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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bdf | Find all positive integers $a < b < c$ so that $a^3 + b^3 + c^3 = (a + b + c)^2$. | [] | Romania | Shortlisted Problems for the 64th NMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (1, 2, 3) | |
0h8x | Positive numbers $a$, $b$, $c$ satisfy the condition $a^2 + b^2 + c^2 + abc = 4$. Prove that the inequality $c + ab \le 2$ holds. | [
"We consider equality from the problem as a quadratic equation with respect to the variable $c$:\n$$\nc^2 + abc + (a^2 + b^2 - 4) = 0 \\Rightarrow c = \\frac{-ab \\pm \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}.\n$$\nSince $c$ is a positive number, then $c = \\frac{-ab + \\sqrt{a^2b^2 + 16 - 4a^2 - 4b^2}}{2}$. Now we can... | Ukraine | 58th Ukrainian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof only | null | |
05nm | Problem:
Soit $ABCD$ un quadrilatère convexe. On suppose que les cercles inscrits aux triangles $ABC$, $BCD$, $CDA$ et $DAB$ ont un point commun. Montrer que $ABCD$ est un losange.
 | [
"Solution:\n\nOn note $\\omega_A$ le cercle inscrit à $DAB$ et ainsi de suite : les cercles $\\omega_B$ et $\\omega_D$ sont situés de part et d'autre de $(AC)$, donc ils ne peuvent s'intersecter que sur $(BC)$. De même, $\\omega_A$ et $\\omega_C$ ne peuvent s'intersecter que sur $(BD)$, donc l'intersection des 4 ce... | France | French Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler li... | null | proof only | null | |
04ss | We are given $n$ points in a plane, $n \ge 3$, no three of them collinear. Consider all the interior angles of all triangles with vertices in given points and denote $\phi$ the size of the smallest angle. For given $n$ find the largest possible $\phi$. | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls"
] | English | proof and answer | 180/n | |
00pl | Let $ABC$ be a triangle, $O$ its circumcenter and $AD$ the bisector of the angle $A$ where $D \in BC$. Let $\ell$ be the line passing through $O$ and parallel to the bisector $AD$. Prove that $\ell$ passes through the orthocenter $H$ of the triangle $ABC$ if and only if $ABC$ is isosceles or $\angle BAC = 120^{\circ}$. | [
"Let $G$ be the center of mass of the triangle $ABC$.\nAssume that $OH \\parallel AD$. Since $G, O, H$ are collinear, then $OG \\parallel AD$. Suppose that $AD$ meets the circumcircle of $ABC$ again at point $M$ and consider the midpoints $K, N$ of $MC, AC$, respectively. Let the lines $OG$ and $BK$ meet at $S$ and... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"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 | |
0g3s | Problem:
Beweise, dass es für jede natürliche Zahl $n \geq 3$ natürliche Zahlen $a_{1}<a_{2}<\ldots<a_{n}$ gibt, sodass
$$
a_{k} \mid\left(a_{1}+a_{2}+\ldots+a_{n}\right)
$$
für jedes $k=1,2, \ldots, n$ gilt. | [
"Solution:\n\nLösung 1: Es wird mit Induktion bewiesen.\n\na) Induktionsanfang, $n=3$ :\nMan bemerke, dass $1<2<3$ die gegebene Bedingung erfüllt. Somit ist die Aussage wahr für $n=3$.\n\nb) Induktionsschritt:\nNach der Induktionsannahme ist die Aussage wahr für alle $m \\leq n-1$. Seien also $a_{1}<a_{2}<\\cdots<a... | Switzerland | Zweite Runde 2021 | [
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0dh9 | Consider the function $f: \mathbb{R}^{+} \to \mathbb{R}^{+}$ and satisfying
$$
f(x + 2y + f(x + y)) = f(2x) + f(3y), \forall x, y > 0.
$$
1. Find all functions $f(x)$ that satisfy the given condition.
2. Suppose that $f(4\sin^4x)f(4\cos^4x) \ge f^2(1)$ for all $x \in (0; \frac{\pi}{2})$. Find the minimum value of $f(20... | [
"1) In the given condition, replace $x \\rightarrow 3x, y \\rightarrow 2y$, we have\n$$\nf(3x + 4y + f(3x + 2y)) = f(6x) + f(6y).\n$$\nBy swapping $x, y$ and comparing the two left sides, we have\n$$\nf(3x + 4y + f(3x + 2y)) = f(4x + 3y + f(2x + 3y)).\n$$\nFrom injectivity, we have\n$$\n3x + 4y + f(3x + 2y) = 4x + ... | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof and answer | All solutions are f(x) = x + c for c ≥ 0; under the inequality condition, the minimum value of f(2022) is 2023. | |
08qa | Problem:
Find all triples $(p, q, r)$ of prime numbers such that all of the following numbers are integers
$$
\frac{p^{2}+2 q}{q+r}, \quad \frac{q^{2}+9 r}{r+p}, \quad \frac{r^{2}+3 p}{p+q}
$$ | [
"Solution:\nWe consider the following cases:\n\n1st Case: If $r=2$, then $\\frac{r^{2}+3 p}{p+q}=\\frac{4+3 p}{p+q}$. If $p$ is odd, then $4+3 p$ is odd and therefore $p+q$ must be odd. From here, $q=2$ and $\\frac{r^{2}+3 p}{p+q}=\\frac{4+3 p}{p+2}=3-\\frac{2}{p+2}$ which is not an integer. Thus $p=2$ and $\\frac{... | JBMO | Junior Balkan Mathematical Olympiad Shortlist | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (2, 3, 7) | |
07lz | Suppose $p$ is the polynomial
$$
p(z) = az^4 + bz^3 + cz^2 + dz + e,
$$
where $a$, $b$, $c$, $d$, $e$ are complex numbers with $a \neq 0$. Prove that there is a complex number $h$ such that $p(h+z) = p(h-z)$ for all complex numbers $z$, if and only if
$$
b^3 - 4abc + 8a^2d = 0.
$$ | [
"We have\n$$\n\\begin{aligned}\n0 &= p(h+z) - p(h-z) \\\\\n &= a((h+z)^4 - (h-z)^4) + b((h+z)^3 - (h-z)^3) \\\\\n &\\quad + c((h+z)^2 - (h-z)^2) + d((h+z) - (h-z)) \\\\\n &= 8a(h^3z + hz^3) + 2b(3h^2z + z^3) + 4chz + 2dz \\\\\n &= z(8ah^3 + 6bh^2 + 4ch + 2d) + z^3(8ah + 2b).\n\\end{aligned}\n$$\nThis is satisfi... | Ireland | Irish Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
03ha | Problem:
For every positive integer $n$, let
$$
h(n) = 1 + \frac{1}{2} + \frac{1}{3} + \cdots + \frac{1}{n}
$$
For example, $h(1) = 1$, $h(2) = 1 + \frac{1}{2}$, $h(3) = 1 + \frac{1}{2} + \frac{1}{3}$. Prove that for $n = 2, 3, 4, \ldots$
$$
n + h(1) + h(2) + h(3) + \cdots + h(n-1) = n h(n)
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0kpz | Problem:
Find the number of pairs of integers $(a, b)$ with $1 \leq a < b \leq 57$ such that $a^{2}$ has a smaller remainder than $b^{2}$ when divided by $57$. | [
"Solution:\nThere are no such pairs when $b=57$, so we may only consider pairs with $1 \\leq a < b \\leq 56$. The key idea is that unless $a^{2} \\bmod 57 = b^{2} \\bmod 57$, $(a, b)$ can be paired with $(57-b, 57-a)$ and exactly one of them satisfies $a^{2} \\bmod 57 < b^{2} \\bmod 57$. Hence if $X$ is the number ... | United States | HMMT November 2022 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | proof and answer | 738 | |
0kyj | Problem:
Let $a_{1}, a_{2}, a_{3}, \ldots, a_{100}$ be integers such that
$$
\frac{a_{1}^{2}+a_{2}^{2}+a_{3}^{2}+\cdots+a_{100}^{2}}{a_{1}+a_{2}+a_{3}+\cdots+a_{100}}=100 .
$$
Determine, with proof, the maximum possible value of $a_{1}$. | [
"Solution:\nWe can rearrange the equation as follows:\n$$\n\\begin{gathered}\na_{1}^{2}+a_{2}^{2}+\\cdots+a_{99}^{2}+a_{100}^{2}=100\\left(a_{1}+a_{2}+\\cdots+a_{99}+a_{100}\\right) \\\\\n\\left(a_{1}^{2}-100 a_{1}\\right)+\\left(a_{2}^{2}-100 a_{2}\\right)+\\left(a_{3}^{2}-100 a_{3}\\right)+\\cdots+\\left(a_{100}^... | United States | HMMT February 2024 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 550 | |
0c7v | Let $f : [0, \infty) \to [0, \infty)$ be a continuous function with $f(0) > 0$, such that for any $0 \le x < y$, we have
$$
x - y < f(y) - f(x) \le 0.
$$
Prove that:
a) there exists a unique number $\alpha \in (0, \infty)$ with the property $(f \circ f)(\alpha) = \alpha$;
b) the sequence $(x_n)_{n \ge 1}$, defined by $... | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
09dh | $1, 2, 3, \ldots, 10^{2012}$ дарааллын бүх цифруудийн тоо нь $1, 2, 3, \ldots, 10^{2013}$ дарааллын бүх тэгүүдийн тоотой тэнцүү гэж батал. | [] | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | Mongolian | proof only | null | |
0api | Problem:
Simplify: $(x-1)^{4} + 4(x-1)^{3} + 6(x-1)^{2} + 4(x-1) + 1$. | [
"Solution:\n$$(x-1)^{4} + 4(x-1)^{3} + 6(x-1)^{2} + 4(x-1) + 1 = [(x-1) + 1]^{4} = x^{4}$$"
] | Philippines | Tenth Philippine Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | x^4 | |
0eij | Problem:
Zapiši enačbo stožnice, na kateri ležijo središča vseh tistih krožnic, ki se hkrati dotikajo krožnice $x^{2}+y^{2}=9$ in krožnice $x^{2}+(y-4)^{2}=1$, a ne vsebujejo točke $T(0,3)$. Enačbo stožnice zapiši v standardni obliki.
