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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0ajh | Solve the equation $1 + x^2 + y^2 = \text{lcm}(x^2, y^2)$ in the set of natural numbers. | [
"Let $d = \\text{gcd}(x, y)$. Then $d | \\text{lcm}(x^2, y^2)$, $d | x^2$ and $d | y^2$, from where we get $d = 1$. The equation is transformed into $1 + x^2 + y^2 = x^2 y^2$, or, equivalently, $(x^2 - 1)(y^2 - 1) = 2$. We get $x^2 - 1 = 1$, $y^2 - 1 = 2$ or $x^2 - 1 = 2$, $y^2 - 1 = 1$, from where it follows that ... | North Macedonia | Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | no solutions | |
0l3h | The product of three integers is $60$. What is the least possible positive sum of the three integers?
(A) 2 (B) 3 (C) 5 (D) 6 (E) 13 | [
"**Answer (B):** Note that $60 = 10 \\cdot (-1) \\cdot (-6)$, and the sum of these factors is $3$. It remains to show that no positive sum can be less than $3$. Such a sum would have to consist of one positive integer and two negative integers with smaller absolute value. If the positive integer is greater than or ... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | B | |
0fpd | Sea $n \ge 1$ y $P(x)$ un polinomio con coeficientes enteros que cumple que los números $P(1), P(2), \dots, P(n)$ son $1, 2, \dots, n$ (no necesariamente en este orden). Demostrar que uno de los números $P(0)$ o $P(n + 1)$ es múltiplo de $n!$. | [
"Si $i$ y $j$ son dos números enteros, se tiene que $i^k - j^k = (i-j)(i^{k-1} + i^{k-2}j + \\dots + ij^{k-2} + j^{k-1})$ es múltiplo de $i-j$. Entonces, si $P(x) = a_m x^m + \\dots + a_2 x^2 + a_1 x + a_0$,\n$$\nP(i) - P(j) = a_m(i^m - j^m) + \\dots + a_2(i^2 - j^2) + a_1(i - j)\n$$\ntambién es múltiplo de $i-j$. ... | Spain | LII Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | Spanish | proof only | null | |
0c2d | Prove that, for any positive integer $n$, the number $18^n$ is a sum of three squares. | [] | Romania | Shortlisted problems for the 69th NMO | [
"Number Theory > Algebraic Number Theory > Quadratic forms"
] | null | proof only | null | |
00vs | Let $\triangle ABC$ be an acute-angled triangle with orthocenter $H$ and let $D$ be an arbitrary point on side $BC$. Let $E$ and $Z$ be the points on the segments $AB$ and $AC$ respectively such that the quadrilaterals $ABDZ$ and $ACDE$ are cyclic, and let segments $BZ$ and $CE$ intersect at $P$. Let $L$ be the interse... | [
"We have $\\angle PCD = \\angle ECD = \\angle EAD$ and $\\angle PBD = \\angle ZBD = \\angle ZAD$, therefore\n$$\n\\angle BPC = 180^\\circ - \\angle EAD - \\angle ZAD = 180^\\circ - \\angle BAC = \\angle BHC,\n$$\nmeaning that $BHPC$ is cyclic.\n\nWe also have $\\angle PZD = \\angle BZD = \\angle BAD = \\angle PCD$ ... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | English | proof only | null | |
0gyt | Distinct real numbers $a$, $b$, $c$ satisfy the condition $a + \frac{1}{b} = b + \frac{1}{c} = c + \frac{1}{a}$. Find which values the product $abc$ can attain. | [
"From the first equality we can get that $a-b = \\frac{1}{c} - \\frac{1}{b}$ or that $(a-b) = \\frac{b-c}{bc}$.\nAnalogously $b-c = \\frac{c-a}{ac}$ and $c-a = \\frac{a-b}{ab}$. Thus we obtain: $a-b = \\frac{a-b}{(abc)^2}$. Since $a$, $b$ are distinct, $(abc)^2 = 1$, which means that $abc = \\pm 1$. Let us show tha... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | {-1, 1} | |
048j | Let $ABCD$ be a cyclic quadrangle. The perpendicular bisector of $\overline{BC}$ meets the side $\overline{AB}$ in point $E$. The circle through $E$, $C$ and the midpoint $F$ of the side $\overline{BC}$ meets the side $\overline{CD}$ in $G$. Prove that $AD \perp FG$. (D. Monk) | [
"Let the lines $AD$ and $FG$ intersect at point $H$.\n\n\n\nAs quadrilaterals $ABCD$ and $EFCG$ are cyclic, it follows that\n$$\n\\begin{align*}\n\\angle HDG &= 180^\\circ - \\angle ADC = \\angle ABC, \\\\\n\\angle HGD &= \\angle FGC = \\angle FEC = \\angle FEB = 90^\\circ - \\angle ABC.\n\... | Croatia | CroatianCompetitions2011 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0cdl | Let $M \ge 1$ be a real number. Determine all natural numbers $n$ for which there exist pairwise distinct natural numbers $a, b, c > M$, such that
$$
n = (a, b) \cdot (b, c) + (b, c) \cdot (c, a) + (c, a) \cdot (a, b)
$$
(where $(x, y)$ denotes the greatest common divisor of natural numbers $x$ and $y$). | [
"Call good a number $n \\in \\mathbb{N}$ for which there exist $a, b, c \\in \\mathbb{N}_{\\ge 1}$ such that\n$$n = (a, b) \\cdot (b, c) + (b, c) \\cdot (c, a) + (c, a) \\cdot (a, b).$$\nWe will prove that, regardless of the value of $M$, the good numbers are those of the form $n = 2^{2t}(2k+1)$, where $t, k \\in \... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - SECOND SELECTION TEST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | All n of the form n = 2^{2t}(2k+1) with t, k ∈ ℕ and k ≥ 1. | |
0hc5 | Find all pairs of positive integers $(m, n)$ that satisfy the equation
$$
m! + n! = m^n + 1.
$$ | [
"**Answer:** $(1, 1)$, $(2, 1)$, $(5, 3)$.\n\nBy checking trivial cases $m = 1$, $n = 1$ and $n = 2$, we find the first two of the solutions listed. From now on, we assume $m \\ge 2$ and $n \\ge 3$.\n\n**Step 1.** We prove that $m > n$.\nSuppose $n \\ge m$. Then, numbers $m!$, $n!$ and $m^n$ are divisible by $m$, h... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | (1, 1), (2, 1), (5, 3) | |
0e1k | A certain high school offers its students the choice of two sports: football and basketball. One fifth of the footballers also plays basketball and one seventh of the basketball players plays football. There are 110 students who practice exactly one of the sports. How many of them practice both? | [
"Denote the number of footballers by $n$, the number of basketball players by $k$ and the number of students practicing both sports by $d$. One fifth of the footballers also plays basketball. So $\\frac{n}{5} = d$. One seventh of the basketball players also plays football, so $\\frac{k}{7} = d$. This implies $n = 5... | Slovenia | National Math Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 11 | |
0l85 | Problem:
Two players, Alice and Bob, play a game on $n$ vertices labelled $1, 2, \dots, n$. Alice and Bob take turns adding edges $\{i, j\}$, with Alice going first. Both players are banned from creating a cycle, and the game ends after $n-1$ total turns.
Let the weight of the edge $\{i, j\}$ be $|i - j|$, and let $W$... | [] | United States | USA TST Selection Test for 67th IMO and 15th EGMO | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | If n is odd, W = 3(n^2 − 1)/8. If n is even, W = (3n^2 + 2n)/8 − 1. | |
0aru | Problem:
Let $a=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}$ and $b=\frac{\sqrt{5}-\sqrt{3}}{\sqrt{5}+\sqrt{3}}$. Find the value of $a^{4}+b^{4}+(a+b)^{4}$. | [
"Solution:\n\n$ab = 1$, $a^{2}b^{2} = 1$.\n\n$(a+b) = \\frac{10+6}{5-3} = 8 \\Rightarrow (a+b)^{2} = 64 \\Rightarrow (a+b)^{4} = 64^{2}$\n\nand $a^{2} + b^{2} = 62$.\n\n$a^{4} + b^{4} = (a^{2} + b^{2})^{2} - 2a^{2}b^{2} = 62^{2} - 2$.\n\nThus\n\n$a^{4} + b^{4} + (a+b)^{4} = 64^{2} + 62^{2} - 2 = 7938$."
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | final answer only | 7938 | |
085c | Problem:
Sia $ABC$ un triangolo e sia $A'$ il simmetrico di $A$ rispetto a $BC$; sia poi $DAA'$ simile ad $ABC$ e sia $D'$ il simmetrico di $D$ rispetto a $AA'$. Sapendo che il prodotto delle aree dei quadrilateri $ABA'C$ e $ADA'D'$ è $16$, si può dire che $AA'$ ...
(A) è $1$
(B) è $2 \sqrt[4]{2}$
(C) è $2$
(D) è $2 ... | [
"Solution:\n\nLa risposta è (D). Sia $H$ il piede dell'altezza da $A$ su $BC$, sia $K$ il piede dell'altezza da $D$ su $AA'$; il prodotto tra le aree dei quadrilateri indicati è $(AH \\cdot BC)(DK \\cdot AA') = 16$. Abbiamo le seguenti uguaglianze: $AA' = 2AH$ per costruzione e $AA' \\cdot AH = BC \\cdot DK$ per si... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | D | |
0cbf | Determine all integers $n \ge 2$, having at least four positive divisors, with the property that for any two distinct divisors $d_1$ and $d_2$ of $n$, such that $1 < d_1 < d_2 < n$, the number $d_2 - d_1$ is also a divisor of $n$.
