triangle YBC$ interse... | [
"Assume the circumcircles of $\\triangle ABY$ and $\\triangle AEX$ meet for the second time at $K$. Since\n$$\n\\angle KEX = \\angle KAX = 180^\\circ - \\angle KAY = 180^\\circ - \\angle KBY = \\angle KBS,\n$$\nwe find out that $KBES$ is concyclic. We have\n$$\n\\angle YKB = \\angle YAB = \\angle BCA = \\angle PYB,... | Iran | 37th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0425 | For each positive integer $n$ and each integer $i$ ($0 \leq i \leq n$), let $C_n^i \equiv c(n,i) \pmod{2}$, where $c(n,i) \in \{0, 1\}$, and define
$$
f(n,q) = \sum_{i=0}^{n} c(n,i)q^i.
$$
Let $m$, $n$ and $q$ be positive integers with $q+1$ not a power of $2$. Suppose that $f(m,q) \mid f(n,q)$. Prove that
$f(m,r) \mid... | [
"For each positive integer $n$, we write $n$ in binary representation as $n = 2^{a_1} + 2^{a_2} + \\dots + 2^{a_k}$, where $0 \\le a_1 < a_2 < \\dots < a_k$. Define a set $T(n) = \\{2^{a_1}, \\dots, 2^{a_k}\\}$, $T(0)$ is considered empty set.\nBy Lucas' theorem, $C_n^i$ is odd if and only if $T(i) \\le T(n)$, henc... | China | China Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
0grc | Find all pairs of positive integers $(m, n)$ satisfying
$$
m^2 + n^2 = 2018(m - n).
$$ | [
"Using the condition given, we have $(m+n)^2 + (2018-m+n)^2 = 2018^2$. Let $m+n = u$, $2018-m+n = v$. It can be seen that $u,v > 0$. Since $u^2 + v^2 = 2018^2 \\equiv 0 \\pmod 4$, we get $u \\equiv v \\equiv 0 \\pmod 2$. Let $u = 2u_1$, $v = 2v_1$. Then we have $u_1^2 + v_1^2 = 1009^2$. Since $(u_1, v_1, 1009)$ is ... | Turkey | Team Selection Test | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | [(728, 390), (1290, 390)] | |
032m | Problem:
Prove that any graph with $10$ vertices and $26$ edges contains at least $4$ triangles. | [
"Solution:\nDenote by $V$ and $E$ the sets of the vertices and the edges of $G$, respectively. For any vertex $x \\in V$, let $\\Gamma(x)$ be the set of the edges of $G$ which are adjacent to $x$ and let $d(x) = |\\Gamma(x)|$. Then for $x, y \\in V$ one has that\n$$\n|\\Gamma(x) \\cap \\Gamma(y)| = |\\Gamma(x)| + |... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
00kn | Let $f: \mathbb{Z}_{>0} \to \mathbb{Z}$ be a function with the following properties:
(i) $f(1) = 0$,
(ii) $f(p) = 1$ for all prime numbers $p$,
(iii) $f(xy) = y f(x) + x f(y)$ for all $x, y$ in $\mathbb{Z}_{>0}$.
Determine the smallest integer $n \ge 2015$ that satisfies $f(n) = n$. | [
"1. We claim that\n$$\nf(q_1 \\cdots q_s) = q_1 \\cdots q_s \\left( \\frac{1}{q_1} + \\cdots + \\frac{1}{q_s} \\right)\n$$\nholds for (not necessarily distinct) prime numbers $q_1, \\dots, q_s$.\nWe prove the claim by induction on $s$. For $s=0$, the claim reduces to $f(1) = 0$, which is true by assumption.\nIf (4)... | Austria | Austrian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 3125 | |
0i4d | Problem:
The mathematician John is having trouble remembering his girlfriend Alicia's 7-digit phone number. He remembers that the first four digits consist of one $1$, one $2$, and two $3$'s. He also remembers that the fifth digit is either a $4$ or $5$. While he has no memory of the sixth digit, he remembers that the... | [
"Solution:\n\nThere are $\\frac{4!}{2!} = 12$ possibilities for the first four digits. There are two possibilities for the fifth digit. There are $10$ possibilities for the sixth digit, and this uniquely determines the seventh digit. So he has to dial $12 \\cdot 2 \\cdot 10 = 240$ numbers."