(20 točk) | [
"Solution:\n\nOznačimo dani krožnici zaporedoma s $\\mathcal{K}_1$ in $\\mathcal{K}_2$. Krožnica $\\mathcal{K}_1$ ima središče v $S_1(0,0)$ in polmer $r_1=3$, krožnica $\\mathcal{K}_2$ pa ima središče v $S_2=(0,4)$ in polmer $r_2=1$. Krožnici se od zunaj dotikata v točki $T(0,3)$. Krožnice, ki se hkrati dotikajo kr... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | x^2/3 - (y - 2)^2 = -1 | |
0ecz | Let $a_1, a_2, a_3, \dots$ be an arithmetic sequence. For an arbitrary positive integer $k$ denote $S_k = a_1 + a_2 + \dots + a_k$. Prove that
$$
S_n - S_m = \frac{n-m}{n+m} S_{n+m}
$$
holds for all positive integers $m$ and $n$. | [
"Denote by $d$ the difference of two consecutive terms of this arithmetic sequence. By the formula for the sum of an arithmetic sequence we have\n$$\nS_k = k a_1 + \\frac{k(k-1)}{2} d.\n$$\nUsing this we calculate\n$$\nS_{n+m} = (n+m)a_1 + \\frac{(n+m)(n+m-1)}{2} d = (n+m)\\left(a_1 + (n+m-1)\\frac{d}{2}\\right)\n$... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0jmi | Problem:
The integers $1,2, \ldots, 64$ are written in the squares of a $8 \times 8$ chess board, such that for each $1 \leq i < 64$, the numbers $i$ and $i+1$ are in squares that share an edge. What is the largest possible sum that can appear along one of the diagonals? | [
"Solution:\n\nAnswer: 432 Our answer is $26+52+54+56+58+60+62+64$.\nOne possible configuration:\n\n| 26 | 25 | 24 | 23 | 18 | 17 | 8 | 7 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| 27 | 52 | 53 | 22 | 19 | 16 | 9 | 6 |\n| 28 | 51 | 54 | 21 | 20 | 15 | 10 | 5 |\n| 29 | 50 | 55 | 56 | 57 |... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 432 | |
0gqf | Let $D$ be the midpoint of the side $BC$ of a triangle $ABC$ and $AD$ intersect the circumcircle of $ABC$ for the second time at $E$. Let $P$ be the point symmetric to the point $E$ with respect to the point $D$ and $Q$ be the point of intersection of the lines $CP$ and $AB$. Prove that if $A, C, D, Q$ are concyclic, t... | [
"\nSince $BD = DC$ and $PD = DE$, we conclude that $BPCE$ is a parallelogram and hence $\\angle PBC = \\angle BCE = \\angle BAE$. Since the points $Q, A, C, D$ are concyclic, $\\angle BAE = \\angle QCB$ and therefore $\\angle PBC = \\angle QCB$. Since $BD = DC$, the lines $PD$ and $BC$ are ... | Turkey | Team Selection Test | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fhb | Problem:
En el plano, donde se ha tomado un sistema de referencia ortonormal, se consideran todos los puntos $(m, n)$ cuyas coordenadas son números enteros. Se suponen trazados todos los segmentos que unen pares cualesquiera de estos puntos y cuya longitud es entera. Probar que no hay dos de esos segmentos que formen ... | [
"Solution:\n\nSupongamos que los puntos $A$, $B$ y $C$ son tales que $AC$ y $AB$ forman un ángulo de $45^{\\circ}$, y $AB$ y $AC$ son enteros; entonces $BC = \\sqrt{v}$, con $v$ entero. Según el teorema del coseno,\n$$\n\\frac{\\sqrt{2}}{2} = \\cos 45^{\\circ} = \\frac{AB^{2} + AC^{2} - v}{2 \\cdot AB \\cdot AC} \\... | Spain | OME 27 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry >... | null | proof only | null | |
08ff | Problem:
Ad una gara matematica partecipano $n=10000$ concorrenti.
Alla festa conclusiva, in successione, il primo prende $1/n$ della torta, il secondo prende $2/n$ della torta rimanente, il terzo prende $3/n$ della torta che rimane dopo che il primo e il secondo si sono serviti, e così via fino all'ultimo, che prende... | [
"Solution:\n\nIl concorrente che prende la fetta più grande è quello che si serve per 100-esimo.\nPer ogni $k=1,2, \\ldots, 10000$, indichiamo con $F_{k}$ la fetta che spetta al concorrente $k$-esimo, e con $T_{k}$ la quantità di torta rimasta dopo che si sono serviti i primi $k$ concorrenti. Poniamo per convenzion... | Italy | XXXVIII Olimpiade Italiana di Matematica, Cesenatico | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 100 | |
0h6v | Solve the equation $x[x] = 2016$.
Here $[x]$ is the integer part of the number $x$, i.e. the largest integer number not greater than $x$. | [
"**Answer:** $x = -\\frac{224}{5}$.\nSuppose that $x \\ge 0$. Let us denote $t = [x] \\ge 0$, and then we can write the following estimations:\n$$\nt \\le x < t+1 \\Rightarrow t^2 \\le x[x] < t^2 + t.\n$$\n$44^2 + 44 = 1980 < 2016 < 45^2$, therefore there are no solutions of the equation among positive numbers.\nSu... | Ukraine | UkraineMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | -224/5 | |
0id2 | Let $a$, $b$ and $c$ be positive real numbers. Prove that
$$
(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \geq (a + b + c)^3.
$$ | [
"**First Solution:** For any positive number $x$, the quantities $x^2 - 1$ and $x^3 - 1$ have the same sign. Thus, we have $0 \\le (x^3 - 1)(x^2 - 1) = x^5 - x^3 - x^2 + 1$, or\n$$\nx^5 - x^2 + 3 \\geq x^3 + 2.\n$$\nIt follows that\n$$\n(a^5 - a^2 + 3)(b^5 - b^2 + 3)(c^5 - c^2 + 3) \\geq (a^3 + 2)(b^3 + 2)(c^3 + 2)... | United States | USA IMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
01j1 | Some points are marked on a circle. Each marked point is coloured red, green or blue. In one step, one can erase two marked points of different colours that have no marked points between them and mark a new point between the locations of the erased points with the third colour. In a final state, all marked points have ... | [
"Answer: All even numbers $n$ greater than 2.\n\nIf $n = 2$ then the colour of the final state is uniquely determined. We show now that required initial states are impossible for odd $n$. Note that if one colour is missing then the numbers of marked points of existing two colours have different parities, i.e., the ... | Baltic Way | Baltic Way 2023 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | All even integers greater than 2 | |
022q | Problem:
Uma das afirmações abaixo é falsa:
(a) André é mais velho do que Bruno;
(b) Cláudia é mais nova do que Bruno
(c) A soma das idades de Bruno e Cláudia é o dobro da idade de André;
(d) Cláudia é mais velha do que André.
Quem é o mais velho? E o mais novo? | [
"Solution:\n\nCláudia e Bruno."