Lucian Petrescu | [
"There are three solutions: $n = 6$, $n = 8$ and $n = 12$.\n\nFirst, notice that if $n$ is odd, all its divisors are also odd, so if $d_1$ and $d_2$ are two divisors such that $1 < d_1 < d_2 < n$, then $d_2 - d_1$ is even, so $d_2 - d_1 \\nmid n$. Therefore, $n$ is even.\n\nConsider $n = 2^a \\cdot b$, where $a, b ... | Romania | THE 73rd NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD - THIRD SELECTION TEST | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 6, 8, 12 | |
098f | Problem:
Fie $ABC$ un triunghi ascuțitunghic, în care $m(\angle BAC)=60^{\circ}$, iar $BC=a$. Fie $M$ mijlocul laturii $BC$, iar $[BB_{1}]$ și $[CC_{1}]$ două înălțimi ale triunghiului. Calculați suma distanțelor de la ortocentrul triunghiului $ABC$ până la laturile triunghiului $MB_{1}C_{1}$. | [
"Solution:\n\n1. În triunghiul dreptunghic $BCB_{1}$ segmentul $MB_{1}$ este mediană, atunci $MB_{1}=\\frac{BC}{2}=\\frac{a}{2}$. Idem $MC_{1}=\\frac{a}{2}$.\n\n2. Considerăm triunghiurile $AB_{1}C_{1}$ și $ABC$. Vom arăta că ele sunt asemenea.\nAceste triunghiuri au unghiul $A$ comun.\nVom arăta că laturile care f... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a*sqrt(3)/4 | |
0hlw | Problem:
A point $P$ lies inside a regular hexagon $A B C D E F$. The distances from $P$ to the sides $A B, B C, C D, D E, E F$, and $F A$ are respectively $1,2,5,7,6$, and $x$. Find $x$. | [
"Solution:\n\nOpposite sides of a regular hexagon are parallel. The sum of the distances from $P$ to $A B$ and to $D E$ is simply the distance between $A B$ and $D E$, which must therefore equal $1+7=8$. Because of the symmetry of the hexagon, the distance between any pair of opposite sides must be $8$. Applying th... | United States | Berkeley Math Circle Monthly Contest 8 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3 | |
0cho | The triangle $ABC$ has $\angle BAC = 54^\circ$ and $\angle ACB = 45^\circ$. Let $D$ and $E$ be points on the segments $BC$ and, respectively, $AD$ so that $AD = AB$ and $BE = BD$. Denote $F$ the common point of the lines $BE$ and $AC$, and let $DM$ be the bisector of the angle $\angle ADF$, where $M \in AC$. The perpen... | [
"a) From the statement, $\\angle ABC = 81^\\circ$ and, in the isosceles triangle $ABD$, $\\angle ADB = \\angle ABD = 81^\\circ$, hence $\\angle DAB = 18^\\circ$ and $\\angle CAD = \\angle CAB - \\angle DAB = 36^\\circ$.\nThe isosceles triangle $BED$ yields $\\angle BED = \\angle BDE = 81^\\circ$, therefore $\\angle... | Romania | 74th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0gdf | 令 $a_0, a_1, a_2, \dots$ 為實數數列使得 $a_0 = 0, a_1 = 1$, 且對每個 $n \ge 2$ 存在 $1 \le k \le n$ 滿足
$$
a_n = \frac{a_{n-1} + \cdots + a_{n-k}}{k}.
$$
試求 $a_{2018} - a_{2017}$ 之最大可能值。 | [
"The claimed maximal value is achieved at\n$$\na_1 = a_2 = \\cdots = a_{2016} = 1, \\quad a_{2017} = \\frac{a_{2016} + \\cdots + a_0}{2017} = 1 - \\frac{1}{2017},\n$$\n$$\na_{2018} = \\frac{a_{2017} + \\cdots + a_1}{2017} = 1 - \\frac{1}{2017^2}.\n$$\n\nNow we need to show that this value is optimal. For brevity, w... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 2016/2017^2 | |
0jdy | Problem:
Find all real numbers $x$ between $0$ and $360$ such that $$\sqrt{3} \cos 10^{\circ} = \cos 40^{\circ} + \sin x^{\circ}.$$ | [
"Solution:\n$70, 110$ (need both, but order doesn't matter)\n\nNote that $\\sqrt{3} = 2 \\cos 30^{\\circ}$, so $\\sin x^{\\circ} = \\cos 20^{\\circ} \\Longrightarrow x \\in \\{70, 110\\}$."
] | United States | HMMT November 2013 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | 70, 110 | |
0d1t | Let $a_{1}, a_{2}, a_{3}, \ldots$ be a sequence of real numbers which satisfy the relation
$$
a_{n+1}=\sqrt{a_{n}^{2}+1}
$$
Suppose that there exists a positive integer $n_{0}$ such that $a_{2 n_{0}}=3 a_{n_{0}}$. Find the value of $a_{46}$. | [
"We have $a_{n+1}^{2}-a_{n}^{2}=1$ for all natural number $n$. Using a telescopic sum we obtain\n$$\nn_{0}=\\sum_{k=n_{0}}^{2 n_{0}-1}\\left(a_{k+1}^{2}-a_{k}^{2}\\right)=a_{2 n_{0}}^{2}-a_{n_{0}}^{2}=9 a_{n_{0}}^{2}-a_{n_{0}}^{2}=8 a_{n_{0}}^{2} .\n$$\nHence\n$$\na_{n_{0}}^{2}=\\frac{n_{0}}{8}\n$$\nOn the other ha... | Saudi Arabia | Preselection tests for the full-time training | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series"
] | English | proof and answer | 19*sqrt(2)/4 | |
08jm | Problem:
Un parlament are $n$ deputati. Aceştia fac parte din 10 partide şi din 10 comisii parlamentare. Fiecare deputat face parte dintr-un singur partid și dintr-o singură comisie.
Determinați valoarea minimă a lui $n$ pentru care, indiferent de componența numerică a partidelor și indiferent de repartizarea în comis... | [] | JBMO | Olimpiada Naţională de Matematică | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 101 | |
0ikc | Let $ABCD$ be a quadrilateral, and let $E$ and $F$ be points on sides $AD$ and $BC$, respectively, such that $AE/ED = BF/FC$. Ray $FE$ meets rays $BA$ and $CD$ at $S$ and $T$, respectively. Prove that the circumcircles of triangles $SAE$, $SBF$, $TCF$, and $TDE$ pass through a common point. | [
"Let $P$ be the second intersection of the circumcircles of triangles $TCF$ and $TDE$. Because the quadrilateral $PEDT$ is cyclic, $\\angle PET = \\angle PDT$, or\n$$\n\\angle PEF = \\angle PDC. \\quad (*)\n$$\nBecause the quadrilateral $PFCT$ is cyclic,\n$$\n\\angle PFE = \\angle PFT = \\angle PCT = \\angle PCD. \... | United States | USAMO | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane G... | null | proof only | null | |
0113 | Problem:
Two squares on an $8 \times 8$ chessboard are called adjacent if they have a common edge or common corner. Is it possible for a king to begin in some square and visit all squares exactly once in such a way that all moves except the first are made into squares adjacent to an even number of squares already visi... | [
"Solution:\n\nNo, it is not possible.\n\nConsider the set $S$ of all (non-ordered) pairs of adjacent squares. Call an element of $S$ treated if the king has visited both its squares. After the first move there is one treated pair. Each subsequent move creates a further even number of treated pairs. So after each mo... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | No | |
0ahn | $n \ge 4$ points in the plane are given such that every three of them are not collinear. Prove that there exists a triangle such that all the points are in its interior, and on each of its sides lies exactly one point of the given points. | [
"The given points are finitely many in number so therefore there exists a disk which contains them in its interior.\nNamely, we are searching for the point which is at a maximal distance from the origin of the coordinate system. If we denote that distance by $R$, then the disk centered at the origin with radius $2R... | North Macedonia | XVI Junior Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | English | proof only | null | |
0ae6 | Определи ги сите цели броеви $x$, за кои $\log_2(x^2 - 4x - 1)$ е исто така цел број. | [
"Цели броеви $x$ за кои $\\log_2(x^2 - 4x - 1)$ е определен се\n$$\nx \\in (-\\infty, 2 - \\sqrt{5}, 2 + \\sqrt{5}, +\\infty) \\cap \\mathbb{Z}.\n$$\nНека $n$ е цел број за кој постои $x \\in \\mathbb{Z}$ така што\n$$\n\\log_2(x^2 - 4x - 1) = n.\n$$\nТогаш\n$$\nx^2 - 4x - (1 + 2^n) = 0 \\quad (1)\n$$\nод каде добив... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | Macedonian, English | proof and answer | -1, 5 | |
0cg4 | Let $f: \mathbb{R} \to \mathbb{R}$ be a continuous function such that $f(x) + \sin(f(x)) \ge x$, for all $x \in \mathbb{R}$. Prove that
$$
\int_{0}^{\pi} f(x) \, dx \geq \frac{\pi^2}{2} - 2.