] | United States | Harvard-MIT Math Tournament | [
"Statistics > Probability > Counting Methods > Permutations",
"Math Word Problems"
] | null | final answer only | 240 | |
01ol | Some businessmen decide to establish a firm and divide all profits in equal parts. However, some day after a good pennyworth, a head of the firm transfers a part of the funds from firm's account on his individual account. This part is three times as many as the part of each of the others if they divided the rest of the... | [
"Answer: 20.\nLet $n$ be the number of co-funders and $d_i$ be the value of the $i$-th director, $i = 1, \\ldots, n$. By condition,\n$$\nd_i = 3 \\cdot \\frac{d_{i+1} + d_{i+2} + \\dots + d_n}{n-i}.\n$$\nSo,\n$$\n\\begin{align*}\nd_{i-1} &= 3 \\cdot \\frac{d_i + d_{i+1} + \\dots + d_n}{n-i+1} = \\\\\n&= 3 \\cdot \\... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 20 | |
0kho | Recall that the conjugate of the complex number $w = a + bi$, where $a$ and $b$ are real numbers and $i = \sqrt{-1}$, is the complex number $\bar{w} = a - bi$. For any complex number $z$, let $f(z) = 4i\bar{z}$. The polynomial $P(z) = z^4 + 4z^3 + 3z^2 + 2z + 1$ has four complex roots: $z_1, z_2, z_3$, and $z_4$. Let $... | [] | United States | AMC 12 A | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | 208 | |
08k0 | Problem:
Let $a$, $b$, $p$, $q$ be positive integers such that $a$ and $b$ are relatively prime, $ab$ is even and $p, q \geq 3$. Prove that
$$
2 a^{p} b - 2 a b^{q}
$$
cannot be a square of an integer number. | [
"Solution:\nWithout loss of generality, assume that $a$ is even and consequently $b$ is odd. Let $a = 2 a'$. Then\n$$\n2 a^{p} b - 2 a b^{q} = 4 a' b \\left(a^{p-1} - b^{q-1}\\right)\n$$\nIf this is a square, then $a'$, $b$ and $a^{p-1} - b^{q-1}$ are pairwise coprime.\n\nOn the other hand, $a^{p-1}$ is divisible b... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
03ck | Prove that $\left(\frac{6}{5}\right)^{\sqrt{3}} > \left(\frac{5}{4}\right)^{\sqrt{2}}$. | [
"We will prove that if $x > -1$, $x \\neq 0$, and $\\alpha \\in (1, 2)$, then\n$$\n(1) \\quad 0 < f(x) = (1+x)^\\alpha - 1 - \\alpha x - \\frac{\\alpha(\\alpha-1)}{2}x^2 - \\frac{\\alpha(\\alpha-1)(\\alpha-2)}{6}x^3.\n$$\nWe have\n$$\nf'(x) = \\alpha\\left[(1+x)^{\\alpha-1} - 1 - (\\alpha-1)x - \\frac{(\\alpha-1)(\... | Bulgaria | Bulgarian National Mathematical Olympiad | [
"Calculus > Differential Calculus > Applications"
] | English | proof only | null | |
00nx | Let $ABC$ be a triangle, and $O$ its circumcenter. The circumcircle of triangle $AOC$ shall intersect the segment $BC$ in points $C$ and $D$ and the segment $AB$ in points $A$ and $E$.
Prove that triangles $BDE$ and $AOC$ have equal circumradii. | [
"In the circumcircle of triangle $ABC$ we have $\\angle COA = 2\\angle CBA$. In the circumcircle of $ADC$ we therefore have $\\angle CDA = \\angle COA = 2\\angle CBA$. The angle $\\angle CDA$ is an external angle in triangle $ABD$, and we therefore obtain $\\angle CBA + \\angle BAD = \\angle CDA = 2\\angle CBA$, an... | Austria | AUT_ABooklet_2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0fby | Problem:
Un campeonato de baloncesto se ha jugado por sistema de liga a dos vueltas (cada par de equipos se enfrentan dos veces) y sin empate (si el partido acaba en empate hay prórrogas hasta que gane uno de los dos). El ganador del partido obtiene 2 puntos y el perdedor 1 punto. Al final del campeonato, la suma de d... | [
"Solution:\n\nSupongamos que el número de equipos es $n$. Entonces, se juegan un total de $2\\left(\\begin{array}{l}n \\\\ 2\\end{array}\\right)=n^{2}-n$ partidos en el campeonato por ser a doble vuelta. En cada partido se dan 3 puntos, por lo que $3 n^{2}-3 n$ es el número total de puntos dados. Si el campeón tien... | Spain | Problemas Primera Sesión | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 39 | |
0fqq | Problem:
Se consideran todos los pares $(x, y)$ de números reales tales que $0 \leq x \leq y \leq 1$. Sea $M(x, y)$ el máximo valor del conjunto
$$
A=\{x y, x y-x-y+1, x+y-2 x y\}
$$
Hallar el mínimo valor que puede tomar $M(x, y)$ para todos estos pares $(x, y)$. | [
"Solution:\n\nHaciendo el cambio de variable $x y=p$, y $x+y=s$ y escribiendo los tres elementos del conjunto $A$ en términos de $s$ y $p$, tenemos\n$$\na=x y=p, \\quad b=x y-x-y+1=(1-x)(1-y)=s-1+p, \\quad c=s-2 p\n$$\nverificándose que $a+b+c=1$. Observemos que $s^{2}-4 p=(x-y)^{2} \\geq 0$.\nAhora consideremos lo... | Spain | LV Olimpiada matemática Española | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4/9 | |
0k1t | Problem:
Suppose there are 100 cookies arranged in a circle, and 53 of them are chocolate chip, with the remainder being oatmeal. Pearl wants to choose a contiguous subsegment of exactly 67 cookies and wants this subsegment to have exactly $k$ chocolate chip cookies. Find the sum of the $k$ for which Pearl is guarante... | [
"Solution:\n\nWe claim that the only values of $k$ are 35 and 36.\n\nWLOG assume that the cookies are labelled 0 through 99 around the circle. Consider the following arrangement: cookies 0 through 17, 34 through 50, and 67 through 84 are chocolate chip, and the remaining are oatmeal. (The cookies form six alternati... | United States | HMMT February 2018 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 71 | |
055u | A quadratic equation $x^2 + px + q = 0$ is written on the blackboard, whereby $p$ and $q$ are real numbers such that real solutions exist to the equation on the blackboard and all the solutions are positive. Two players change in turns the coefficients in the equation according to the following rules. The first player ... | [
"Suppose that the first player always decreases the constant term by the smaller solution. We show that this is a winning strategy. Assume the opposite, i.e., that the play lasts infinitely. As the combined effect of the first and the second player's moves, the coefficient at the linear term either increases by 1 o... | Estonia | Estonian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | English | proof only | null | |
04mz | Determine all pairs $(x, y)$ of real numbers such that
$$
\begin{aligned}
x^2 + x y - 4 y^2 &= -1 \\
4 x^2 + x y - 11 y^2 &= -2.