] | Brazil | Desafios | [
"Discrete Mathematics > Logic"
] | null | final answer only | Cláudia e Bruno. | |
0fr3 | Problem:
Sea $ABC$ un triángulo con $AB < AC$ y sea $I$ su incentro. El incírculo es tangente al lado $BC$ en el punto $D$. Sea $E$ el único punto que satisface que $D$ es el punto medio del segmento $BE$. La línea perpendicular a $BC$ que pasa por $E$ corta a $CI$ en el punto $P$. Demostrar que $BP$ es perpendicular ... | [
"Solution:\n\nComo $P$ está en la bisectriz de $\\angle ACB$, el círculo $\\gamma$ de centro $P$ y radio $PE$ es tangente al lado $AC$ en un punto $J$. De esta manera $AJ = AC - CJ = AC - CE = AC - BC + 2BD = AC - BC + (AB + BC - AC) = AB$. Esto implica que el cuadrilátero $ABDP$ tiene diagonales perpendiculares, y... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Quadrilaterals with perpendicular diagonals",
"Geometry > Plane Geometry > A... | null | proof only | null | |
0fh9 | Problem:
Tres puntos $A'$, $B'$, $C'$ están situados, respectivamente, sobre los lados $BC$, $CA$ y $AB$ de un triángulo dado $ABC$ de área $S$, de forma que
$$
\frac{AC'}{AB} = \frac{BA'}{BC} = \frac{CB'}{CA} = p
$$
siendo $p$ un parámetro variable, $0 < p < 1$. Determinar
1) El área del triángulo $A'B'C'$ en funció... | [
"Solution:\n\nSi ponemos $a = BC$, $b = CA$, $c = AB$, se tiene:\n$$\n\\begin{array}{ll}\nAC' = cp, & BC' = c(1-p) \\\\\nBA' = ap, & CA' = a(1-p) \\\\\nCB' = bp, & AB' = b(1-p)\n\\end{array}\n$$\nEntonces\n$$\nS_{A'BC'} = \\frac{1}{2} c(1-p) a p \\sen B = p(1-p) S\n$$\ny lo mismo ocurre con las áreas de los triángu... | Spain | OME 26 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 1) Area equals S times bracket one minus three times p times one minus p, i.e., S[1 − 3p(1 − p)]. 2) The minimum occurs at p = 1/2. 3) The locus is the median from A to the midpoint of BC. | |
0a7e | Problem:
Show that
$$
\frac{1}{2^{2}}+\frac{1}{3^{2}}+\ldots+\frac{1}{n^{2}}<\frac{2}{3}
$$ | [
"Solution:\nSince\n$$\n\\frac{1}{j^{2}}<\\frac{1}{j(j-1)}=\\frac{1}{j-1}-\\frac{1}{j}\n$$\nwe have\n$$\n\\begin{gathered}\n\\sum_{j=k}^{n} \\frac{1}{j^{2}}<\\left(\\frac{1}{k-1}-\\frac{1}{k}\\right)+\\left(\\frac{1}{k}-\\frac{1}{k+1}\\right)+\\cdots+\\left(\\frac{1}{n-1}-\\frac{1}{n}\\right) \\\\\n=\\frac{1}{k-1}-\... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 5 | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0kmh | Problem:
Let $n$ be the answer to this problem. Box $B$ initially contains $n$ balls, and Box $A$ contains half as many balls as Box $B$. After 80 balls are moved from Box $A$ to Box $B$, the ratio of balls in Box $A$ to Box $B$ is now $\frac{p}{q}$, where $p, q$ are positive integers with $\operatorname{gcd}(p, q)=1$... | [
"Solution:\n\nOriginally, box $A$ has $n / 2$ balls and $B$ has $n$ balls. After moving, box $A$ has $n / 2-80$ balls and $B$ has $n+80$ balls. The answer to the problem is thus\n$$\n\\frac{100(n / 2-80)+(n+80)}{\\operatorname{gcd}(n / 2-80, n+80)}=\\frac{51 n-80 \\cdot 99}{\\operatorname{gcd}(n / 2-80, n+80)} \\st... | United States | HMMT November 2021 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | final answer only | 720 | |
0fwh | Problem:
Sei $A B C D E$ ein konvexes Fünfeck mit
$$
\angle B A C=\angle C A D=\angle D A E \quad \text{ und } \quad \angle A B C=\angle A C D=\angle A D E
$$
Die Diagonalen $B D$ und $C E$ treffen sich in $P$. Zeige, dass die Gerade $A P$ die Seite $C D$ in deren Mittelpunkt schneidet. | [
"Solution:\n\nDie Diagonalen $A C$ und $B D$ schneiden sich in $X$, die Diagonalen $A D$ und $C E$ schneiden sich in $Y$ und die Gerade $A P$ schneide $C D$ in $M$. Wir wollen $C M = M D$ beweisen. Die Idee ist als erstes zu zeigen, dass $X Y$ und $C D$ parallel sind.\n\n\n\nAus den gegeben... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof only | null | |
0dl5 | Amina has a deck of cards numbered from $1$ to $101$. Initially, she holds $50$ cards in her hand, while the remaining cards are laid face up on the table. In one move, she can swap all $50$ cards in her hand for any $50$ cards from the table. Amina wants to obtain all the cards numbered from $1$ to $50$ in her hand in... | [] | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 51 | |
0ijy | Problem:
Write down at least one, and up to ten, different 3-digit prime numbers. If you somehow fail to do this, we will ignore your submission for this problem. Otherwise, you're entered into a game with other teams. In this game, you start with 10 points, and each number you write down is like a bet: if no one else... | [
"Solution:\n\nThere are 143 three-digit primes. None of the following necessarily applies to the actual contest, but it might be useful to think about. Suppose that you're trying to maximize your expected score on this problem. Then you should write down a number if you think the probability that someone else is wr... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Expected values",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | final answer only | null | |
0bpe | Problem:
Fie $m$ şi $n$ numere naturale, cu $m \geq 2$ şi $n \geq 3$. Demonstraţi că există $m$ numere naturale nenule distincte $a_{1}, a_{2}, a_{3}, \ldots, a_{m}$, toate divizibile cu $n-1$, astfel încât
$$
\frac{1}{n}=\frac{1}{a_{1}}-\frac{1}{a_{2}}+\frac{1}{a_{3}}-\ldots+(-1)^{m-1} \frac{1}{a_{m}}
$$ | [
"Solution:\n\nDacă $\\left(a_{n}\\right)_{n \\geq 1}$ este o progresie geometrică având raţia $q=n-1$, atunci\n$$\n\\frac{1}{a_{1}}-\\frac{1}{a_{2}}+\\ldots+(-1)^{p-1} \\frac{1}{a_{p}}=\\frac{1/a_{1}+1/q \\cdot(-1)^{p-1} 1/a_{p}}{1+1/q}, \\forall p \\in \\mathbb{N}^{*}\n$$\nRelaţia precedentă se scrie\n$$\n\\frac{1... | Romania | Olimpiada Naţională de Matematică | [
"Number Theory > Other",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0l4v | Sixteen chairs are arranged in a row. Eight people each select a chair in which to sit so that no person sits next to two other people. Let $N$ be the number of subsets of the 16 chairs that could be selected. Find the remainder when $N$ is divided by 1000. | [
"The problem is equivalent to counting the rearrangements of PPPPPPPPEEEEEEEE (standing for people and empty seats) where no three Ps appear together. Suppose such an arrangement contains $q$ pairs $PP$ and $8 - 2q$ Ps not adjacent to another P, where $0 \\le q \\le 4$. There are 9 spaces before, after, or between ... | United States | AIME II | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 907 | |
088a | Problem:
Quanti sono i numeri primi che possono essere espressi nella forma $n^{n+1}+1$, con $n$ intero positivo?
(A) 0
(B) 1
(C) 2
(D) più di 2, ma in numero finito
(E) infiniti. | [
"Solution:\n\nLa risposta è (B). Consideriamo il caso in cui $n$ sia dispari. Allora una qualunque sua potenza è dispari, e il numero successivo è pari. L'unico numero primo pari è $2$, ed è possibile ottenerlo solo nel caso $n=1$. Consideriamo ora il caso in cui $n$ sia pari. In questo caso $n+1$ è dispari, ed è d... | Italy | Progetto Olimpiadi di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | B | |
0dsl | Let $n$ be a positive integer. Show that there exists an integer $m$ such that
$$
2018m^2 + 20182017m + 2017
$$
is divisible by $2^n$. | [
"We shall show more generally that $am^2 + bm + c \\equiv 0 \\pmod{2^n}$ has a solution for all $n$ whenever $b$ is odd and $a$ or $c$ is even. For $n = 1$, take $m = 0$ if $c$ is even and $m = 1$ if $c$ is odd. Now suppose the claim is true for $n$. If $c$ is even, then by assumption, the congruence $2at^2 + bt + ... | Singapore | Singapore Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0262 | Problem:
Os números $a$ e $b$ são inteiros positivos e satisfazem $96 a^{2} = b^{3}$. Qual é o menor valor de $a$? | [
"Solution:\n\nFatorando $96$ temos: $2^{5} \\times 3 \\times a^{2} = b^{3}$. Para que $2^{5} \\times 3 \\times a^{2}$ seja um cubo, o número $a$ deve ser da forma: $a = 2^{n} \\times 3^{m}$. Assim,\n$$\n2^{5} \\times 3 \\times a^{2} = 2^{5} \\times 3 \\times (2^{n} \\times 3^{m})^{2} = 2^{5+2n} \\times 3^{1+2m} = b... | Brazil | Lista 4 | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 12 | |
0bx0 | Let $A_1$, $B_1$ and $C_1$ be the feet of the altitudes in the acute triangle $ABC$. On the line segments $B_1C_1$, $C_1A_1$, $A_1B_1$ one considers the points $X$, $Y$, and $Z$, respectively, such that
$$
\frac{C_1X}{XB_1} = \frac{b \cos C}{c \cos B}, \quad \frac{A_1Y}{YC_1} = \frac{c \cos A}{a \cos C} \quad \text{and... | [
"Since $b \\cos C = A_1C$ and $c \\cos B = BA_1$, we deduce that $\\frac{C_1X}{XB_1} = \\frac{CA_1}{A_1B}$ or $\\frac{C_1X}{C_1B_1} = \\frac{CA_1}{CB}$ or, equivalently, $\\frac{C_1X}{CA_1} = \\frac{C_1B_1}{CB}$. (1)\n\nTriangles $AC_1B_1$ and $ACB$ being similar, we obtain $\\frac{C_1B_1}{CB} = \\frac{AC_1}{AC}$, ... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasin... | English | proof only | null | |
0jt4 | Problem:
Let $\mathcal{V}$ be the volume enclosed by the graph
$$
x^{2016}+y^{2016}+z^{2}=2016
$$
Find $\mathcal{V}$ rounded to the nearest multiple of ten. | [
"Solution:\nAnswer: 360\nLet $R$ be the region in question. Then we have\n$$\n[-1,1]^2 \\times[-\\sqrt{2014}, \\sqrt{2014}] \\subset R \\subset[-\\sqrt[2016]{2016}, \\sqrt[2016]{2016}]^2 \\times[-\\sqrt{2016}, \\sqrt{2016}]\n$$\nWe find some bounds: we have\n$$\n\\sqrt{2016}<\\sqrt{2025}=45\n$$\nBy concavity of $\\... | United States | HMMT November | [
"Geometry > Solid Geometry > Volume",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | final answer only | 360 | |
0jhm | Problem:
Let $ABC$ be an obtuse triangle with circumcenter $O$ such that $\angle ABC = 15^{\circ}$ and $\angle BAC > 90^{\circ}$. Suppose that $AO$ meets $BC$ at $D$, and that $OD^{2} + OC \cdot DC = OC^{2}$. Find $\angle C$. | [
"Solution:\n\nLet the radius of the circumcircle of $\\triangle ABC$ be $r$.\n\n$$\n\\begin{gathered}\nOD^{2} + OC \\cdot DC = OC^{2} \\\\\nOC \\cdot DC = OC^{2} - OD^{2} \\\\\nOC \\cdot DC = (OC + OD)(OC - OD) \\\\\nOC \\cdot DC = (r + OD)(r - OD)\n\\end{gathered}\n$$\n\nBy the power of the point at $D$,\n\n$$\n\\... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 35° | |
08ul | Suppose for a triangle $ABC$ the following condition is satisfied:
A circle $O$ going through the vertices $B, C$ intersects the line segments $AB$ and $AC$ (excluding the end-points) at the points $D$ and $E$, respectively, and $AD + AE = BC$ is satisfied.