$$ | [
"Let $g : \\mathbb{R} \\to \\mathbb{R}$ defined by $g(x) = x + \\sin(x)$, for any $x \\in \\mathbb{R}$. Obviously, $g$ is differentiable on $\\mathbb{R}$, with $g'(x) > 0$, for all $x \\in ((2k-1)\\pi, (2k+1)\\pi)$, for $k \\in \\mathbb{Z}$. Thus $g$ is increasing on any interval $[(2k-1)\\pi, (2k+1)\\pi]$, with $k... | Romania | 74th Romanian Mathematical Olympiad | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Functions"
] | English | proof only | null | |
0lg4 | Problem:
Let $\alpha, \beta, \gamma$ be the angles of a triangle opposite to the sides $a, b, c$ respectively. Prove the inequality
$$
2\left(\cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma\right) \geq \frac{a^2}{b^2 + c^2} + \frac{b^2}{a^2 + c^2} + \frac{c^2}{a^2 + b^2}
$$ | [
"Solution:\nBy the Law of Sines, RHS equals $\\frac{\\sin^2 \\alpha}{\\sin^2 \\beta + \\sin^2 \\gamma} + \\frac{\\sin^2 \\beta}{\\sin^2 \\alpha + \\sin^2 \\gamma} + \\frac{\\sin^2 \\gamma}{\\sin^2 \\alpha + \\sin^2 \\beta}$. Applying Cauchy-Bunyakowski inequality we have\n$$\n\\sin^2 \\alpha = \\sin^2(\\beta + \\ga... | Zhautykov Olympiad | IZhO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0equ | Zola and Ron play a game by alternately moving a single ten cent coin on a circular board. The game starts with the ten cent coin already on the board as shown. A player may move the coin either clockwise one position or one position toward the centre, but cannot move to a position that has been previously occupied.
T... | [
"Since the game must end (finite number of blocks and can't play on a previously occupied block) and only one person can move last, there will be a winner and hence a winning strategy.\n\nWe claim that the first player (Zola) has the winning strategy and it is to always move clockwise. There are $7$ positions left ... | South Africa | South African Mathematics Olympiad Third Round | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | Zola | |
0knz | Problem:
How many ways are there to write all numbers from $1$ to $9$ in the cells of a $3 \times 3$ grid so that for all integers $1 \leq n < 9$, the cell labeled $n$ shares an edge with the cell labeled $n+1$? | [
"Solution:\n\nApply a checkerboard coloring to the grid so that the corners and the center are black. Then by parity, the $5$ odd numbers must be in the $5$ black cells and the $4$ even numbers in the $4$ white cells.\n\nIf the center cell is $1$, there are $4$ ways to pick $2$ and $2$ ways to pick $3$, at which po... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 40 | |
0gef | 設 $n$ 為正整數，並令 $N = n^{2021}$。平面上有 $2021$ 個 (半徑兩兩相異的) 同心圓，圓心為 $O$。以 $O$ 為始點作間隔角度皆相同的 $N$ 條射線。在同心圓與射線的 $2021N$ 個交點中，有某些點被塗上紅色，其餘點不著色。
已知：不論如何在每一個圓上恰取一個交點，都可以找到一個角度 $\theta$，使得這些選點在以 $O$ 為中心一起旋轉 $\theta$ 角後皆疊合在紅點上。試證：紅點數量的最小可能值為 $2021n^{2020}$。
Let $n$ be a positive integer and $N = n^{2021}$. There are 2021 concentric ... | [
"讓我們將同心圓依序編號為 $1, 2, \\dots, 2021$，將射線依序編號為 $0, 1, \\dots, N-1$，並令 $A_i \\in \\{0, 1, \\dots, N-1\\} := \\mathcal{N}$ 為在第 $i$ 個同心圓上有紅點的射線編號集合。題設的重合條件，等價於對於所有 $(x_1, \\dots, x_{2021}) \\in \\mathcal{N}^{2021}$，皆存在 $(y_1, \\dots, y_{2021}) \\in A_1 \\times \\dots \\times A_{2021}$ 使得\n$$\nx_1 - y_1 \\equiv \\dots \\e... | Taiwan | 2021 數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > QM-AM-G... | null | proof and answer | 2021 n^{2020} | |
0bp3 | Problem:
Arătaţi că nu putem alege 45 de elemente distincte ale mulţimii $\{\sqrt{1}, \sqrt{2}, \sqrt{3}, \ldots, \sqrt{2015}\}$, astfel încât numerele selectate să fie în progresie aritmetică. | [
"Solution:\n\nDacă $m, n, p \\in \\mathbb{N}^{*}$ și $\\sqrt{m}, \\sqrt{n}, \\sqrt{p}$ sunt trei numere în progresie aritmetică, atunci $p + m + 2 \\sqrt{p m} = 4 n$, deci $\\sqrt{p m}$ este rațional. Rezultă $m = a^{2} d$, $p = c^{2} d$, $n = 2 b^{2} d$, cu $a, b, c, d \\in \\mathbb{N}$ și $a + c = 2b$.\n\nAstfel,... | Romania | Olimpiada Naţională de Matematică, Etapa Naţională | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0ey0 | Problem:
Given a triangle $ABC$. Suppose the point $P$ in space is such that $PH$ is the smallest of the four altitudes of the tetrahedron $PABC$. What is the locus of $H$ for all possible $P$? | [
"Solution:\nAnswer: the triangle $DEF$ with $FAE$ parallel to $BC$, $DBF$ parallel to $CA$ and $DCE$ parallel to $AB$.\n\nLet $\\alpha$ be the angle between planes $ABC$ and $PBC$. Let $h$ be the perpendicular distance from $H$ to the line $BC$, and let $h_A$ be the perpendicular distance from $A$ to the line $BC$.... | Soviet Union | 6th ASU | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | The locus is the interior of the triangle formed by the three lines through the vertices parallel to the opposite sides (triangle DEF). | |
0cr2 | В сейфе $n$ ячеек с номерами от $1$ до $n$. В каждой ячейке первоначально лежала карточка с её номером. Вася переложил карточки в некотором порядке так, что в $i$-й ячейке оказалась карточка с числом $a_i$. Петя может менять местами любые две карточки с номерами $x$ и $y$, платя за это $2|x-y|$ рублей. Докажите, что Пе... | [
"Пусть $(b_1, \\dots, b_n)$ — произвольное расположение карточек (здесь $b_i$ — число на карточке в $i$-й ячейке). Назовём его ценой число $|b_1 - 1| + |b_2 - 2| + \\dots + |b_n - n|$.\n\n**Лемма.** Для любого расположения $(b_1, \\dots, b_n)$, в котором не все карточки лежат на своих местах, можно поменять местами... | Russia | XL Russian mathematical olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
035d | Problem:
Prove that any triangle of area $\sqrt{3}$ can be placed into an infinite band of width $\sqrt{3}$. | [
"Solution:\n\nSuppose that it is not possible to cut a triangle $T$ of area $\\sqrt{3}$ from the band. Then it is clear that the lengths of its altitudes are greater than $\\sqrt{3}$. Hence the lengths of its sides are less than $2$.\n\nLet $\\alpha$ be the smallest angle of $T$. Then $\\alpha \\leq 60^{\\circ}$ an... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ac4 | $ABCD$ is a convex quadrilateral with area $P$. Let the point $A_1$ be on the ray $AB$ such that $\overline{AB} = \overline{BA_1}$, $B_1$ on the ray $BC$ such that $\overline{BC} = \overline{CB_1}$, $C_1$ on the ray $CD$ such that $\overline{CD} = \overline{DC_1}$ and $D_1$ on the ray $DA$ such that $\overline{DA} = \o... | [
"Connecting the points $A$ to $C_1$, $B$ to $D_1$, $C$ to $A_1$ and $D$ to $B_1$, we have\n$$\n\\begin{align*}\nP_{\\Delta ABC} &= P_I = P_{II} \\\\\nP_{\\Delta BCD} &= P_{III} = P_{IV} \\\\\nP_{\\Delta CDA} &= P_V = P_{VI} \\\\\nP_{\\Delta DAB} &= P_{VII} = P_{VIII}\n\\end{align*}\n$$\nwhere\n![](attached_image_1.... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5P | |
0doj | An integer is assigned to each unit square of a finite set of unit squares on an infinite chessboard such that the sum of numbers lying on each row as well as each column is divisible by $2002$. Prove that every number $a$ can be replaced by some number $a'$ divisible by $2002$ such that $|a - a'| < 2002$ and the sums ... | [
"Consider any number $a_1$ which is not divisible by $2002$. Then there exists a number $a_2$ not divisible by $2002$ lying on the same row with $a_1$. In turn, there exists a number $a_3$ not divisible by $2002$ lying on the same column with $a_2$. We continue this process till we first reach a number $a_k$ which ... | Silk Road Mathematics Competition | Silk Road Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Divisibility / Factorization"
] | English | proof only | null | |
0jep | Problem:
If Alex does not sing on Saturday, then she has a $70\%$ chance of singing on Sunday; however, to rest her voice, she never sings on both days. If Alex has a $50\%$ chance of singing on Sunday, find the probability that she sings on Saturday. | [
"Solution:\n\nLet $p$ be the probability that Alex sings on Saturday. Then, the probability that she sings on Sunday is $0.7(1-p)$; setting this equal to $0.5$ gives $p=\\frac{2}{7}$."