\end{aligned}
$$ | [] | Croatia | Croatia_2018 | [
"Algebra > Intermediate Algebra > Other"
] | English | proof and answer | (3, 2), (-3, -2), (-1/2, 1/2), (1/2, -1/2) | |
0j6r | Let $p$ be a prime. We say that a sequence of integers $\{z_n\}_{n=0}^{\infty}$ is a $p$-pod if for each $e \ge 0$, there is an $N \ge 0$ such that whenever $m \ge N$, $p^e$ divides the sum
$$
\sum_{k=0}^{m} (-1)^k \binom{m}{k} z_k.
$$
Prove that if both sequences $\{x_n\}_{n=0}^{\infty}$ and $\{y_n\}_{n=0}^{\infty}$ ... | [
"Let\n$$\nX_n = \\sum_{i=0}^{n} (-1)^i \\binom{n}{i} x_i \\quad \\text{and} \\quad Y_n = \\sum_{i=0}^{n} (-1)^i \\binom{n}{i} y_i.\n$$\nFor nonnegative integers $i \\le j$, consider the expression\n$$\n\\sum_{k=i}^{j} (-1)^k \\binom{k}{i} \\binom{j}{k}.\n$$\nViewing $\\binom{j}{k}$ as the number of ways to choose a... | United States | Team Selection Test | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Algebraic properties of ... | null | proof only | null | |
0djd | Let $ABC$ be a triangle with incenter $I$ and ex-center $J$. Denote $K$ as the reflection of $A$ over $BC$ and take $X$, $Y$ on the opposite rays of $BA$, $CA$ such that $XB = BC = CY$. Let $T$ be the circumcenter of $AXY$. Prove that:
a) $OJ \perp XY$ and $AT$ is tangent to $(AIK)$.
b) $R_{(T)} = OJ$ and $TJ \perp B... | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"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 > Transformations > Homothety",
... | English | proof only | null | |
0bjf | Let $f$ be a primitive polynomial with integral coefficients (their highest common factor is $1$) such that $f$ is irreducible in $\mathbb{Q}[X]$, and $f(X^2)$ is reducible in $\mathbb{Q}[X]$. Show that $f = \pm(u^2 - Xv^2)$ for some polynomials $u$ and $v$ with integral coefficients.
For instance, if $a$ and $b$ are ... | [
"Unless otherwise stated, we work in $\\mathbb{Q}[X]$. Since the case $\\deg f = 1$ is easily dealt with, let $\\deg f \\ge 2$ and write $f(X^2) = gh$, where $g$ and $h$ both have a positive degree, and $g$ is irreducible. Next, write $g = a(X^2) + Xb(X^2)$ and $h = c(X^2) + Xd(X^2)$ to infer (from $f(X^2) = gh$ by... | Romania | THE 2014 Twelfth IMAR MATHEMATICAL COMPETITION | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
027s | Problem:
Pirajuba possui 10 cidades, chamadas $H_{1}, H_{2}, \ldots, H_{10}$, e algumas delas são ligadas por estradas de mão dupla. Sabe-se que é possível chegar de $H_{1}$ a $H_{10}$. Mostre que uma das situações abaixo ocorre:
(i) Existe um caminho ligando $H_{1}$ a $H_{10}$ utilizando no máximo 3 estradas.
(ii) Ex... | [
"Solution:\n\nSe $H_{1}$ ou $H_{10}$ estão ligadas a no máximo 2 cidades, a condição (i) é claramente satisfeita, pois todo caminho ligando $H_{1}$ a $H_{10}$ deve passar por elas.\n\nSuponha então que $H_{1}$ está ligada a $A_{1}, A_{2}$ e $A_{3}$ e que $H_{10}$ está ligada a $B_{1}, B_{2}$ e $B_{3}$ (eventualment... | Brazil | null | [
"Discrete Mathematics > Graph Theory > Menger's theorem / max-flow, min-cut",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0b6r | The tetrahedron $ABCD$ has $AD + BC + AB\sqrt{3} = 3$, volume $\frac{1}{12}$, and $\angle ABC = 60^\circ$. Compute the lengths of its edges. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Solid Geometry > Volume",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | AB = 1/√3, BC = 1, AC = √((4 − √3)/3), AD = 1, BD = 2/√3, CD = √((7 − √3)/3) | |
05ua | Problem:
Trouver tous les nombres réels $x$ et $y$ tels que
$$
\left(x-\sqrt{x^{2}+1}\right)\left(y-\sqrt{y^{2}+1}\right)=1 \text{ et }\left(x^{2}+y+2\right)\left(y^{2}+x+2\right)=8 .