Let $I$ be the in-center of the triangle $ABC$ and suppose tha... | [
"From $AD + AE = BC$ it follows that we can choose a point $F$ on the side $BC$ in such a way that both $AD = CF$ and $AE = BF$ are satisfied. We have $DP = CP$, since the angles the cords $DP$ and $CP$ subtend on the circle $O$ are equal. We also have $\\angle PDA = \\angle PCB = \\angle PCF$, since the points $P,... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
0jqy | Problem:
Find, with proof, the largest possible value of
$$
\frac{x_{1}^{2}+\cdots+x_{n}^{2}}{n}
$$
where real numbers $x_{1}, \ldots, x_{n} \geq -1$ are satisfying $x_{1}^{3}+\cdots+x_{n}^{3}=0$. | [
"Solution:\nFor any $i$, we have $0 \\leq (x_{i}+1)(x_{i}-2)^{2} = x_{i}^{3} - 3x_{i}^{2} + 4$. Adding all of these we deduce that\n$$\n\\sum_{i=1}^{n} x_{i}^{2} \\leq \\frac{1}{3} \\sum_{i=1}^{n} (x_{i}^{3} + 4) = \\frac{4}{3} n.\n$$\nEquality occurs, for example, when $n=9$, $x_{1}=\\cdots=x_{8}=-1$ and $x_{9}=2$... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | 4/3 | |
0ebh | Problem:
Poišči vse pare naravnih števil $a$ in $b$, za katere je $2 a^{b}=a b+3$. | [
"Solution:\n\n1. način. Ker je $b$ naravno število, $a$ deli $2 a^{b}$ in $a b$, torej $a$ deli tudi $3$. Ker je $3$ praštevilo, je $a=1$ ali $a=3$. Če je $a=1$, dobimo enačbo $2 = b + 3$, ki pa nima rešitev v naravnih številih. Torej je $a=3$, od koder sledi $2 \\cdot 3^{b} = 3b + 3$ oziroma $2 \\cdot 3^{b-1} = b ... | Slovenia | 59. matematično tekmovanje srednješolcev Slovenije | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | a=3, b=1 | |
0b9m | Show that $\frac{1}{\pi} \int_{\sin \frac{\pi}{13}}^{\cos \frac{\pi}{13}} \sqrt{1-x^2} \, dx$ is a rational number. | [
"Thus $F(t) = -t + k,\\ k \\in \\mathbb{C}$. Since $F(\\pi/4) = 0$, it follows that $k = \\pi/4$, implying $F(t) = \\pi/4 - t$. Consequently, $F(\\pi/13)/\\pi = (\\pi/4 - \\pi/13)/\\pi = 9/52$, which is a rational number."
] | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Calculus > Integral Calculus > Techniques > Single-variable",
"Calculus > Differential Calculus > Derivatives",
"Precalculus > Trigonometric functions"
] | null | proof only | null | |
003t | Sean $a_1, a_2, \dots, a_n$ números positivos no necesariamente distintos. La suma de todos los productos tomados dos a dos ($a_i a_j$ con $i < j$) es igual a $1$. Demostrar que existe un número entre ellos tal que la suma de los números restantes es menor que $\sqrt{2}$. | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | Español | proof only | null | |
07iq | Find all the functions $f : \mathbb{Q}[x] \to \mathbb{Q}[x]$ such that
i. For all $P, Q \in \mathbb{Q}[x]$ we have $f(P + Q) = f(P) + f(Q)$;
ii. For all $P \in \mathbb{Q}[x]$, we have $\text{gcd}(P, f(P)) = 1$ if and only if $P$ is square-free.
(We call a polynomial $P \in \mathbb{Q}[x]$ square-free if there is no n... | [
"Take any polynomial $Q(x)$ with rational coefficients then the function $f$ meets the condition of the problem if and only if $g(P) = f(P) - P \\cdot Q$ satisfies. Where $P(x)$ is a polynomial with rational coefficients. We shall prove this claim in one direction. In the opposite direction, just change $Q$ by $-Q$... | Iran | 41th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Abstract Algebra > Ring Theory",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All functions are f(P) = C·P'(x) + P(x)·Q(x), where C is a nonzero rational constant and Q(x) is an arbitrary polynomial in Q[x]. | |
09ms | Let $BE$ and $CF$ be altitudes of an acute-angled triangle $ABC$. The segment $AD$ is the diameter of the circumcircle of $ABC$. Let $M$ be the midpoint of side $BC$. The internal common tangents of the incircles of triangles $BMF$ and $CME$ intersect at point $K$. Prove that $K$, $M$, and $D$ are collinear. | [
"Denote the incircles of $BMF$, $CME$ by $\\omega_1$, $\\omega_2$, respectively, with their centers denoted by $I_1, I_2$. Points $P$ and $Q$ are the midpoints of $BF$ and $CE$, respectively. Point $N$ is the intersection of external common tangents of $\\omega_1$ and $\\omega_2$. The orthocenter of triangle $ABC$ ... | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Polar tr... | English | proof only | null | |
046j | Find all functions $f : \mathbb{Z} \to \mathbb{Z}$ that satisfy the following equation for any integers $a, b, c$:
$$
2f(a^2 + b^2 + c^2) - 2f(ab + bc + ca) = (f(a - b))^2 + (f(b - c))^2 + (f(c - a))^2.