] | United States | HMMT | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 2/7 | |
0azk | Problem:
Suppose that $\log_{a} 125 = \log_{5} 3$ and $\log_{b} 16 = \log_{4} 7$. Find the value of $a^{\left(\log_{5} 3\right)^{2}} - b^{\left(\log_{4} 7\right)^{2}}$. | [
"Solution:\nNote that $a^{\\log_{5} 3} = 125$ and $b^{\\log_{4} 7} = 16$. Thus\n$$\na^{\\left(\\log_{5} 3\\right)^{2}} - b^{\\left(\\log_{4} 7\\right)^{2}} = \\left(a^{\\log_{5} 3}\\right)^{\\log_{5} 3} - \\left(b^{\\log_{4} 7}\\right)^{\\log_{4} 7} = 125^{\\log_{5} 3} - 16^{\\log_{4} 7} = 3^{3} - 7^{2} = -22.\n$$"... | Philippines | 20th Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | final answer only | -22 | |
00jq | Find all integers $n > 1$ such that the sum of $n$ and its second-largest divisor is 2013. | [
"The second-largest divisor of $n$ is of the form $\\frac{n}{p}$ where $p$ is the smallest prime that divides $n$.\nThe given condition gives $2013 = n + \\frac{n}{p} = \\frac{n}{p}(p+1)$. Therefore, $p+1$ is a divisor of $2013$ and thus odd. So, $p$ is $2$, the only even prime.\nThe equation now becomes $2013 = \\... | Austria | AustriaMO2013 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 1342 | |
08d1 | Problem:
$ABCD$ è un trapezio isoscele con lati di lunghezze $AB=15$, $BC=DA=5$, $CD=7$. Sui lati di $ABCD$, esternamente ad $ABCD$, vengono costruiti quattro quadrati. Con riferimento alla figura, qual è l'area del poligono $A_1A_2B_1B_2C_1C_2D_1D_2$?
(A) $297+35\sqrt{5}$
(E) $200\sqrt{5}$
(B) 396
(C) 423
(D) $1... | [
"Solution:\n\nLa risposta è $\\mathbf{(C)}$. Osserviamo dapprima che l'altezza del trapezio $ABCD$ è pari a $3$: infatti, detto $H$ il piede dell'altezza condotta da $C$ ad $AB$, il triangolo $HB C$ è retto in $H$ e valgono le uguaglianze $BC=5$ e $HB=\\frac{1}{2}(AB-CD)=4$, per cui (dal teorema di Pitagora) segue ... | Italy | GARA di FEBBRAIO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | C | |
0068 | Se dan 10 números reales $a_1, a_2, ..., a_{10}$, y se forman las 45 sumas de dos de estos números $a_i + a_j$, $1 \le i < j \le 10$. Se sabe que no todas estas sumas son números enteros. Determinar el mínimo valor de $k$ tal que es posible que entre las 45 sumas haya $k$ que no son números enteros y $45 - k$ que son n... | [] | Argentina | Argentina 2008 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Other"
] | Spanish | proof and answer | 9 | |
0g0l | Problem:
Finde alle Polynome $P$ mit reellen Koeffizienten, sodass folgende Gleichung für alle $x \in \mathbb{R}$ gilt:
$$
(x-2) P(x+2)+(x+2) P(x-2)=2 x P(x)
$$ | [
"Solution:\nEinsetzen von $x=0$ und $x=2$ liefert $P(-2)=P(0)=P(2)$, also können wir $P(x)$ schreiben als $Q(x)(x-2)(x)(x+2)+a$. Einsetzen in die ursprüngliche Gleichung liefert\n$$\n(x-4) Q(x-2)+(x+4) Q(x+2)=2 x Q(x)\n$$\nFalls wir ein $s<4$ finden, sodass $Q(s)=Q(s+2)=k$, dann folgt\n$$\nQ(s-2)=\\frac{2 s k-(s+4)... | Switzerland | IMO-Selektion | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | P(x) = b(x-2)x(x+2) + a for real constants a, b | |
044t | In a plane rectangular coordinate system $xOy$, the graph of parabola $y = ax^2 - 3x + 3$ ($a \neq 0$) and that of parabola $y^2 = 2px$ ($p > 0$) are symmetric with respect to line $y = x + m$. Then the product of real numbers $a, p, m$ is ______. | [
"For any point $(x_0, y_0)$ on parabola $y = ax^2 - 3x + 3$ ($a \\neq 0$), there is\n$$\ny_0 = a x_0^2 - 3x_0 + 3. \\qquad \\textcircled{1}\n$$\nSuppose the symmetric point of $(x_0, y_0)$ with respect to line $y = x + m$ is $(x_1, y_1)$.\nBy $\\frac{y_1 + y_0}{2} = \\frac{x_1 + x_0}{2} + m$, $x_1 + y_1 = x_0 + y_0... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | -3 | |
018g | A $2011 \times 2011$ square grid is divided into triangles by the diagonals of the squares. What is the total number of isosceles triangles in the figure? | [
"[The natural setting of the problem and its solution is of course an $n \\times n$ grid. There is a numerical challenge in doing just the $n = 2011$ case, but the task is not overwhelming; doing the computations by hand might be an educating task for the electronics oriented generation.]\n\nThere are two kinds of ... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 24416382548 | |
0095 | There are 50 clubs in an island. Every inhabitant of the island is member of 1 or 2 clubs. Each club has a maximum of 55 members, and for any pair of clubs, there is an inhabitant who is a member of both clubs. How many inhabitants can the island have? Give all the possibilities. | [] | Argentina | XXIX Olimpíada Matemática Argentina National Round | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All integers from 1225 to 1525 inclusive | |
0gth | In a school with $2021$ students each student has exactly $k$ friends and among any three students there are two which are not friends. Find the largest possible value of $k$. | [
"**Answer: 808.**\n\nLet us reformulate the problem in terms of graph theory: a $k$-regular graph $G$ on $2021$ vertices does not contain any triangle. What is the largest value of $k$? Below we give an example for $k = 808$:\n\n\n\nNow we show that $k \\le 808$. First of all, suppose that ... | Turkey | Turkish Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Turán's theorem",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 808 | |
09rv | Problem:
Bepaal alle paren $(a, b)$ van positieve gehele getallen waarvoor
$$
a^{2}+b \mid a^{2} b+a \quad \text{en} \quad b^{2}-a \mid a b^{2}+b .
$$ | [
"Solution:\nUit $a^{2}+b \\mid a^{2} b+a$ volgt\n$$\na^{2}+b \\mid\\left(a^{2} b+a\\right)-b\\left(a^{2}+b\\right)=a-b^{2}\n$$\nUit $b^{2}-a \\mid a b^{2}+b$ volgt\n$$\nb^{2}-a \\mid\\left(a b^{2}+b\\right)-a\\left(b^{2}-a\\right)=b+a^{2} .\n$$\nWe zien dus dat $a^{2}+b\\mid a-b^{2}\\mid a^{2}+b$. Dat betekent dat ... | Netherlands | IMO-selectietoets | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (a, a+1) for all positive integers a | |
0fep | Problem:
Se suponen conocidas las raíces reales de las $n$ ecuaciones de segundo grado que se indican en el siguiente cuadro:
| Ecuación | Raíces |
| :---: | :---: |
| $x^{2}+a_{1} x+b_{1}=0$ | $x_{0}, x_{1}$ |
| $x^{2}+a_{2} x+b_{2}=0$ | $x_{0}, x_{2}$ |
| $\cdots$ | $\cdots$ |
| $x^{2}+a_{n} x+b_{n}=0$ | $x_{0}, x_... | [
"Solution:\n\nPor hipótesis, se tiene\n$$\n\\begin{aligned}\n& x_{0}^{2}+a_{1} x_{0}+b_{1}=0 \\\\\n& \\cdots \\cdots \\cdots \\cdots \\cdots \\cdots \\\\\n& x_{0}^{2}+a_{n} x_{0}+b_{n}=0\n\\end{aligned}\n$$\nSumando estas igualdades, resulta\n$$\nn x_{0}^{2}+\\left(a_{1}+\\cdots+a_{n}\\right) x_{0}+\\left(b_{1}+\\c... | Spain | TANDA I | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | Roots: x0 and (x1 + x2 + ... + xn)/n | |
05nv | Problem:
On considère 7 îles $A_{1}, \ldots, A_{7}$. On est autorisé à construire des ponts, soit entre une île $A_{i}$ et l'île suivante $A_{i+1}$ (pour $i \in \{1,2, \ldots, 6\}$ ), soit entre une île $A_{i}$ et la dernière $A_{7}$ (pour $i \in \{1,2, \ldots, 6\}$ ). De combien de manières peut-on réaliser ces const... | [
"Solution:\n\nOn dira qu'une configuration est bonne si elle satisfait les conditions de l'énoncé.\nNotons $a_{n}$ le nombre de bonnes configurations avec $n$ îles. On a $a_{1}=a_{2}=1$ et $a_{3}=3$.\n\nPartant d'une bonne configuration avec $n$ îles telle que $A_{n-1}$ et $A_{n}$ ne soient pas reliées, alors $A_{n... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 144 | |
02ud | Problem:
Resolva em $\mathbb{R}$ a equação $\sqrt{x^{2}+9}+\sqrt{x^{2}-6 x+10}=5$. | [
"Solution:\nPrimeira Solução: Podemos eliminar radicais elevando membros da equação abaixo ao quadrado:\n$$\n\\begin{aligned}\n\\sqrt{x^{2}+9}+\\sqrt{x^{2}-6 x+10} & =5 \\\\\n\\left(\\sqrt{x^{2}-6 x+10}\\right)^{2} & =\\left(5-\\sqrt{x^{2}+9}\\right)^{2} \\\\\nx^{2}-6 x+10 & =25-10 \\sqrt{x^{2}+9}+x^{2}+9 \\\\\n10 ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 9/4 | |
0ft4 | Problem:
In einem konvexen Fünfeck ist jede Diagonale parallel zu einer Seite. Zeige, dass das Verhältnis zwischen den Längen der Diagonalen und der dazu parallelen Seite für alle Diagonalen dasselbe ist. Bestimme den Wert dieses Verhältnisses. | [
"Solution:\n\nEine Diagonale kann nur zu einer Seite parallel sein, mit der sie keinen gemeinsamen Punkt hat. Es ist also für jede Diagonale eindeutig, zu welcher Seite sie parallel ist. Wir führen nun einige Bezeichnungen ein\n$$\n\\begin{gathered}\nBD \\cap CE = \\{P\\} \\quad CE \\cap DA = \\{Q\\} \\quad DA \\ca... | Switzerland | IMO - Selektion | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | (1 + sqrt(5)) / 2 | |
08el | Problem:
Una mappa del tesoro è costituita dalla cartina di un'isola (infinitamente estesa in tutte le direzioni) a cui è sovrapposta una griglia quadrata. Le caselle della griglia sono numerate a spirale come in figura.