$$ | [
"Solution:\nSoit $(x, y)$ une solution éventuelle. Les égalités\n$$\n\\left\\{\n\\begin{array}{l}\n\\left(x-\\sqrt{x^{2}+1}\\right)\\left(x+\\sqrt{x^{2}+1}\\right)=x^{2}-\\left(x^{2}+1\\right)=-1 \\\\\n\\left(y-\\sqrt{y^{2}+1}\\right)\\left(y+\\sqrt{y^{2}+1}\\right)=y^{2}-\\left(y^{2}+1\\right)=-1\n\\end{array}\n\\... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (x, y) = (1, -1) and (x, y) = (-1, 1) | |
0h33 | Let $f(x) = x^2 - 8x + 20$. Solve the equation $f(f(f(x))) = 5$. | [
"Неважко помітити, що $f(x) = (x - 4)^2 + 4$. Тоді рівняння $f(f(f(x))) = 5$ рівносильне рівнянню $(x - 4)^8 + 4 = 5$, звідки $x - 4 = \\pm 1$, тобто $x = 3$ або $x = 5$.\n\nВідповідь: $x = 3, x = 5$."
] | Ukraine | Ukrainian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 3, 5 | |
0k7w | Problem:
Let $S(x)$ denote the sum of the digits of a positive integer $x$. Find the maximum possible value of $S(x+2019)-S(x)$. | [
"Solution:\n\nWe note that $S(a+b) \\leq S(a)+S(b)$ for all positive $a$ and $b$, since carrying over will only decrease the sum of digits. (A bit more rigorously, one can show that $S\\left(x+a \\cdot 10^{b}\\right)-S(x) \\leq a$ for $0 \\leq a \\leq 9$.) Hence we have $S(x+2019)-S(x) \\leq S(2019)=12$, and equali... | United States | HMMT February 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 12 | |
0hsk | Problem:
A stacking of circles in the plane consists of a base, or some number of unit circles centered on the $x$-axis in a row without overlap or gaps, and circles above the $x$-axis that must be tangent to two circles below them (so that if the ends of the base were secured and gravity were applied from below, then ... | [
"Solution:\n$C(4)=14$."
] | United States | null | [
"Discrete Mathematics > Combinatorics > Catalan numbers, partitions",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 14 | |
04i6 | Does there exist a positive integer $n$ such that $n^2 + 2n + 2015$ is a perfect square? | [] | Croatia | Croatia Mathematical Competitions | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
01sc | Let $M$ be the subset of all numbers from $\{1, 2, \dots, 2015\}$ which are not perfect squares.
a) Prove that $\{\sqrt{n}\} > 0.011$ for any $n \in M$.
b) Prove that there exists a number $n \in M$ such that $\{\sqrt{n}\} < 0.0115$.
(Here $\{y\}$ stands for the fractional part of $y$.) | [
"**a.)** To prove the required statement it suffices to find the number $n \\in M$ such that $\\{\\sqrt{n}\\} = \\min_{k \\in M}\\{\\sqrt{k}\\}$. Any number $n \\in M$ can be uniquely presented as\n$$\nn = k^2 + r, \\qquad (1)\n$$\nwhere $1 \\le r \\le 2k$, and $k = [\\sqrt{n}]$ ($\\cdot$ is the whole part of a num... | Belarus | FINAL ROUND | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Other"
] | English | proof only | null | |
06sx | We are given an infinite deck of cards, each with a real number on it. For every real number $x$, there is exactly one card in the deck that has $x$ written on it. Now two players draw disjoint sets $A$ and $B$ of $100$ cards each from this deck. We would like to define a rule that declares one of them a winner. This r... | [
"Answer. $100$.\n\nSolution 1. We prove a more general statement for sets of cardinality $n$ (the problem being the special case $n=100$, then the answer is $n$). In the following, we write $A>B$ or $B<A$ for \"$A$ beats $B$\".\n\nPart I. Let us first define $n$ different rules that satisfy the conditions. To this ... | IMO | 55th International Mathematical Olympiad Shortlist | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Logic"
] | null | proof and answer | 100 | |
03r5 | All the grids of an $m \times n$ chessboard ($m \ge 3, n \ge 3$) are colored either red or blue. Two adjacent grids (with a common side) are called a *good couple* if they are of different colors. Suppose that there are $S$ good couples, explain how to determine whether $S$ is odd or even. Does it depend on certain spe... | [
"**Solution I**\nClassify all grids into three parts: the grids at the four corners, the grids along the borderlines (not including four corners), and the other grids. Fill all red grids with label number $1$, all blue grids with label number $-1$. Denote the label numbers filled in the grids in the first part by $... | China | China Western Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | The parity of S depends only on the non-corner border squares: S is odd if and only if there are an odd number of blue border (non-corner) squares; otherwise S is even. | |
0fql | Problem:
El trapecio isósceles $ABCD$ tiene lados paralelos $AB$ y $CD$. Sabemos que $\overline{AB}=6$, $\overline{AD}=5$ y $\angle DAB=60^{\circ}$. Se lanza un rayo de luz desde $A$ que rebota en $CB$ en el punto $E$ e interseca en $AD$ en el punto $F$. Si $AF=3$, calcula el área del triángulo $AFE$. | [