$$ | [
"Taking $a = b = c = 0$, we have $3(f(0))^2 = 0$, which implies $f(0) = 0$.\n\nTaking $a = 1, b = 0, c = 0$, we have $2f(1) = (f(1))^2 + (f(-1))^2$. Thus, $(f(1) - 1)^2 + (f(-1))^2 = 1$. This means either $f(1) = 1$ and $f(-1) = \\pm 1$, or $f(-1) = 0$ and $f(1) = 0$ or $2$.\n\nTaking $a = 1, b = 1, c = 0$, we have... | China | China-TST-2023B | [
"Algebra > Algebraic Expressions > Functional Equations"
] | English | proof and answer | Two functions: f(n) = 0 for all integers n, and f(n) = n for all integers n. | |
04hc | Višnja decided to write all integers from $1$ to $2014$ on a board in some order. Her brother Marijan will write the absolute value of the difference of each pair of adjacent numbers, and then erase the previously written numbers. Marijan will repeat this procedure until there is only one number left on the board. Dete... | [] | Croatia | Mathematica competitions in Croatia | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other"
] | English | proof and answer | 2012 | |
0gxe | Compare two numbers:
$$
\sqrt{2008 + \sqrt{2009}} + \sqrt{2009 + \sqrt{2008}} \text{ and } \sqrt{2008 + \sqrt{2008}} + \sqrt{2009 + \sqrt{2009}}
$$ | [
"Reformulate this task in the general case: for the positive distinct real numbers $a, b$ compare two numbers: $A = \\sqrt{a+\\sqrt{b}} + \\sqrt{b+\\sqrt{a}}$ and $B = \\sqrt{a+\\sqrt{a}} + \\sqrt{b+\\sqrt{b}}$. Prove that $A > B$.\n\nIs equivalent to $A^2 > B^2 \\Leftrightarrow \\sqrt{a+\\sqrt{b}} \\cdot \\sqrt{b+... | Ukraine | 49th Mathematical Olympiad in Ukraine | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | sqrt(2008 + sqrt(2009)) + sqrt(2009 + sqrt(2008)) > sqrt(2008 + sqrt(2008)) + sqrt(2009 + sqrt(2009)) | |
0080 | Find the sum of all products $a_1a_2 \cdots a_{50}$ where $a_1, a_2, \ldots, a_{50}$ are distinct positive integers not exceeding $101$ and such that no two of them have sum $101$. | [
"We distinguish between two cases for an admissible $50$-tuple $a_1, a_2, \\ldots, a_{50}$.\n\na) If no $a_i$ is equal to $101$ then $a_1, a_2, \\ldots, a_{50}$ contains exactly one number from every pair $(i, 101-i)$, $1 \\le i \\le 50$. Hence there are $2^{50}$ choices for $a_1, a_2, \\ldots, a_{50}$, and each re... | Argentina | National Olympiad of Argentina | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 51 · 101^50 | |
0b5k | Of the vertices of a cube, 7 of them have assigned the value $0$, and the eighth the value $1$. A *move* is selecting an edge and increasing the numbers at its ends by an integer value $k > 0$. Prove that after any finite number of moves, the g.c.d. of the $8$ numbers at vertices is equal to $1$. | [
"Let us alternately colour the vertices in black and white. After any move, the difference between the sums of the numbers at the black and the white vertices remains $1$, therefore the g.c.d. of the $8$ numbers is equal to $1$ (as it is dividing the difference of the sums mentioned above)."
] | Romania | Local Mathematical Competitions | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
01lj | Let $S(n)$ be the sum of all digits in the decimal representation of $n \in \mathbb{N}$. Do there exist integers $n$ such that
$$
a)\quad n - S(n) = 3 \cdot 2010? \qquad b)\quad n - S(n) = 3 \cdot 2011?
$$
(I. Gorodnin) | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | a) Yes, for example n = 6042. b) No. | |
08lx | Problem:
Let $a, b, c, d, e, f$ are nonzero digits such that the natural numbers $\overline{a b c}$, $\overline{d e f}$ and $\overline{a b c d e f}$ are squares.
a) Prove that $\overline{a b c d e f}$ can be represented in two different ways as a sum of three squares of natural numbers.
b) Give an example of such a num... | [
"Solution:\na) Let $\\overline{a b c}=m^{2}$, $\\overline{d e f}=n^{2}$ and $\\overline{a b c d e f}=p^{2}$, where $11 \\leq m \\leq 31, 11 \\leq n \\leq 31$ are natural numbers. So, $p^{2}=1000 \\cdot m^{2}+n^{2}$. But $1000=30^{2}+10^{2}=18^{2}+26^{2}$. We obtain the following relations\n$$\n\\begin{gathered}\np^... | JBMO | 2008 Shortlist JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Other"
] | null | proof and answer | 225625 | |
0k0w | Problem:
In an $n \times n$ square array of $1 \times 1$ cells, at least one cell is colored pink. Show that you can always divide the square into rectangles along cell borders such that each rectangle contains exactly one pink cell. | [
"Solution:\n\nWe claim that the statement is true for arbitrary rectangles. We proceed by induction on the number of marked cells. Our base case is $k=1$ marked cell, in which case the original rectangle works.\n\nTo prove it for $k$ marked cells, we split the rectangle into two smaller rectangles, both of which co... | United States | HMMT February | [
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0i25 | Twenty-one girls and twenty-one boys took part in a mathematical competition. It turned out that
a. each contestant solved at most six problems, and
b. for each pair of a girl and a boy, there was at least one problem that was solved by both the girl and the boy.
Prove that there is a problem that was solved by at l... | [
"**First Solution.** (By Zhiqiang Zhang, China) We proceed indirectly. Assume that no problem is both girl-easy and boy-easy.\n\n**Lemma 1.** Let $A = \\{p_1, p_2, \\dots, p_k\\}$ denote the set of all girl-hard problems. Let $B = \\{p_{k+1}, p_{k+2}, \\dots, p_{k+m}\\}$ denote the set of all boy-hard problems that... | United States | USA IMO | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
0d2v | Determine whether it is possible to place the integers $1,2, \ldots, 2012$ in a circle in such a way that the 2012 products of adjacent pairs of numbers leave pairwise distinct remainders when divided by 2013. | [
"Assume that it is possible to place the integers $1,2, \\ldots, 2012$ in a circle in such a way that the 2012 products of adjacent pairs of numbers leave pairwise distinct remainders when divided by 2013. Let $a_{1}, a_{2}, \\ldots, a_{2012}$ be such a reordering of the integers $1,2, \\ldots, 2012$ on the circle.... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)"
] | English | proof and answer | Not possible | |
0kkr | Problem:
For positive integers $n$, let $f(n)$ be the product of the digits of $n$. Find the largest positive integer $m$ such that
$$
\sum_{n=1}^{\infty} \frac{f(n)}{m^{\left\lfloor\log_{10} n\right\rfloor}}
$$
is an integer. | [
"Solution:\n\nWe know that if $S_{\\ell}$ is the set of all positive integers with $\\ell$ digits, then\n$$\n\\begin{aligned}\n& \\sum_{n \\in S_{\\ell}} \\frac{f(n)}{k^{\\left\\lfloor\\log_{10}(n)\\right\\rfloor}} = \\sum_{n \\in S_{\\ell}} \\frac{f(n)}{k^{\\ell-1}} = \\frac{(0+1+2+\\ldots+9)^{\\ell}}{k^{\\ell-1}}... | United States | HMMT November 2021 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 2070 | |
0gtc | 2022 points of a circle are uniformly marked (arc distances between neighbouring marked points coincide). $k$ arcs with different lengths and with endpoints at marked points are chosen so that no arc lies inside another one. Find the largest possible value of $k$. | [
"Answer: 1011.\nWe show that the answer for $2n$ points is $n$. Let us numerate marked points in clockwise order by $1, 2, \\dots, 2n$. The arc starting at point $a$ and ending at $b$ in clockwise order will be denoted by $[a, b]$.\n\nThe $n$ arcs chosen as $[1, 2], [2, 4], [3, 6], [4, 8], \\dots, [n-1, 2n-2], [n, ... | Turkey | Team Selection Test | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 1011 | |
0ju4 | Problem:
How many equilateral hexagons of side length $\sqrt{13}$ have one vertex at $(0,0)$ and the other five vertices at lattice points?
(A lattice point is a point whose Cartesian coordinates are both integers. A hexagon may be concave but not self-intersecting.) | [
"Solution:\n\nWe perform casework on the point three vertices away from $(0,0)$. By inspection, that point can be $( \\pm 8, \\pm 3),( \\pm 7, \\pm 2),( \\pm 4, \\pm 3),( \\pm 3, \\pm 2),( \\pm 2, \\pm 1)$ or their reflections across the line $y=x$. The cases are as follows:\n\nIf the third vertex is at any of $( \... | United States | HMMT February | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 216 | |
077a | Problem:
Let $\Gamma_{1}$ and $\Gamma_{2}$ be two circles of unequal radii, with centres $O_{1}$ and $O_{2}$ respectively, in the plane intersecting in two distinct points $A$ and $B$. Assume that the centre of each of the circles $\Gamma_{1}$ and $\Gamma_{2}$ is outside the other. The tangent to $\Gamma_{1}$ at $B$ i... | [
"Solution:\n\n\nLet $\\angle C B A=\\alpha$ and $\\angle D B A=\\beta$. Then $\\angle B D A=\\alpha$ and $\\angle B C A=\\beta$. We also observe that $\\angle A O_{1} O_{2}=\\left(\\angle A O_{1} B / 2\\right)=\\alpha$ and, simiarly, $\\angle A O_{2} O_{1}=\\beta$. Hence\n$$\n\\angle O_{1} ... | India | INMO | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
03n1 | Let $A$, $B$, and $F$ be positive integers, and assume $A < B < 2A$. A flea is at the number $0$ on the number line. The flea can move by jumping to the right by $A$ or by $B$. Before the flea starts jumping, Lavaman chooses finitely many intervals $\{m+1, m+2, \dots, m+A\}$ consisting of $A$ consecutive positive integ... | [
"Let $B = A + C$ where $\\frac{A}{n + 1} \\leq C < \\frac{A}{n}$.\n\nFirst, here is an informal sketch of the proof.\n\nLavaman's strategy: Use only safe intervals with $nA + C - 1$ integers. The flea will start at position $[1, C]$ from the left, which puts him at position $[nA, nA + C - 1]$ from the right. After ... | Canada | Kanada | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smo... | null | proof and answer | F = (n - 1)A + B, where n is the positive integer with A/(n+1) ≤ B − A < A/n. | |
0atf | Problem:
Solve the equation $\left(2-x^{2}\right)^{x^{2}-3 \sqrt{2} x+4}=1$. | [] | Philippines | Philippine Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof and answer | x = -1, 1, 2√2 | |
0czs | Find all quadruples $ (x, y, z, w) $ of integers satisfying the system of equations
$$
x + y + z + w = x y + y z + z x + w^{2} - w = x y z - w^{3} = -1.