I pirati hanno nascosto il loro tesoro suddividendolo in forzieri che hanno seppellito in alcune d... | [
"Solution:\n\nFissiamo un intero positivo $n$ e consideriamo una casella, numerata $C$, che si trovi al di fuori del quadrato centrato sulla casella 1 di lato $2n-1$ caselle, ma che abbia almeno un vertice in comune col perimetro di tale quadrato. Per via della numerazione a spirale, le caselle all'interno del quad... | Italy | Italian Mathematical Olympiad - February Round | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Number Theory > Other"
] | null | proof only | null | |
0fbe | Problem:
En un espacio euclidiano de tres dimensiones se designan por $u_{1}, u_{2}, u_{3}$ los tres vectores unitarios ortogonales sobre los ejes $x, y, z$, respectivamente.
a) Probar que el punto $P(t)=(1-t) u_{1}+(2-3 t) u_{2}+(2 t-1) u_{3}$, donde $t$ toma todos los valores reales, describe una recta (que designar... | [
"Solution:\n\na) Se trata de la recta que pasa por el punto $(1,2,-1)$ y tiene como vector director $(-1,-3,2)$.\n\nb) Una semirrecta contenida en $L$, de origen $(1,2,-1)$ y que pasa por el punto $(0,5,1)$.\n\nc) $(-1,-3,2)$.\n\nd) $2(1-t)+3(2-3 t)+2(2 t-1)+1=0$, de donde $t=1$.\n\ne) $P(3)=(-2,-7,5)$, el plano pe... | Spain | OME 10 | [
"Algebra > Linear Algebra > Vectors",
"Geometry > Solid Geometry > Other 3D problems"
] | null | proof and answer | a) Line through (1, 2, −1) with direction (−1, −3, 2).
b) The ray on L corresponding to nonnegative parameter values, starting at (1, 2, −1); for example it passes through (0, −1, 1).
c) (−1, −3, 2).
d) t = 1.
e) 2x + 3y + 2z + 15 = 0.
f) −x − 3y + 2z − 19 = 0. | |
04kr | Let $a$ and $b$ be positive integers such that $a > b$ and
$$
a - b = 5b^2 - 4a^2.
$$
Prove that $a-b$ is a square of some positive integer. (Gazeta Matematică 2016) | [
"The given equality can be written more conveniently as:\n$$\n(a - b)(1 + 4(a + b)) = b^2.\n$$\nIt suffices to show that the bracketed expressions are coprime, since that implies that $a-b$ and $1+4(a+b)$ are squares of positive integers.\nDenote by $d$ the greatest common divisor of $a-b$ and $1+4(a+b)$. We have\n... | Croatia | Mathematical competitions in Croatia | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0i2p | Problem:
A kingdom consists of 12 cities located on a one-way circular road. A magician comes on the 13th of every month to cast spells. He starts at the city which was the 5th down the road from the one that he started at during the last month (for example, if the cities are numbered 1-12 clockwise, and the direction ... | [
"Solution:\nNumber the cities from $0$ to $11$. Encode the current state of affairs by starting at city $11$, writing \"1\" if the city is not under a spell, and a \"0\" if it is under a spell. Do the same thing for each city, from city $10$ down to city $0$. Interpret the resulting list as the binary expansion of ... | United States | 3rd Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Other"
] | null | proof only | null | |
089w | Problem:
Ogni numero intero viene colorato con uno di due colori, rosso o blu. Sappiamo che, per ogni insieme finito $A$ di interi consecutivi, il valore assoluto della differenza tra il numero degli interi rossi e il numero degli interi blu nell'insieme $A$ è al più 1000. Dimostrare che esiste un insieme di 2000 inte... | [
"Solution:\n\nDato un intero $n$, chiamiamo $I_{n}$ l'insieme dei 2000 numeri consecutivi $\\{n, n+1, \\ldots, n+1999\\}$ di lunghezza 2000, e chiamiamo $R_{n}$ il numero di interi rossi e $B_{n}$ il numero di interi blu in $I_{n}$: osserviamo che, siccome $R_{n}+B_{n}=2000$ è pari, anche $D_{n}=R_{n}-B_{n}$ è pari... | Italy | Cesenatico | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0bhy | Let $ABC$ be a triangle, let $A'$, $B'$, $C'$ be the orthogonal projections of the vertices $A$, $B$, $C$ on the lines $BC$, $CA$ and $AB$, respectively, and let $X$ be a point on the line $AA'$. Let $\gamma_B$ be the circle through $B$ and $X$, centered on the line $BC$, and let $\gamma_C$ be the circle through $C$ an... | [
"Let $H$ be the orthocenter of the triangle $ABC$. The line $AH$ is the radical axis of the circles $\\gamma_B$ and $\\gamma_C$, hence $HM' \\cdot HB = HN' \\cdot HC$ and $AM \\cdot AB = AN \\cdot AC$, so the lines $M'N'$ and $MN$ are both antiparallel to $BC$.\n\nThe circle $\\gamma_B$ meets the line $BC$ again at... | Romania | 65th NMO Selection Tests for BMO and IMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Co... | null | proof only | null | |
02ab | Problem:
Divisão de potências - Qual é o quociente de $50^{50}$ por $25^{25}$?