"Solution:\n\nPuesto que el trapecio es isósceles y $\\angle DAB=60^{\\circ}$, podemos alargar los lados $AD$ y $BC$ que intersectan en $G$, formando así un triángulo equilátero $ABG$.\n\n\n\nLlamando $\\alpha=\\angle EAB$, tendremos que $\\angle AEB=120-\\alpha$. Como el rayo sale simétric... | Spain | OME fase local | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 3√3/2 | |
0f1k | Problem:
Show that there are $2^{n + 1}$ numbers each with $2^{n}$ digits, all $1$ or $2$, so that every two numbers differ in at least half their digits. | [] | Soviet Union | ASU | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
00ui | In a given community of people, each person has at least two friends within the community. Whenever some people from this community sit on a round table such that each adjacent pair of people are friends, it happens that no non-adjacent pair of people are friends. Prove that there exist two people in this community suc... | [
"Let $G$ be the accompanying simple graph: its vertices are the members of the community and each pair of friends is connected by an edge. Thus, the minimum degree $\\delta(G) \\ge 2$ and every cycle in $G$ is induced (i.e. chord-free).\n\nConsider a path $P : v_1v_2\\dots v_n$ of maximum length (clearly $n \\ge 3$... | Balkan Mathematical Olympiad | BMO 2023 Short List | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0225 | Problem:
Quantos pares de inteiros positivos $(x, y)$ são soluções da equação $3x + 5y = 501$? | [
"Solution:\n\nA equação dada é equivalente a $y = \\frac{3(167 - x)}{5}$. Como $y$ é um inteiro positivo, $167 - x$ deve ser um múltiplo positivo de $5$, ou seja:\n$$\n167 - x = 5k \\quad \\Rightarrow \\quad x = 167 - 5k\n$$\n\n$$\nx = 5 \\times 33 + 2 - 5k \\quad \\Rightarrow \\quad x = 5(33 - k) + 2\n$$\n\nonde $... | Brazil | Nível 2 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 33 | |
02pl | Problem:
A progressão geométrica $121,242,484,968,1936, \ldots$ possui três termos inteiros entre 200 e 1200.
a) Encontre uma progressão geométrica crescente que possui quatro termos inteiros entre 200 e 1200.
b) Encontre uma progressão geométrica crescente que possui seis termos inteiros entre 200 e 1200. | [
"Solution:\n\nObservemos que para obter termos inteiros, a razão entre os termos inteiros deve ser um racional $\\frac{p}{q}$ e para obter mais de três termos a razão tem que ser menor do que 2, já que para uma progressão de razão maior ou igual a 2, com o primeiro termo maior ou igual a 200, o quarto termo é maior... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) 256, 384, 576, 864 (common ratio 3/2).
b) 243, 324, 432, 576, 768, 1024 (common ratio 4/3). | |
01ih | Find the smallest possible value of
$$
\frac{x^{2023} + 203}{17x^7 + 7x^{17}}
$$
over all positive real numbers $x$. | [
"Answer: $\\frac{17}{2}$.\n\nNote that $2023 = 7 \\cdot 17^2 = 7 \\cdot 289 = 17 \\cdot 119$. Using AGM on the numbers $x^{2023}, 1, 1, \\dots, 1$, where there are 288 ones, yields\n$$\nx^{2023} + 288 \\geq 289 \\sqrt[289]{x^{2023}} = 289x^7.\n$$\nAnalogously, using AGM on the numbers $x^{2023}, 1, 1, \\dots, 1$, w... | Baltic Way | Baltic Way 2023 Shortlist | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 17/2 | |
0hmr | Problem:
A gadget has four dials in a row, each of which can be turned to point to one of three numbers: $0$ (left), $1$ (up) or $2$ (right). Initially the dials are in the respective positions $2,0,1,0$, so that the gadget reads "2010." You may perform the following operation: choose two adjacent dials pointing at di... | [
"Solution:\n\nThe answer is no. We notice that initially the sum of the numbers on the dials is $3$. We claim that after each operation, the sum of the numbers on the dials remains a multiple of $3$. To see this, consider the three possible types of moves:\n(a) Changing a $0$ and a $2$ to two $1$'s does not change ... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | No | |
0dz4 | In a village with $12k$ inhabitants each person knows $3k + 6$ other people and the acquaintances are mutual. There exists a positive integer $n$, such that for any two villagers the number of villagers who know both is $n$. How many villagers are there in the village? | [
"Consider an arbitrary villager $a$. Let $A$ denote the set of all villagers who know $a$ and let $B$ contain all the rest. There are $3k + 6$ villagers in $A$ and $9k - 7$ villagers in $B$. Let $x$ be a villager from $a$. The villagers who know $a$ as well as $x$ form a subset of $A$. Let $n$ be the number of vill... | Slovenia | Slovenija 2008 | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 36 | |
0beg | Let $n$ be a positive integer and let $a_1, a_2, \ldots, a_n$ be real numbers with $a_1 + a_2 + \ldots + a_k \le k$ for all $k \in \{1, 2, \ldots, n\}$. Show that
$$
\frac{a_1}{1} + \frac{a_2}{2} + \ldots + \frac{a_n}{n} \le \frac{1}{1} + \frac{1}{2} + \ldots + \frac{1}{n}.