$$ | [
"The system is equivalent to\n$$\n\\left\\{\n\\begin{array}{l}\nx + y + z = -(w + 1) \\\\\nx y + y z + z x = -\\left(w^{2} - w + 1\\right) \\\\\nx y z = w^{3} - 1\n\\end{array}\n\\right.\n$$\nMultiplying the first two equations of the system we get\n$$\n(x + y + z)(x y + y z + z x) = w^{3} + 1.\n$$\nEliminating $w$... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | (1, 1, -2, -1), (1, -2, 1, -1), (-2, 1, 1, -1) | |
0fir | Problem:
Sean $a$, $b$ y $c$ números reales no nulos (con suma no nula) tales que:
$$
\frac{1}{a} + \frac{1}{b} + \frac{1}{c} = \frac{1}{a+b+c}
$$
Prueba que también se verifica:
$$
\frac{1}{a^{1999}} + \frac{1}{b^{1999}} + \frac{1}{c^{1999}} = \frac{1}{a^{1999} + b^{1999} + c^{1999}}
$$ | [
"Solution:\nDel enunciado, deducimos: $\\frac{1}{a} + \\frac{1}{b} + \\frac{1}{c} = \\frac{1}{a+b+c} \\Leftrightarrow \\frac{bc + ac + ab}{abc} = \\frac{1}{a+b+c}$\nOperando en la última ecuación, tenemos:\n$$\n\\begin{gathered}\n(a+b+c)(bc + ac + ab) = abc \\\\\n(a+b)(bc + ac + ab) + c(bc + ac + ab) = abc \\\\\n(a... | Spain | XXXV Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
00tg | Let $I$ and $O$ be the incenter and the circumcenter of a triangle $ABC$, respectively, and let $s_a$ be the exterior bisector of angle $\angle BAC$. The line through $I$ perpendicular to $IO$ meets the lines $BC$ and $s_a$ at points $P$ and $Q$, respectively. Prove that $IQ = 2IP$.
**Proposed by Serbia** | [
"Denote by $I_b$ and $I_c$ the respective excenters opposite to $B$ and $C$. Also denote the midpoint of side $BC$ by $D$, the midpoint of the necessarily major arc $BAC$ by $M$, and the midpoint of segment $AM$ by $N$. Recall that $M$ is on the perpendicular bisector of $BC$, i.e. on line $OD$. Points $I$, $O$, $D... | Balkan Mathematical Olympiad | 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 > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0isz | Problem:
Sarah is deciding whether to visit Russia or Washington, DC for the holidays. She makes her decision by rolling a regular 6-sided die. If she gets a $1$ or $2$, she goes to DC. If she rolls a $3$, $4$, or $5$, she goes to Russia. If she rolls a $6$, she rolls again. What is the probability that she goes to DC... | [
"Solution:\n\n$\\boxed{\\dfrac{2}{5}}$ On each roll, the probability that Sarah decides to go to Russia is $\\dfrac{3}{2}$ times the probability she decides to go to DC. So, the total probability that she goes to Russia is $\\dfrac{3}{2}$ times the total probability that she goes to DC. Since these probabilities su... | United States | 1st Annual Harvard-MIT November Tournament | [
"Statistics > Probability > Counting Methods > Other",
"Math Word Problems"
] | null | final answer only | 2/5 | |
09h0 | For a finite set $A$ of integers, denote by $|A|$ the cardinality of $A$, and denote by $\Sigma(A)$ the sum of all elements of $A$. Let $p$ be a prime number, and $A$ be a given set of positive integers such that $|A| > p$.
Let $N_{\text{тэгш}}$ denote the number of subsets $B$ of $A$ such that $|B|$ is even and $\Sigm... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof only | null | |
0id8 | Problem:
A pair of successive numbers in the same row is called a switch pair if one number in the pair is even and the other is odd.
Prove that the number of switch pairs in row $n$ is at most twice the number of odd numbers in row $n$. | [
"Solution:\nEach switch pair contains an odd number, and each odd number can belong to at most two switch pairs (since it has only two neighbors)."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0j60 | Problem:
Josh takes a walk on a rectangular grid of $n$ rows and 3 columns, starting from the bottom left corner. At each step, he can either move one square to the right or simultaneously move one square to the left and one square up. In how many ways can he reach the center square of the topmost row? | [
"Solution:\n\nAnswer: $2^{n-1}$\n\nNote that Josh must pass through the center square of each row. There are 2 ways to get from the center square of row $k$ to the center square of row $k+1$. So there are $2^{n-1}$ ways to get to the center square of row $n$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 2^{n-1} | |
04ul | Find all pairs of positive integers $a, b$ such that $a^{a-b} = b^a$. | [] | Czech Republic | First Round | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | All pairs (k^k, k^{k-1}) for positive integers k. | |
0c6o | Let $ABCD$ be a square and $E$ an arbitrary point on the side $(CD)$. On the outside of the triangle $ABE$ one considers the squares $ENMA$ and $EBQP$. Prove that:
a) $ND = PC$;
b) $ND \perp PC$. | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0b6a | Given a triangle $ABC$, let $D$ be the point where the incircle of the triangle $ABC$ touches the side $BC$. A circle through the vertices $B$ and $C$ is tangent at point $E$ to the incircle of the triangle $ABC$. Show that the line $DE$ passes through the excentre of the triangle $ABC$ corresponding to the vertex $A$. | [
"\n\nLet $I$ be the incentre of the triangle $ABC$, let $I_A$ be the excentre corresponding to the vertex $A$, and notice that the vertices $B$ and $C$ both lie on the circle of diameter $II_A$. The line $DI_A$ meets again the latter circle at point $K$, and the lines $BC$ and $IK$ meet at ... | Romania | 2010 Eighth IMAR MATHEMATICAL COMPETITION | [
"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"
] | null | proof only | null | |
010p | Problem:
In a triangle $A B C$ it is given that $2|A B|=|A C|+|B C|$. Prove that the incentre of $A B C$, the circumcentre of $A B C$, and the midpoints of $A C$ and $B C$ are concyclic. | [
"Solution:\n\nLet $N$ be the midpoint of $B C$ and $M$ the midpoint of $A C$. Let $O$ be the circumcentre of $A B C$ and $I$ its incentre (see Figure 8). Since $\\angle C M O=\\angle C N O=90^\\circ$, the points $C, N, O$ and $M$ are concyclic (regardless of whether $O$ lies inside the triangle $A B C$). We now hav... | Baltic Way | Baltic Way | [
"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 > Analytic / Coordinate M... | null | proof only | null | |
02aa | Problem:
Divisão de números grandes - Determine o valor de $123456123456 \div 10000001$. | [
"Solution:\n\nÉ claro que com números tão grandes, o objetivo da questão não é efetuar a divisão. Em vez disso, decompomos o número em partes convenientes.\n$$\n\\begin{aligned}\n123456123456 & = 123456000000 + 123456 = 123456 \\times 1000000 + 123456 \\\\\n& = 123456 \\times (1000000 + 1) = 123456 \\times 1000001\... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 123456 | |
0l6g | Problem:
Determine whether there exist infinitely many pairs of distinct positive integers $m$ and $n$ such that $2^{m} + n$ divides $2^{n} + m$. | [
"Solution:\n\nLet $k$ be a positive integer, and set $m = 2^{k}$ and $n = p - 2^{2^{k}}$ for prime $p$ to be chosen later. We want $2^{m} + n = p$ to divide $2^{p - 2^{2^{k}}} + 2^{k}$, which is equivalent to having\n\n$$\n0 \\equiv 2^{p - 2^{2^{k}}} + 2^{k} \\equiv 2^{1 - 2^{2^{k}}} + 2^{k} \\equiv 2^{1-2^{2^{k}}}... | United States | HMIC | [
"Number Theory > Divisibility / Factorization",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | Yes, infinitely many such pairs exist. | |
0g6j | 已知多項式函數 $f, g$ 為實數映至實數。試求出所有的多項式函數對 $(f(x), g(x))$ 使得: 對任意實數 $x$,
$$
f(f(f(f(x)))) = g(g(g(g(x))))
$$
成立。 | [
"所有滿足題意的多項式為:\n(i) $f(x) = g(x)$\n(ii) $f(x) = S(x+a) - a, g(x) = -S(x+a) - a$, 其中 $a$ 是任意實數, $S$ 是任意奇多項式函數 (即 $S(x) = \\sum a_i x^{2k+1}$).\n\n易知 $f(x)$ 與 $g(x)$ 的次數相同。令 $F(x) = f(f(x)) = \\sum_{i=0}^{m} a_m x^{m-i}$,\n$G(x) = g(g(x)) = \\sum_{i=0}^{m} b_m x^{m-i}$。則多項式 $F(x), G(x)$ 滿足 $F(F(x)) = G(G(x))$。記 $H(x) ... | Taiwan | 二〇一二數學奧林匹亞競賽第三階段選訓營 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = g(x) | |
0b9n | In a square of side length $60$, $121$ distinct points are given. Show that among them there exists three points which are vertices of a triangle with an area not exceeding $30$. | [
"Divide the square into $60$ rectangles $5 \\times 12$. By the pigeonhole principle, there are three points among the given ones inside one of the rectangles. The area of this triangle does not exceed half of the area of the rectangle, that is $5 \\cdot 12/2 = 30$, as needed."