(a) $25^{25}$
(b) $10^{25}$
(c) $100^{25}$
(d) $2^{25}$
(e) $2 \times 25^{25}$ | [
"Solution:\n\nSolução 1:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{\\left(2 \\times 5^{2}\\right)^{50}}{\\left(5^{2}\\right)^{25}} = \\frac{2^{50} \\times 5^{100}}{5^{50}} = 2^{50} \\times 5^{50} = \\left(2^{2} \\times 5^{2}\\right)^{25} = 100^{25}\n$$\n\n\nSolução 2:\n$$\n\\frac{50^{50}}{25^{25}} = \\frac{(2 \\times ... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Other",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | MCQ | c | |
07pf | Suppose $0 \le x < 1$ and let $a_n = \frac{1-x^n}{n(1-x)}$ for all integers $n \ge 1$. Prove that
$$
|a_{n+1} - a_n| \le \frac{1}{n+1}, \quad n = 1, 2, 3, \dots
$$ | [
"First of all, note that $a_n = \\frac{1}{n}(1 + x + \\cdots + x^{n-1})$, hence\n$$\n\\begin{align*}\na_{n+1} - a_n &= \\frac{n(1 + x + \\cdots + x^n) - (n+1)(1 + x + \\cdots + x^{n-1})}{n(n+1)} \\\\\n&= \\frac{1}{n(n+1)} \\left( n x^n - \\sum_{k=0}^{n-1} x^k \\right) \\\\\n&= \\frac{1}{n(n+1)} \\sum_{k=0}^{n-1} (x... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0kfk | Problem:
In the Cartesian plane, a perfectly reflective semicircular room is bounded by the upper half of the unit circle centered at $(0,0)$ and the line segment from $(-1,0)$ to $(1,0)$. David stands at the point $(-1,0)$ and shines a flashlight into the room at an angle of $46^{\circ}$ above the horizontal. How man... | [
"Solution:\n\nNote that when the beam reflects off the $x$-axis, we can reflect the entire room across the $x$-axis instead. Therefore, the number of times the beam reflects off a circular wall in our semicircular room is equal to the number of times the beam reflects off a circular wall in a room bounded by the un... | United States | HMMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 65 | |
0ae7 | Аритметичката прогресија се состои од цели броеви. Збирот на првите $n$ членови на прогресијата е степен на бројот 2. Докажи дека и $n$ е степен на бројот 2. | [
"Нека првиот член на аритметичката прогресија е $a$, $n$-тиот член е $b$ а разликата е $d$. Нека $S$ е збирот на првите $n$ членови на прогресијата. Тогаш\n$$\n\\begin{align*}\nS &= a + (a+d) + (a+2d) + \\dots + (a+(n-1)d) = na + d(1+2+\\dots+n-1) = \\\\\n&= na + \\frac{(n-1)n}{2}d = \\frac{n}{2}[2a + (n-1)d] = \\f... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | Macedonian, English | proof only | null | |
004w | Ocho niños, todos de distintas estaturas, deben formar una fila ordenada de menor a mayor. Diremos que la fila tiene exactamente un error si hay un niño que está inmediatamente detrás de otro más alto que él, y todos los demás (salvo el primero de la fila) están inmediatamente detrás de uno más bajo. ¿De cuántas manera... | [] | Argentina | XIIIª OLIMPÍADA de MAYO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | Español | proof and answer | 247 | |
00zh | Problem:
Let $ABCD$ be a unit square and let $P$ and $Q$ be points in the plane such that $Q$ is the circumcentre of triangle $BPC$ and $D$ is the circumcentre of triangle $PQA$. Find all possible values of the length of segment $PQ$. | [
"Solution:\n\nAs $Q$ is the circumcentre of triangle $BPC$, we have $|PQ| = |QC|$ and $Q$ lies on the perpendicular bisector $s$ of $BC$. On the other hand, as $D$ is the circumcentre of triangle $PQA$, $Q$ lies on the circle centred at $D$ and passing through $A$. Thus $Q$ must be one of the two intersection point... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | sqrt(2 - sqrt(3)) and sqrt(2 + sqrt(3)) | |
05h7 | Problem:
Soient $m$ et $n$ deux entiers tels que $0 \leqslant m \leqslant 2n$. Prouver que le nombre entier
$$
2^{2n+2} + 2^{m+2} + 1
$$
est un carré parfait si, et seulement si, $m = n$. | [
"Solution:\n\nRemarquons que si $m = n$, alors $2^{2n+2} + 2^{m+2} + 1 = \\left(2^{m+1} + 1\\right)^2$ est bien un carré parfait.\n\nRéciproquement, supposons que $2^{2n+2} + 2^{m+2} + 1$ est un carré parfait. Comme il est strictement plus grand que $2^{2n+2} = \\left(2^{n+1}\\right)^2$, il est supérieur ou égal au... | France | OLYMPIADES FRANÇAISES DE MATHÉMATIQUES, ENVOI No. 3 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
09fn | Given a triangle $ABC$, let the bisectors of the angles $\angle BAC$, $\angle CBA$, $\angle ACB$ intersect the circumcircle of the triangle $ABC$ at the points $M$, $N$, $K$ respectively. Let $P$ be the intersection of the segments $AB$ and $MK$ and $Q$ be the intersection of the segments $AC$ and $MN$. Prove that the ... | [
"From $\\vec{AN} = \\vec{NC}$, we see that $MQ$ is a bisector of $\\triangle AMC$.\n\n\n\nBy the angle bisector theorem, we have\n$$\n\\frac{AQ}{QC} = \\frac{AM}{MC}. \\qquad (1)\n$$\nSimilarly, we have\n$$\n\\frac{AP}{PB} = \\frac{AM}{MB}. \\qquad (2)\n$$\nSince $\\vec{BM} = \\vec{MC}$, we... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06w9 | Given a positive integer $n$, find the smallest value of
$$
\left\lfloor\frac{a_{1}}{1}\right\rfloor+\left\lfloor\frac{a_{2}}{2}\right\rfloor+\cdots+\left\lfloor\frac{a_{n}}{n}\right\rfloor
$$
over all permutations $(a_{1}, a_{2}, \ldots, a_{n})$ of $(1,2, \ldots, n)$. | [
"Answer: The minimum of such sums is $\\left\\lfloor\\log _{2} n\\right\\rfloor+1$; so if $2^{k} \\leqslant n<2^{k+1}$, the minimum is $k+1$.\n\nSolution 1. Suppose that $2^{k} \\leqslant n<2^{k+1}$ with some nonnegative integer $k$. First we show a permutation $(a_{1}, a_{2}, \\ldots, a_{n})$ such that\n$$\n\\left... | IMO | IMO 2021 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof and answer | floor(log_2 n) + 1 | |
09rq | Problem:
De verzamelingen $A$ en $B$ zijn deelverzamelingen van de positieve gehele getallen. De som van elke twee verschillende elementen uit $A$ is een element van $B$. Het quotiënt van elke twee verschillende elementen van $B$ (waarbij we de grootste door de kleinste delen) is een element van $A$. Bepaal het maxima... | [
"Solution:\n\nStel dat $A$ minstens drie elementen bevat, zeg $a < b < c$. Dan bevat $B$ de drie verschillende elementen $a + b < a + c < b + c$. Dus $A$ bevat in elk geval het element $\\frac{b + c}{a + c}$. Deze breuk is kennelijk geheel, dus $a + c \\mid b + c$. Maar dan volgt $a + c \\mid (b + c) - (a + c) = b ... | Netherlands | IMO-selectietoets II | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 5 | |
0c9s | Problem:
Aflaţi numerele naturale nenule $x, y$ cu proprietatea că $x \leq y$, astfel încât
$$
\frac{(x+y)(x y-1)}{x y+1}=p
$$
unde $p$ este un număr prim. | [] | Romania | Primul test de selecţie pentru OBMJ | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All solutions with x ≤ y are: (1, q+1) for any prime q (giving p = q), and the pair (3, 5) (giving p = 7). | |
07kh | Suppose $a$ and $b$ are real numbers such that the quadratic polynomial
$$
f(x) = x^2 + a x + b
$$
has no nonnegative real roots. Prove that there exist two polynomials $g, h$, whose coefficients are nonnegative real numbers, such that
$$
f(x) = \frac{g(x)}{h(x)},
$$
for all real numbers $x$. | [
"First Solution. Since $f(x) = 0$ has no nonnegative real solutions, $b = f(0) > 0$. If $a \\ge 0$, then take $g(x) = f(x), h(x) = 1$. Suppose that $a < 0$. Since $f(x) = 0$ has no nonnegative roots, we must have $a^2 < 4b$. Replacing $f(x)$ by $b^{-1} f(x \\sqrt{b})$, we may assume that $b = 1$.\n\nPut $c_1 = -a$ ... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0h9v | Given real numbers $a$, $b$, $c$ such that: $ab - c = 1000$, $bc - a = 1018$ and $ca - b = -2018$. Show that $a + b + c \neq 0$. | [
"By contradiction, assume that $a + b + c = 0$. Thus,\n$$\nab + bc + ca = a + b + c = 0.\n$$\n\nSuppose that $abc = 0$, so without loss of generality let $c = 0$. Hence $ab = 0$. Therefore, two variables are zeros. By the first condition on numbers in the problem, we obtain a contradiction. Therefore, $abc \\neq 0$... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
0aqu | Problem:
Find the sum of the coefficients of the polynomial $\cos \left(2 \arccos \left(1-x^{2}\right)\right)$. | [] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | -1 | |
0l8u | Solve the equation:
$$
\sqrt{4 - 3\sqrt{10 - 3x}} = x - 2.
$$ | [
"The set of definition of the equation is $X = [74/27; 10/3]$.\nOn $X$, we have:\n$$\n\\text{given equation} \\Leftrightarrow 4 - 3\\sqrt{10 - 3x} = x^2 - 4x + 4 \\Leftrightarrow 9(10 - 3x) = x^2(4 - x)^2 \\Leftrightarrow \\\\\nx^4 - 8x^3 + 16x^2 + 27x - 90 = 0 \\Leftrightarrow (x - 3)(x + 2)(x^2 - 7x + 15) = 0 \\L... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 3 | |
01rd | Prove that
$$
\frac{1}{x+y+1} - \frac{1}{(x+1)(y+1)} < \frac{1}{11}
$$
for all positive $x$ and $y$. | [
"By the Cauchy inequality,\n$$\n(x+1)(y+1) \\le \\frac{((x+1) + (y+1))^2}{4} = \\frac{(x+y+2)^2}{4}.\n$$\nSo the required inequality\n$$\n\\frac{1}{x+y+1} - \\frac{1}{(x+1)(y+1)} < \\frac{1}{11}, \\quad \\text{if } x > 0 \\text{ and } y > 0, \\quad (1)\n$$\nis a consequence of the inequality\n$$\n\\frac{1}{x+y+1} -... | Belarus | Final Round | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null | |
0jpt | Problem:
Eric is taking a biology class. His problem sets are worth 100 points in total, his three midterms are worth 100 points each, and his final is worth 300 points. If he gets a perfect score on his problem sets and scores $60\%$, $70\%$, and $80\%$ on his midterms respectively, what is the minimum possible perce... | [
"Solution:\n\nAnswer: $60\\%$\n\nWe see there are a total of $100 + 3 \\times 100 + 300 = 700$ points, and he needs $70\\% \\times 700 = 490$ of them. He has $100 + 60 + 70 + 80 = 310$ points before the final, so he needs 180 points out of 300 on the final, which is $60\\%$."