$$ | [
"We induct on $n$. The case $n=1$ is trivial. Suppose the claim holds for $n$ numbers. If $a_{n+1} \\le 1$ then $\\frac{a_{n+1}}{n+1} \\le \\frac{1}{n+1}$ and the conclusion follows. If $a_{n+1} > 1$, then $\\frac{a_1}{1} + \\dots + \\frac{a_n}{n} + \\frac{a_{n+1}}{n+1} \\le \\frac{a_1}{1} + \\dots + \\frac{a_n + a... | Romania | 64th Romanian Mathematical Olympiad - District Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0cjk | Determine all the natural numbers $a, b, c$ such that $ab + bc + ca$ is a prime number $p$ and $p$ divides the number $a^2b^2 + b^2c^2 + c^2a^2$. | [
"From the identity $a^2b^2 + b^2c^2 + c^2a^2 = (ab+bc+ca)^2 - 2abc(a+b+c)$ it follows that $p$ divides $abc(a+b+c)$. Because $p$ is a prime number, we get $p \\mid a$, $p \\mid b$, $p \\mid c$ or $p \\mid (a+b+c)$.\n\nSince $a, b, c < p$, the first three cases are impossible. The fourth situation can be true if and... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | a = b = c = 1 | |
0999 | $P$ нь $ABC$ гурвалжны дотоод цэг. $AP$, $BP$ ба $CP$ шулуунууд $ABC$ гурвалжныг багтаасан $\Gamma$ тойргийг хоёр дахь удаагаа харгалзан $K$, $L$ ба $M$ цэгүүдэд огтолно. $C$-цэгийг дайрсан $\Gamma$ тойргийн шүргэгч $AB$ шулууныг $S$ цэгт огтолно. $SC = SP$ бол $MK = ML$ гэж батал. | [
"$CA > CB$ гэж үзье. Тэгвэл $S$ нь $AB$ цацраг дээр оршино. $\\triangle PKM \\sim \\triangle PCA$ ба $\\triangle PLM \\sim \\triangle PCB$ гурвалжнуудын төсөөгөөс $\\frac{PM}{KM} = \\frac{PA}{CA}$ ба $\\frac{LM}{PM} = \\frac{CB}{PB}$ болно. Эдгээрийг үржүүлбэл\n$$\n\\frac{LM}{KM} = \\frac{CB}{CA} \\cdot \\frac{PA}{... | Mongolia | International Mathematical Olympiad 51 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometr... | Mongolian | proof only | null | |
037l | Problem:
Let $CL$ and $CK$ be the inner and the outer bisectors of angle $ACB$ in $\triangle ABC$, $AC > BC$ and let $CM$ be its median. A point $P$ on $CM$ is such that the points $C, A_{1}, B_{1}$ and $P$ are concyclic, where $A_{1} = AP \rightarrow \cap BC$ and $B_{1} = BP \rightarrow \cap AC$. Prove that the points... | [
"Solution:\nIt follows from Ceva's theorem that $\\frac{AM \\cdot BA_{1} \\cdot CB_{1}}{MB \\cdot A_{1}C \\cdot B_{1}A} = 1$, i.e. $\\frac{CB_{1}}{B_{1}A} = \\frac{CA_{1}}{A_{1}B}$. Thus, $A_{1}B_{1} \\parallel AB$ and $\\Varangle A_{1}B_{1}C = \\Varangle BAC$. We have $\\Varangle APM = \\Varangle A_{1}PC = \\frac{... | Bulgaria | Team selection test for 23. BMO | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0eei | Problem:
Določi vse pare realnih števil $a$ in $b$, ki ustrezajo neenakosti
$$
a^{2}(2a-b) + b^{2}(2b-a) \geq 0
$$ | [
"Solution:\n\nLevo stran neenakosti zmnožimo, da dobimo $2a^{3} - a^{2}b - ab^{2} + 2b^{3}$, in jo razstavimo\n$$\n\\begin{aligned}\n2a^{3} - a^{2}b - ab^{2} + 2b^{3} &= 2(a^{3} + b^{3}) - ab(a + b) = 2(a + b)(a^{2} - ab + b^{2}) - ab(a + b) = \\\\\n&= (a + b)(2a^{2} - 3ab + 2b^{2})\n\\end{aligned}\n$$\nDrugi fakto... | Slovenia | 60. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a + b ≥ 0 | |
084o | Problem:
Durante una festa, tre ragazze e tre ragazzi si siedono casualmente ad un tavolo rotondo. Qual è la probabilità che non ci siano due persone dello stesso sesso sedute a fianco?