] | Romania | 62nd ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
03il | Problem:
For any real number $t$, denote by $[t]$ the greatest integer which is less than or equal to $t$. For example: $[8]=8$, $[\pi]=3$ and $[-5 / 2]=-3$. Show that the equation
$$
[x]+[2x]+[4x]+[8x]+[16x]+[32x]=12345
$$
has no real solution. | [
"Solution:\nLet $x$ be a real number. For any integer $n \\geq 0$, $[2^n x]$ is the greatest integer less than or equal to $2^n x$.\n\nLet $S(x) = [x] + [2x] + [4x] + [8x] + [16x] + [32x]$.\n\nLet $x = a + y$, where $a$ is an integer and $0 \\leq y < 1$.\nThen $[2^k x] = [2^k a + 2^k y] = 2^k a + [2^k y]$ for $k = ... | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
045s | Let $C = \{z \in \mathbb{C} \mid |z| = 1\}$ be the unit circle in the complex plane. 240 complex numbers $z_1, z_2, \dots, z_{240} \in C$ (can be repeated) satisfy the following conditions:
(1) for any open arc $\Gamma$ of length $\pi$ on $C$, there are at most 200 $j$'s $(1 \le j \le 240)$ such that $z_j \in \Gamma$;
... | [
"The maximum is $80 + 40\\sqrt{3}$. Take 80 of 1's, 40 of each of $\\exp(\\frac{\\pi}{6}i)$, $\\exp(-\\frac{\\pi}{6}i)$, $i$, $-i$. It is straightforward to check that they satisfy (1), (2), and their sum equals $80 + 40\\sqrt{3}$. We need to show this is the maximum value.\n\nLet $z_1, z_2, \\dots, z_{240}$ satisf... | China | China National Team Selection Test | [
"Algebra > Intermediate Algebra > Complex numbers",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 80 + 40*sqrt(3) | |
05s7 | Problem:
Trouver tous les triplets $\left(a, b, c\right)$ d'entiers strictement positifs tels que
$$
3^{a}-5^{b}=c^{2}
$$ | [
"Solution:\nComme $3^{a}$ et $5^{b}$ sont tous deux impairs, on a $c$ pair donc $c^{2} \\equiv 0[4]$. Comme $5^{b} \\equiv 1[4]$, on doit avoir $3^{a} \\equiv 1[4]$, donc $a$ est pair. On écrit $a=2 a'$, et l'équation devient\n$$\n5^{b}=3^{2 a'}-c^{2}=\\left(3^{a'}-c\\right)\\left(3^{a'}+c\\right) .\n$$\nNotons que... | France | Préparation Olympique Française de Mathématiques - ENVOI 4 : POT-POURRI | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | (2, 1, 2) | |
0hhd | For a positive integer $n$ we write out all its divisors $1 = d_1 < d_2 < \cdots < d_k = n$. A divisor $d_i$ is called a good divisor if $d_{i-1}d_{i+1}$ is not divisible by $d_i$, $2 \le i \le k-1$. Find all $n$ for which the number of their good divisors is smaller than the number of their different prime divisors.
... | [
"**Answer.** $p^\\beta$ and $p^\\alpha q$, where $q > p^\\alpha$ where $p, q$ are prime numbers.\n\nSolution:\n\nFirst, let's prove that if a number has at least three different prime divisors, it is not suitable for us. Let $m$ be the number of different prime divisors and $p < q$ be the two smallest prime divisor... | Ukraine | 62nd Ukrainian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | All n of the form p^β and p^α q with q > p^α, where p and q are prime numbers. | |
0iim | Problem:
Suppose $ABC$ is a scalene right triangle, and $P$ is the point on hypotenuse $\overline{AC}$ such that $\angle ABP = 45^{\circ}$. Given that $AP = 1$ and $CP = 2$, compute the area of $ABC$. | [
"Solution:\n\nNotice that $\\overline{BP}$ bisects the right angle at $B$. Thus, we write $AB = 2x$, $BC = x$. By the Pythagorean theorem, $5x^{2} = 9$, from which the area $\\frac{1}{2}(x)(2x) = x^{2} = \\frac{9}{5}$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 9/5 | |
021m | Problem:
É dada uma circunferência $\mathcal{C}$. Construir, usando somente compasso, o centro de $\mathcal{C}$. | [] | Brazil | Nível 3 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0klt | Problem:
Let $AEF$ be a triangle with $EF=20$ and $AE=AF=21$. Let $B$ and $D$ be points chosen on segments $AE$ and $AF$, respectively, such that $BD$ is parallel to $EF$. Point $C$ is chosen in the interior of triangle $AEF$ such that $ABCD$ is cyclic. If $BC=3$ and $CD=4$, then the ratio of areas $\frac{[ABCD]}{[AEF]... | [
"Solution:\nRotate $\\triangle ABC$ around $A$ to $\\triangle AB' C'$, such that $B'$ is on segment $AF$. Note that as $BD \\parallel EF$, $AB=AD$. From this, $AB'=AB=AD$, and $B'=D$. Note that\n$$\n\\angle ADC' = \\angle ABC = 180 - \\angle ADC\n$$\nbecause $ABCD$ is cyclic. Therefore, $C$, $D$, and $C'$ are colli... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 5300 | |
0ern | How many grams of pure gold must be added to five grams of a mixture that consists of 10% gold and 90% of some other metal so that the mixture will contain 20% gold?
(A) 0.125 (B) 0.25 (C) 0.375 (D) 0.5 (E) 0.625 | [
"The initial $5$ g of mixture contains $10\\% \\times 5 = 0.5$ g of gold. If we add $x$ g of pure gold, then the mass of gold is $(0.5 + x)$ g and the total mass is $(5 + x)$ g. Therefore $0.5 + x = 20\\% \\times (5 + x) = 1 + 0.2x$, so $0.8x = 0.5$ and $x = 5/8 = 0.625$."
] | South Africa | South African Mathematics Olympiad First Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Decimals",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | MCQ | E | |
0fbc | Problem:
Se considera el conjunto de todos los polinomios de grado menor o igual que $4$ con coeficientes racionales.
a) Probar que tiene estructura de espacio vectorial sobre el cuerpo de los números racionales.
b) Probar que los polinomios $1$, $x-2$, $(x-2)^2$, $(x-2)^3$ y $(x-2)^4$ forman una base de este espaci... | [
"Solution:\n\n1) Es obvio\n\n2) Si consideramos\n$$\n\\lambda_{0} \\cdot 1+\\lambda_{1}(x-2)+\\lambda_{2}(x-2)^{2}+\\lambda_{3}(x-2)^{3}+\\lambda_{4}(x-2)^{4}=0 ; \\quad \\lambda_{i} \\in \\mathbb{Q} \\quad i=0, \\ldots, 4\n$$\nse tiene que $\\lambda_{i}=0 \\quad \\forall i$, es decir los polinomios $1$, $x-2$, $(x... | Spain | OME 10 | [
"Algebra > Linear Algebra > Vectors",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | −121 − 82(x−2) + 27(x−2)^2 + 24(x−2)^3 + 3(x−2)^4 | |
07m8 | Three real numbers $a$, $b$, $c$ satisfy the equations
$$
\begin{align*}
a + 2b + 3c &= 12, \\
2ab + 3ac + 6bc &= 48.
\end{align*}
$$
Solve for $a$, $b$, $c$. | [
"From $(a + 2b + 3c)^2 = a^2 + 4b^2 + 9c^2 + 2(2ab + 3ac + 6bc)$ we obtain\n$$\na^2 + 4b^2 + 9c^2 = 48.\n$$\nHence $(a-2b)^2 + (2b-3c)^2 + (a-3c)^2 = 2(a^2+4b^2+9c^2) - 2(2ab+3ac+6bc) = 96 - 96 = 0$. Therefore $a = 2b = 3c$, and so $a = 4$, $b = 2$ and $c = 4/3$."
] | Ireland | Irish Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | a = 4, b = 2, c = 4/3 | |
0405 | Let $P_0, P_1, P_2, \dots, P_n$ be $n+1$ points on a plane, and the minimum distance between each two points of them is $d$ ($d > 0$). Prove
$$
| P_0 P_1 | \cdot | P_0 P_2 | \cdots | P_0 P_n | > \left( \frac{d}{3} \right)^n \sqrt{(n+1)!}
$$ | [
"We may assume that $|P_0P_1| \\le |P_0P_2| \\le \\dots \\le |P_0P_n|$.\nAt first, we will prove that $|P_0P_k| > \\frac{d}{3}\\sqrt{k+1}$ for any positive integer $n$.\nObviously, $|P_0P_k| \\ge d \\ge \\frac{d}{3}\\sqrt{k+1}$ for $k = 1, 2, \\dots, 8$, and the second equality holds only when $k = 8$. Then we only... | China | China Mathematical Competition (Complementary Test) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle inequalities"
] | English | proof only | null | |
0cp2 | Points $A$, $B$, $C$ lie on a circle. Line $PB$ touches this circle at point $B$. Drop the perpendiculars $PA_1$ and $PC_1$ from $P$ onto the lines $AB$ and $BC$, respectively (points $A_1$ and $C_1$ belong to segments $AB$ and $BC$, respectively). Prove that $A_1C_1 \perp AC$.