] | United States | HMMT November 2015 | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 60% | |
007g | A collection of weights can be divided into 4 groups with equal masses, into 5 groups with equal masses, and into 9 groups with equal masses. Give an example of such a collection with the least possible number of weights. (Non-integer masses are allowed.) | [
"The answer is 14. First we prove that no collection of 13 weights is admissible. Assume on the contrary that 13 weights $a_1, a_2, \\dots, a_{13}$ can be divided into 9, 5 and 4 groups with equal masses (divisions 1, 2 and 3 respectively). Multiplying all $a_i$ by a positive number yields an admissible collection ... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 14; for example masses: 3, 4, 5, 7, 9, 11, 13, 15, 16, 17, 20, 20, 20, 20 | |
0kg4 | For every positive integer $N$, let $\sigma(N)$ denote the sum of the positive integer divisors of $N$. Find all integers $m \ge n \ge 2$ satisfying
$$
\frac{\sigma(m) - 1}{m - 1} = \frac{\sigma(n) - 1}{n - 1} = \frac{\sigma(mn) - 1}{mn - 1}.
$$ | [
"The answer is that $m$ and $n$ should be powers of the same prime number. These all work because for a prime power we have\n$$\n\\frac{\\sigma(p^e) - 1}{p^e - 1} = \\frac{(1 + p + \\cdots + p^e) - 1}{p^e - 1} = \\frac{p(1 + \\cdots + p^{e-1})}{p^e - 1} = \\frac{p}{p-1}.\n$$\nSo we now prove these are the only ones... | United States | USA TSTST | [
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | m = p^a and n = p^b for some prime p and integers a ≥ b ≥ 1 | |
0dm1 | The sequence of positive integers $\{a_n\}_{n \ge 1}$ is such that
$$
a_n \le 1000^{1000^{\sqrt{\log_2 n}}}, \quad \forall n = 1, 2, 3, \dots
$$
It is known that any number appears in the sequence at most $5000$ times. Prove that there are infinitely many primes, each of which divides at least one member of the sequenc... | [
"Let $S$ be the set of all numbers that appear in the sequence $\\{a_n\\}_{n \\ge 1}$. Since each number appears at most $5000$ times, for $N$ terms of the sequence, the number of distinct elements in $S$ is at least $N/5000$.\n\nSuppose, for contradiction, that only finitely many primes divide the elements of $S$.... | Saudi Arabia | Saudi Booklet | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | proof only | null | |
0bm8 | Let $(a_n)_{n \ge 0}$ and $(b_n)_{n \ge 0}$ be sequences of real numbers such that $a_0 > 1/2$, $a_{n+1} \ge a_n$, and $b_{n+1} = a_n(b_n + b_{n+2})$, for all non-negative integers $n$. Show that the sequence $(b_n)_{n \ge 0}$ is bounded. | [
"Use the relation in the statement and the fact that no $a_k$ is $1/2$, to write\n$$\nb_{k+2}^2 - b_k^2 = \\frac{(b_{k+1} - b_k)^2 - (b_{k+2} - b_{k+1})^2}{2a_k - 1}, \\quad k \\in \\mathbb{N},\n$$\nfix an integer $n > 2$, and sum over $0, 1, \\ldots, n-2$, to obtain\n$$\nb_n^2 + b_{n-1}^2 - b_1^2 - b_0^2 = \\frac{... | Romania | 66th NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation"
] | null | proof only | null | |
0dt2 | In a convex quadrilateral $ABCD$, the diagonals intersect at $O$, $M$ and $N$ are points on the segments $OA$ and $OD$ respectively. Suppose $MN$ is parallel to $AD$ and $NC$ is parallel to $AB$. Prove that $\angle ABM = \angle NCD$. | [
"We use the notation $[PQR]$ to denote the area of $\\triangle PQR$.\n$$\nMN \\parallel AD \\Rightarrow [AMN] = [DMN] \\Rightarrow [AON] = [DOM].\n$$\n$$\nNC \\parallel AB \\Rightarrow [ACN] = [BCN] \\Rightarrow [AON] = [BOC].\n$$\n$$\n\\text{Thus } [DOM] = [BOC] \\Rightarrow [MCD] = [BCD] \\Rightarrow MB \\paralle... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0j9y | Two circles $\omega_1$ and $\omega_2$ intersect at points $A$ and $B$. Line $l$ is tangent to $\omega_1$ at $P$ and to $\omega_2$ at $Q$ so that $A$ is closer to $l$ than $B$. Let $X$ and $Y$ be points on major arcs $\widehat{PA}$ (on $\omega_1$) and $\widehat{AQ}$ (on $\omega_2$), respectively, such that $AX/PX = AY/Q... | [
"\nLet $O$ denote the circumcenter of triangle $ASR$. Let $O_1$ denote the image of $O$ under $\\chi$. Since $\\chi$ takes triangle $OAS$ to triangle $O_1QP$, we have $\\triangle O_1QP \\sim \\triangle OAS \\sim \\triangle ORA$. Hence, $\\phi$ takes triangle $ORA$ to triangle $O_1QP$. This ... | United States | Team Selection Test Selection Test | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circl... | null | proof only | null | |
02f7 | Find a real-valued function $f(x)$ on the non-negative reals such that $f(0) = 0$, and $f(2x + 1) = 3f(x) + 5$ for all $x$. | [
"Let $I_n$ be the interval $[2^n - 1, 2^{n+1} - 1)$ for $n = 0, 1, 2, \\dots$. Then the $I_n$ are disjoint and cover the non-negative reals. Also $x \\to 2x + 1$ maps $I_n$ onto $I_{n+1}$. Thus $f$ is determined by the values it takes on $I_0 = [0, 1)$. These can be arbitrary, but the simplest is to take $f(x) = 0$... | Brazil | XV OBM | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | final answer only | One example is: define f(x)=0 for x in [0,1), and for each integer n ≥ 1 define f(x) = 5(3^n − 1)/2 for x in [2^n − 1, 2^{n+1} − 1). | |
09u3 | Problem:
Bewijs dat er voor elke positieve gehele $n$ hoogstens twee paren $(a, b)$ van positieve gehele getallen bestaan met de volgende twee eigenschappen:
(i) $a^{2}+b=n$,
(ii) $a+b$ is een tweemacht, d.w.z. er is een gehele $k \geq 0$ met $a+b=2^{k}$. | [
"Solution:\n\nStel dat er drie of meer zulke paren bestaan bij dezelfde $n$. Dan zijn er wegens het ladenprincipe minstens twee, zeg $(a, b)$ en $(c, d)$, zodat $a \\equiv c \\bmod 2$. Schrijf $a+b=2^{k}$ en $c+d=2^{\\ell}$ met $k$ en $\\ell$ positieve gehele getallen. Neem zonder verlies van algemeenheid aan dat $... | Netherlands | Selectietoets | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0a35 | Pjotr has a bath with a stopper in the bottom that he can fill with two identical taps. The water flows out of both taps at the same constant speed, and the bath empties (if the stopper is not there) at a constant speed. On Monday, Pjotr fills the bathtub to the brim by turning one tap fully open, then pulls out the st... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 10 | |
02j4 | Problem:
O gráfico da parábola $y = x^{2} - 5x + 9$ é rodado de $180^{\circ}$ em torno da origem. Qual é a equação da nova parábola?
A) $y = x^{2} + 5x + 9$
B) $y = x^{2} - 5x - 9$
C) $y = -x^{2} + 5x - 9$
D) $y = -x^{2} - 5x + 9$
E) $y = -x^{2} - 5x - 9$ | [
"Solution:\n\nUma rotação de $180^{\\circ}$ também é conhecida como meia-volta. Neste problema, temos uma meia-volta em torno da origem. O desenho ao lado ilustra o que esta meia-volta faz com as coordenadas dos pontos do plano. Por exemplo, o ponto $A^{\\prime}$ é o resultado da meia-volta aplicada ao ponto $A$; e... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | MCQ | E | |
00ds | Consider the following arrangement of 25 points.
• • • • •
Gastón must choose 4 of these points.
• • • • •
The chosen points must be the vertices of a square.
• • • • •
In how many ways can this be done?
• • • • • | [
"The following figures show how many squares of each type Gastón can choose.\n\n(a) 1 square\n\n(b) 16 squares\n\n(c) 9 squares\n\n(d) 4 squares\n\n(e) 9 squares\n\n(... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | English | proof and answer | 50 | |
02kv | Problem:
Quatro prefeitos decidem construir uma rodovia circular que passe em suas cidades, entretanto, as quatro cidades não estão sobre um mesmo círculo. Eles contratam uma empresa para elaborar um projeto para a construção da rodovia circular eqüidistante das quatro cidades. Qual o maior número de projetos geografi... | [
"Solution:\n\nO número de rodovias é igual ao número de pontos que podem ser centros da circunferência formada pelas rodovias.\n\nObservemos por outra parte que podemos ter dois tipos de configuração. Na primeira configuração a circunferência divide o conjunto das 4 cidades em dois conjuntos: um conjunto com 3 cida... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 7 | |
008r | Find the largest number of rectangular $4 \times 1$ pieces that can be placed onto a $10 \times 10$ checkerboard so that two pieces whatsoever do not touch each other by their sides or their vertices. | [] | Argentina | XXI Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 10 | |
0auj | Problem:
Triangle $ABC$ has a right angle at $B$, with $AB = 3$ and $BC = 4$. If $D$ and $E$ are points on $AC$ and $BC$, respectively, such that $CD = DE = \frac{5}{3}$, find the perimeter of quadrilateral $ABED$. | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 28/3 | |
0k6q | Problem:
Prove that if $a, b, c, d$ are positive integers then $a b + b c + c d + d a$ is not prime. | [
"Solution:\n\nNote that $a b + b c + c d + d a = (a + c)(b + d)$ is the product of two integers each greater than $1$."