(A) $\frac{1}{6}$
(B) $\frac{1}{10}$
(C) $\frac{3}{20}$
(D) $\frac{1}{12}$
(E) $\frac{11}{36}$. | [
"Solution:\nLa risposta è (B). Se fissiamo un ragazzo abbiamo che ci sono $5!$ possibili permutazioni degli altri 5 convitati, di queste quelle in cui ragazzi e ragazze sono alternati sono $3 \\cdot 2 \\cdot 2$ perché la persona alla destra del ragazzo fissato può essere una qualunque delle 3 ragazze, la persona an... | Italy | Progetto Olimpiadi di Matematica | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | MCQ | B | |
0hri | Problem:
On a distant planet, there are $2014$ cities, some pairs of which are connected by two-way roads. It turns out that the population of each city is the average of the populations of the cities to which it is connected by a single road, and moreover that it is possible to travel from every city to every other c... | [
"Solution:\n\nConsider the city $C_{\\max}$ with the maximal population $M$ (breaking ties arbitrarily). Then $M$ is the average of the populations of the neighboring cities, say $p_{1}, p_{2}, \\ldots, p_{n}$, meaning that\n$$\n\\frac{p_{1}+p_{2}+\\cdots+p_{n}}{n}=M\n$$\nBut $p_{1}, p_{2}, \\ldots, p_{n} \\leq M$,... | United States | Berkeley Math Circle Monthly Contest 2 | [
"Discrete Mathematics > Graph Theory",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0h6k | Does there exist real $x$, such that both $x + \sqrt{2}$ and $x^4 + \sqrt{2}$ are rational? | [
"Let us suggest that there exist rational $a, b$, such that: $a = x + \\sqrt{2}$ and $b = x^4 + \\sqrt{2}$. Thus $x = a - \\sqrt{2}$. After substitution in another equality we obtain:\n$$\nb = a^4 - 4a^3\\sqrt{2} + 12a^2 - 8a\\sqrt{2} + 4 + \\sqrt{2}.\n$$\nRight part of equality has to be rational, thus the sum of ... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein"
] | null | proof and answer | No | |
0eiv | Problem:
Dani sta realni funkciji $f(x)=x+1$ in $g(x)=x^{2}+3$.
a) Funkcija $h$ je podana s predpisom $h(x)=\frac{f(x)+g(x)}{g(x)}$. Izračunaj stacionarne točke funkcije $h$. Zapiši enačbo vodoravne asimptote grafa funkcije $h$ in izračunaj presečišče grafa z vodoravno asimptoto.
b) Izračunaj, za katere $a \in \mathbb... | [
"Solution:\n\na) Zapišemo predpis funkcije $h(x)=\\frac{x^{2}+x+4}{x^{2}+3}$. Izračunamo odvod funkcije $h'(x)=\\frac{-x^{2}-2 x+3}{\\left(x^{2}+3\\right)^{2}}$. Ničle odvoda funkcije $h$ so rešitve enačbe $-x^{2}-2 x+3=0$. Rešitvi enačbe sta $x_{1}=1$, $x_{2}=-3$, to sta stacionarni točki funkcije $h$.\n\nEnačba v... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a) Stationary points: x = 1 and x = −3; horizontal asymptote: y = 1; intersection with the asymptote: (−1, 1). b) The function has a real zero for a ≤ −2√3 or a ≥ 2√3. | |
0hux | Problem:
Let $\omega$ be a circle with diameter $A B$. A circle $\gamma$, whose center $C$ lies on $\omega$, is tangent to $A B$ at $D$ and cuts $\omega$ at $E$ and $F$. Prove that triangles $C E F$ and $D E F$ have the same area. | [
"Solution:\nLet segments $C D$ and $E F$ intersect at $M$. Extend $C D$ to meet $\\gamma$ at $G$ and $\\omega$ at $H$, noting that $G C = C D = D H$. By Power of a Point,\n$$\n\\begin{gathered}\nM G \\cdot M D = M E \\cdot M F = M C \\cdot M H \\\\\n(C G + M C) \\cdot M D = M C \\cdot (M D + D H) \\\\\nC G \\cdot M... | United States | Berkeley Math Circle Monthly Contest 5 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0543 | A right triangle $ABC$ has the right angle at vertex $A$. Circle $c$ passes through vertices $A$ and $B$ of the triangle $ABC$ and intersects the sides $AC$ and $BC$ correspondingly at points $D$ and $E$. The line segment $CD$ has the same length as the diameter of the circle $c$. Prove that the triangle $ABE$ is isosc... | [
"Since $\\angle BAD = 90^\\circ$ (Fig. 1), $BD$ is the diameter of circle $c$ and therefore $CD = BD$. Since $BD$ is diameter, also $\\angle BED = 90^\\circ$, so $DE$ is an altitude of the isosceles triangle $BDC$, bisecting its base $BC$. Hence $E$ is the midpoint of the hypotenuse $BC$ of the triangle $ABC$. Sinc... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
07nj | Let $ABC$ be a triangle whose side lengths are, as usual, denoted by $a = |BC|$, $b = |CA|$, $c = |AB|$. Denote by $m_a$, $m_b$, $m_c$, respectively, the lengths of the medians which connect $A$, $B$, $C$, respectively, with the centres of the corresponding opposite sides.