Пусть точки $A$, $B$, $C$ лежат на окруж... | [
"Поскольку $\\angle PC_1B = \\angle PA_1B = 90^\\circ$, четырёхугольник $PA_1C_1B$ вписан (см. рис. 4).\n\nЗначит, $\\angle CC_1A_1 = 180^\\circ - \\angle A_1C_1B = \\angle A_1PB = 90^\\circ - \\angle A_1BP$.\n\nС другой стороны, $\\angle A_1BP = \\angle ACB = \\frac{1}{2} AB$.\n\nПоэтому $\\angle CC_1A_1 = \\angle... | Russia | Regional round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English; Russian | proof only | null | |
0ar7 | Problem:
The quotient of the sum and difference of two integers is $3$, while the product of their sum and difference is $300$. What are the integers? | [
"Solution:\nLet the two integers be $a$ and $b$ with $a > b$.\n\nLet $S = a + b$ and $D = a - b$.\n\nWe are told:\n\n$$\n\\frac{S}{D} = 3\n$$\n\nand\n\n$$\nS \\cdot D = 300.\n$$\n\nFrom the first equation, $S = 3D$.\n\nSubstitute into the second equation:\n\n$$\n(3D) \\cdot D = 300 \\\\\n3D^2 = 300 \\\\\nD^2 = 100 ... | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 20 and 10; -20 and -10 | |
0cka | Find the positive integers $a$ and $b$ fulfilling
$$
\frac{a}{(a,b)} = b + \frac{48 \cdot (a,b)}{[a,b]} \quad \text{and} \quad \frac{b}{(a,b)} = a - \frac{312 \cdot (a,b)}{[a,b]}
$$
where $(a, b) = \gcd(a, b)$ and $[a, b] = \text{lcm}(a, b)$. | [
"We start noticing that $a \\ge \\frac{a}{(a,b)} > b$.\nDenote $d = (a, b)$. Then $a = dx$, $b = dy$, $[a, b] = dxy$ and $(x, y) = 1$, $x > y$. This gives $x = dy + \\frac{48}{xy}$ and $y = dx - \\frac{312}{xy}$ (1).\nIt follows that $xy \\mid 48$ and $xy \\mid 312$, hence $xy \\mid (48, 312) = 24$.\nRelation (1) y... | Romania | 75th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | a = 16, b = 6 | |
05i4 | Problem:
Soit $\mathbb{R}_{+}^{*}$ l'ensemble des réels strictement positifs. Trouver toutes les fonctions $f: \mathbb{R}_{+}^{*} \longrightarrow \mathbb{R}_{+}^{*}$ telles que pour tous $x, y>0$, on ait
$$
f\left(\frac{f(x)}{y f(x)+1}\right)=\frac{x}{x f(y)+1}
$$ | [
"Solution:\n\nOn remarque tout d'abord que pour tout $z>0$, l'application $x \\in \\mathbb{R}^{+*} \\longmapsto \\frac{x}{x z+1} \\in ] 0 ; z^{-1}[$ est une bijection strictement croissante.\n\nMaintenant, si $f(x)=f\\left(x'\\right)$, on trouve que\n$$\n\\frac{x}{x f(1)+1}=f\\left(\\frac{f(x)}{1 \\times f(x)+1}\\r... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = x for all x > 0 | |
01af | Circles $S_1$ and $S_2$ intersect in points $P$ and $Q$ and lay inside an inscribed quadrilateral $ABCD$. $S_1$ touches the sides $AB$, $BC$ and $AD$; $S_2$ touches the sides $CD$, $BC$ and $AD$. The lines $PQ$, $AB$, $CD$ meet in one point. Prove that $BC \parallel AD$. | [
"\n\nProof by contradiction. Let $X$ be the intersection point of lines $AB$ and $CD$, $Y$ be the intersection point of lines $BC$ and $AD$. Let $O_1$ and $O_2$ be the centers of the circles. Then points $O_1$ and $O_2$ belong to the bisector of angle $BYA$. Therefore $PQ$ is perpendicular ... | Baltic Way | Baltic Way 2013 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0a94 | Problem:
A triangle, a line and three rectangles, with one side parallel to the given line, are given in such a way that the rectangles completely cover the sides of the triangle. Prove that the rectangles must completely cover the interior of the triangle. | [
"Solution:\nTake any point $P$ inside the triangle and draw through $P$ the line parallel to the given line as well as the line perpendicular to it. These lines meet the sides of the triangle in four points. Of these four, two must be in one of the three rectangles. Now if the two points are on the same line, then ... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 21 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0jap | Problem:
Given points $a$ and $b$ in the plane, let $a \oplus b$ be the unique point $c$ such that $a b c$ is an equilateral triangle with $a, b, c$ in the clockwise orientation.
Solve $(x \oplus (0,0)) \oplus (1,1) = (1,-1)$ for $x$. | [
"Solution:\n\nAnswer: $\\left(\\frac{1-\\sqrt{3}}{2}, \\frac{3-\\sqrt{3}}{2}\\right)$\n\nIt is clear from the definition of $\\oplus$ that $b \\oplus (a \\oplus b) = a$ and if $a \\oplus b = c$ then $b \\oplus c = a$ and $c \\oplus a = b$. Therefore $x \\oplus (0,0) = (1,1) \\oplus (1,-1) = (1-\\sqrt{3}, 0)$. Now t... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | ((1 - sqrt(3))/2, (3 - sqrt(3))/2) | |
0fyr | Problem:
Finde alle Paare von Primzahlen $(p, q)$ mit $3 \neq p+1$ so dass
$$
\frac{p^{3}+1}{q}
$$
das Quadrat einer natürlichen Zahl ist. | [
"Solution:\n\nEs lässt sich $p^{3}+1=(p+1)\\left(p^{2}-p+1\\right)$ faktorisieren. Da $q$ prim ist muss gelten $q \\mid p+1$ oder $q \\mid p^{2}-p+1$. Ferner ist $\\operatorname{ggT}\\left\\{p^{2}-p+1, p+1\\right\\}=\\operatorname{ggT}\\left\\{p^{2}-p+1-(p+1)^{2}+3(p+1), p+1\\right\\}=\\operatorname{ggT}\\{3, p+1\\... | Switzerland | IMO Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (3, 7) | |
0ke9 | Problem:
The points $(0,0)$, $(1,2)$, $(2,1)$, $(2,2)$ in the plane are colored red while the points $(1,0)$, $(2,0)$, $(0,1)$, $(0,2)$ are colored blue. Four segments are drawn such that each one connects a red point to a blue point and each colored point is the endpoint of some segment. The smallest possible sum of ... | [
"Solution:\n\n\n\nIf $(2,2)$ is connected to $(0,1)$ or $(1,0)$, then the other 6 points can be connected with segments of total length $3$, which is minimal. This leads to a total length of $3+\\sqrt{5}$.\n\nOn the other hand, if $(2,2)$ is connected to $(0,2)$ or $(2,0)$, then connecting ... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 305 | |
02q9 | Problem:
Alberto, Bernardo e Carlos disputaram uma corrida, na qual cada um deles correu com velocidade constante durante todo o percurso. Quando Alberto cruzou a linha de chegada, Bernardo e Carlos estavam 36 e 46 metros atrás dele, respectivamente. Quando Bernardo cruzou a linha de chegada, Carlos estava 16 metros a... | [
"Solution:\n\nSeja $x$ o comprimento em metros da pista. A distância entre Bernardo e Carlos era de 10 metros quando Alberto cruzou a linha de chegada, e era de 16 metros quando Bernardo cruzou a linha de chegada. Vemos assim que, durante o intervalo de tempo no qual Alberto e Bernardo completaram a corrida, Bernar... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | MCQ | A | |
011y | Problem:
Let $a$ be an odd integer. Prove that $a^{2^{n}} + 2^{2^{n}}$ and $a^{2^{m}} + 2^{2^{m}}$ are relatively prime for all positive integers $n$ and $m$ with $n \neq m$. | [
"Solution:\n\nRewriting $a^{2^{n}} + 2^{2^{n}} = a^{2^{n}} - 2^{2^{n}} + 2 \\cdot 2^{2^{n}}$ and making repeated use of the identity\n$$\na^{2^{n}} - 2^{2^{n}} = \\left(a^{2^{n-1}} - 2^{2^{n-1}}\\right) \\cdot \\left(a^{2^{n-1}} + 2^{2^{n-1}}\\right)\n$$\nwe get\n$$\n\\begin{gathered}\na^{2^{n}} + 2^{2^{n}} = \\lef... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null |
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