] | United States | Berkeley Math Circle | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0ckj | Let $\triangle ABC$ be an isosceles triangle with $\angle BAC > 90^\circ$, and let $C$ be the circle centered at $A$ with radius $AB$. Let $M$ be the midpoint of side $AC$. The line $BM$ intersects the circle $C$ a second time at point $D$. Let $E$ be a point on the circle $C$ such that $BE \perp AC$, and suppose that ... | [
"From the condition $AC \\perp BE$, it follows that $AC$ is the perpendicular bisector of segment $BE$, so $MB = ME$. Since $AB = AE$, we conclude that $\\triangle MAB \\equiv \\triangle MAE$ (congruent triangles). It follows that $\\angle MAB = \\angle MAE$ (1), and $\\angle MBA = \\angle MEA$ (2).\nFrom $AB = AD$... | Romania | 75th NMO Selection Tests | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0chm | Let us consider the functions $f, g : \mathbb{R} \to \mathbb{R}$, where $f$ is continuous. Assume that, for all real numbers $a < b < c$, there is a sequence $(x_n)_{n \ge 1}$ which converges to $b$ such that the limit $\lim_{n \to \infty} g(x_n)$ exists and we have
$$
f(a) < \lim_{n \to \infty} g(x_n) < f(c).
$$
a) Gi... | [
"a) Let us consider the functions $f(x) = x$, for all $x \\in \\mathbb{R}$, and\n$$\ng(x) = \\begin{cases} x, & x \\in \\mathbb{Q} \\\\ x+1, & x \\in \\mathbb{R} \\setminus \\mathbb{Q} \\end{cases}.\n$$\nThe function $g$ is discontinuous at any real point. Let $a < b < c$ be three arbitrary real numbers. For any se... | Romania | 74th Romanian Mathematical Olympiad | [
"Precalculus > Functions",
"Precalculus > Limits"
] | English | proof and answer | a) Example: f(x) = x for all real x, and g(x) = x if x is rational, g(x) = x + 1 if x is irrational. Then g is discontinuous at every real point, and for any a < b < c, a rational sequence x_n → b gives lim g(x_n) = b, which lies in (f(a), f(c)).
b) If g is monotone, then g = f on all of R. | |
0de1 | Let $n$ be a positive integer. What is the smallest value of $m$ with $m > n$ such that the set $M = \{n, n+1, ..., m\}$ can be partitioned into subsets so that in each subset, there is a number which equals to the sum of all other numbers of this subset? | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | m = 7n - 4 | |
04cd | If the sides of a triangle satisfy the equality $\frac{a+b}{b+c} = \frac{c}{a-b}$, determine the largest angle in the triangle. | [] | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof and answer | 120 degrees | |
0ckp | Prove that, for a function $f : \mathbb{R} \to \mathbb{R}$, the following affirmations are equivalent:
(i) $f$ is differentiable on $\mathbb{R}$ with continuous derivative;
(ii) for any $a \in \mathbb{R}$ and two sequences $(x_n)_{n \ge 1}$ and $(y_n)_{n \ge 1}$, with limit $a$, such that $x_n \neq y_n$, for all $n \in... | [
"(i) $\\Rightarrow$ (ii)\n\nSuppose $f: \\mathbb{R} \\to \\mathbb{R}$ is continuously differentiable on $\\mathbb{R}$.\nLet $a \\in \\mathbb{R}$, and two sequences $(x_n)_{n \\ge 1}$ and $(y_n)_{n \\ge 1}$, convergent to $a$, such that $x_n \\ne y_n$, for $n \\in \\mathbb{N}^*$. By Lagrange theorem, there are point... | Romania | 75th Romanian Mathematical Olympiad | [
"Calculus > Differential Calculus > Derivatives",
"Calculus > Differential Calculus > Applications",
"Precalculus > Limits"
] | English | proof only | null | |
0i5i | Problem:
Call a positive integer "mild" if its base-3 representation never contains the digit $2$. How many values of $n$ ($1 \leq n \leq 1000$) have the property that $n$ and $n^{2}$ are both mild? | [
"Solution:\n\nSuch a number, which must consist entirely of $0$'s and $1$'s in base $3$, can never have more than one $1$. Indeed, if $n = 3^{a} + 3^{b} +$ higher powers where $b > a$, then $n^{2} = 3^{2a} + 2 \\cdot 3^{a+b} +$ higher powers which will not be mild. On the other hand, if $n$ does just have one $1$ i... | United States | Harvard-MIT Math Tournament | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 7 | |
0klo | Problem:
Are there integers $a$ and $b$ for which $a^{2} = b^{15} + 1004$? | [
"Solution:\n\nThe answer is no. Work modulo $31$: the equation reads $a^{2} \\equiv b^{15} + 12$. (The choice of $31$ is motivated by $2 \\cdot 15 + 1 = 31$.) We must have $b^{15} \\in \\{-1, 0, 1\\}$ $(\\bmod\\ 31)$ since the square of $b^{15} (\\bmod\\ 31)$ is $b^{30} (\\bmod\\ 31)$ which is always $0$ or $1$.\n\... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Quadratic residues"
] | null | proof and answer | No | |
0chi | Let $a, b, c$ be nonzero complex numbers of the same modulus for which the numbers $A = a+b+c$ and $B = abc$ are real. Prove that, for any nonnegative integer number $n$, the number $C_n = a^n + b^n + c^n$ is real. | [
"Let $\\alpha, \\beta, \\gamma$ be the reduced arguments of the complex numbers $a, b, c$. Since $abc \\in \\mathbb{R}$, it follows that $\\sin(\\alpha + \\beta + \\gamma) = 0$, i.e., $\\alpha + \\beta + \\gamma = k\\pi$, where $k \\in \\mathbb{Z}$. Therefore, we can write $\\gamma = k\\pi - \\alpha - \\beta$ for s... | Romania | 74th Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
0gft | 令 $n, s, t$ 為三個正整數, 並令 $A_1, \dots, A_s, B_1, \dots, B_t$ 為 $\{1, 2, \dots, n\}$ 的子集 (不必然相異)。對於 $\{1, \dots, n\}$ 的任何子集 $S$, 我們定義 $f(S)$ 為滿足 $S \subseteq A_i$ 的 $i \in \{1, \dots, s\}$ 的數量; $g(S)$ 為滿足 $S \subseteq B_j$ 的 $j \in \{1, \dots, t\}$ 的數量。假設對任意的 $1 \le x < y \le n$, $f(\{x, y\}) = g(\{x, y\})$ 都成立.
證明：若 $t < ... | [
"Suppose for the sake of contradiction that $t < n$ and $f(\\{x\\}) < g(\\{x\\})$ for any $1 \\le x \\le n$. Let $c_1, \\dots, c_n$ be some real numbers, and for any $S \\subseteq [n]$ let $c(S)$ be the expression\n$$\n\\sum_{i \\in S} c_i .\n$$\nThen clearly\n$$\n\\sum_{i=1}^{s} c(A_i)^2 = \\sum_{j=1}^{t} c(B_j)^2... | Taiwan | 2022 數學奧林匹亞競賽第三階段選訓營, 國際競賽實作(一) | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Linear Algebra > Vectors"
] | Chinese; English | proof only | null | |
0em3 | Let $\{a_1, a_2, ..., a_m\}$ and $\{b_1, b_2, ..., b_n\}$ be two sets of reals with the same sum. Prove that you can find an $m \times n$ array so that the row sums are $a_1, a_2, ..., a_m$ and the column sums are $b_1, b_2, ..., b_n$. | [
"Let $S = \\sum a_i = \\sum b_i$, and let $\\omega = a_n + b_n - S$. Create the table\n\n| 0 | 0 | ... | 0 | $a_1$ |\n|-----|-----|-----|-----|-------|\n| 0 | 0 | ... | 0 | $a_2$ |\n| ... | ... | ... | ... | ... |\n| $b_1$ | $b_2$ | ... | $b_{n-1}$ | $\\omega$ |\n\nThe sum of the first $m-1$ rows and ... | South Africa | South-Afrika 2011-2013 | [
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
0kcl | Problem:
A sheet of graph paper has perpendicular grid lines spaced 1 unit apart. On the paper, you draw a polygon all of whose edges lie along the grid lines. Determine all possible perimeters for this polygon. | [
"Solution:\nThe possible perimeters are all even numbers greater than $2$. To show that these are all possible, note that a $1 \\times (n-1)$ rectangle has perimeter $1 + (n-1) + 1 + (n-1) = 2n$, so there is a shape with perimeter $2n$ for every $n \\geq 2$. To show that no shape can have odd perimeter, imagine wal... | United States | Berkeley Math Circle: Monthly Contest 5 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | All even integers greater than two |
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