a. Prove that $2m_a < b + c$. Deduce that $m_... | [
"Denote by $D$ the mid-point of $BC$. We offer two ways of doing part (a).\n\nFirst way:\nContinue the line segment $AD$ through $D$ to the point $A'$ chosen so that $|A'D| = |DA| = m_a$. Consider the triangles $A'DC$ and $ADB$. Note that $\\angle A'DC = \\angle ADB$, $|BD| = |DC|$ and $|A'D| = |AD|$, by constructi... | Ireland | Ireland | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | null | |
07uk | The four vertices of quadrilateral $ABCD$ lie on the circle with diameter $AB$. The diagonals of $ABCD$ intersect at $E$, and the lines $AD$ and $BC$ intersect at $F$. Line $FE$ meets $AB$ at $K$ and line $DK$ meets the circle again at $L$. Prove that $CL$ is perpendicular to $AB$. | [
"\nNote that $\\angle ADB = 90^\\circ = \\angle ACB$ (angles in a semicircle). It follows that $E$ is the orthocentre of $\\triangle FAB$. Therefore $FK$ is perpendicular to $AB$. It follows that $ADEK$ is cyclic. Therefore $\\angle AEK = \\angle ADK$. Since $\\angle ADK = \\angle ADL = \\a... | Ireland | IRL_ABooklet | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0h4e | Let $I$ be the incenter of triangle $ABC$. Points $M$ and $N$ are chosen on sides $AB$ and $AC$ respectively such that $M \neq B$, $N \neq C$ and the points $A$, $I$, $M$, $N$ are cyclic. Prove that $BM + CN = BC$. | [
"Візьмемо на стороні $BC$ таку точку $K$, що $BM = BK$. Легко бачити, що $\\Delta BMI = \\Delta BKI$. Оскільки навколо чотирикутника $ANIM$ можна описати коло, і $\\angle MAI = \\angle NAI$, то $MI = NI = KI$. Зауважимо, що $\\angle BKI = \\angle BMI = 180^\\circ - \\angle AMI = \\angle ANI$, $\\angle CNI = \\angle... | Ukraine | Ukrainian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | English | proof only | null | |
0k30 | Problem:
There are two prime numbers $p$ so that $5p$ can be expressed in the form $\left\lfloor\frac{n^{2}}{5}\right\rfloor$ for some positive integer $n$. What is the sum of these two prime numbers? | [
"Solution:\n\nNote that the remainder when $n^{2}$ is divided by $5$ must be $0$, $1$, or $4$. Then we have that $25p = n^{2}$ or $25p = n^{2} - 1$ or $25p = n^{2} - 4$. In the first case there are no solutions. In the second case, if $25p = (n-1)(n+1)$, then we must have $n-1 = 25$ or $n+1 = 25$ as $n-1$ and $n+1$... | United States | HMMT February | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 52 | |
0h7p | For positive $a$, $b$, $c$, that satisfy the condition $ab + bc + ca = 3$, prove an inequality:
$$
\frac{1}{2a^3+1} + \frac{1}{2b^3+1} + \frac{1}{2c^3+1} \ge 1.
$$ | [
"Let us make such transformation:\n$$\n1 = \\frac{ab+bc+ca}{3} \\ge \\sqrt[3]{(abc)^2} \\Leftrightarrow abc \\le a \\le \\frac{1}{bc},\\ b \\le \\frac{1}{ac},\\ c \\le \\frac{1}{ab},\n$$\nhence $a+b+c \\le \\frac{1}{ab} + \\frac{1}{bc} + \\frac{1}{ca}$. Then we use well-known inequality:\n$$\n\\frac{a_1^2}{b_1} + \... | Ukraine | UkraineMO | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0caa | Problem:
Fie $A$ o mulţime finită de numere naturale. Determinaţi toate funcţiile $f: \mathbb{N} \rightarrow A$ cu proprietatea că $f(|x-y|)=|f(x)-f(y)|$, pentru orice $x, y \in \mathbb{N}$. | [] | Romania | Olimpiada Naţională GAZETA MATEMATICĂ | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | Necessary: 0 ∈ A. If 0 ∉ A, there are no solutions. If 0 ∈ A, all solutions are exactly:
- The zero function f(n) = 0 for all n.
- For any fixed c ∈ A, the parity function f(n) = 0 for n even and f(n) = c for n odd. | |
03ep | Given a natural number $n \ge 3$. To find the smallest real number $k > 0$ with the following property: If $G$ is a connected graph with $n$ vertices and $m$ edges, then it is always possible to delete no-more than $k \cdot \left(m - \left\lfloor \frac{n}{2} \right\rfloor\right)$ edges so that vertices can be colored i... | [
"**Lemma:** Let $G$ be a connected graph with at least $3$ vertices. Then either there exist two vertices connected by an edge whose removal (along with the outgoing edges) leaves $G$ connected, or there exist two vertices of degree $1$ (i.e., \"leaves\").\nConsider an arbitrary \"covering tree\" of $G$ and take as... | Bulgaria | Bulgarian Winter Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1/2 | |
0h14 | Natural numbers $a, b, c, d$ satisfy
$$
0 < |ad - bc| < \min\{c, d\}.
$$
Prove that for any coprime natural numbers $x, y > 1$ the number $x^a + y^b$ is not divisible by $x^c + y^d$. | [
"We will prove this by contradiction. Denote the sum $x^c + y^d$ by $s$. Then, obviously, $x^c = -y^d \\pmod{s}$ and also if $(x^a + y^b):s$, then $x^a = -y^b \\pmod{s}$. This implies that $x^{ad} = (-1)^d y^{bd} \\pmod{s}$ and $x^{bc} = (-1)^b y^{bd} \\pmod{s}$, hence $(-1)^d x^{ad} = y^{bd} = (-1)^b x^{bc} \\pmod... | Ukraine | 51st Ukrainian National Mathematical Olympiad, 4th Round | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0dzw | At the National Mathematical Olympiad the students were given 4 problems. Each solution was awarded with an integral number of points between 0 and 7. There were 42 participants and exactly half of them achieved at least 50% of the points. To win the award one had to get at least 22 points and one sixth of the contesta... | [
"First note that 7 contestants won the award (one sixth of 42). The upper half consisted of 21 contestants, so $21 - 7 = 14$ got between 14 and 21 points.\nEach of the 7 contestants who won the award had to get at least 22 points, so together they had at least $22 \\cdot 7 = 154$ points.\nLet $x$ represent the numb... | Slovenia | Slovenija 2008 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null |
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