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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0bef | The function $f: \mathbb{R} \to \mathbb{R}$ has the property that every point of local minimum has a neighbourhood $(\alpha, \beta)$ so that $f$ is strictly convex on $(\alpha, \beta)$. Prove that the set of the points of local minimum is countable. | [
"Let $S$ be the set of points of local minimum of $f$.\n\nFor each $x \\in S$, by hypothesis, there exists an open interval $(\\alpha_x, \\beta_x)$ containing $x$ such that $f$ is strictly convex on $(\\alpha_x, \\beta_x)$.\n\nRecall that a strictly convex function on an interval has at most one point of local mini... | Romania | Shortlisted Problems for the 64th NMO | [
"Discrete Mathematics > Other"
] | null | proof only | null | |
074w | Problem:
Let $ABC$ be an acute-angled triangle with altitude $AK$. Let $H$ be its orthocentre and $O$ be its circumcentre. Suppose $KOH$ is an acute-angled triangle and $P$ its circumcentre. Let $Q$ be the reflection of $P$ in the line $HO$. Show that $Q$ lies on the line joining the mid-points of $AB$ and $AC$. | [
"Solution:\nLet $D$ be the mid-point of $BC$; $M$ that of $HK$; and $T$ that of $OH$. Then $PM$ is perpendicular to $HK$ and $PT$ is perpendicular to $OH$. Since $Q$ is the reflection of $P$ in $HO$, we observe that $P, T, Q$ are collinear, and $PT = TQ$. Let $QL$, $TN$ and $OS$ be the perpendiculars drawn respecti... | India | INMO | [
"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 only | null | |
01ez | Prove that for arbitrary positive integer $n$ the following inequality holds
$$
\frac{1}{1^2 + 2019} + \frac{1}{2^2 + 2019} + \dots + \frac{1}{n^2 + 2019} < \frac{1}{22}.
$$ | [] | Baltic Way | Baltic Way 2019 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
073c | Problem:
If $x, y, z$ are positive real numbers, prove that
$$
(x+y+z)^{2}(y z+z x+x y)^{2} \leq 3\left(y^{2}+y z+z^{2}\right)\left(z^{2}+z x+x^{2}\right)\left(x^{2}+x y+y^{2}\right)
$$ | [
"Solution:\nWe begin with the observation that\n$$\nx^{2}+x y+y^{2}=\\frac{3}{4}(x+y)^{2}+\\frac{1}{4}(x-y)^{2} \\geq \\frac{3}{4}(x+y)^{2}\n$$\nand similar bounds for $y^{2}+y z+z^{2}, z^{2}+z x+x^{2}$. Thus\n$$\n3\\left(x^{2}+x y+y^{2}\\right)\\left(y^{2}+y z+z^{2}\\right)\\left(z^{2}+z x+x^{2}\\right) \\geq \\fr... | India | INMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geome... | null | proof only | null | |
013h | Problem:
Consider a rectangle with side lengths $3$ and $4$, and pick an arbitrary inner point on each side. Let $x$, $y$, $z$ and $u$ denote the side lengths of the quadrilateral spanned by these points. Prove that $25 \leq x^{2}+y^{2}+z^{2}+u^{2} \leq 50$. | [
"Solution:\n\nLet $a$, $b$, $c$ and $d$ be the distances of the chosen points from the midpoints of the sides of the rectangle (with $a$ and $c$ on the sides of length $3$). Then\n$$\n\\begin{aligned}\nx^{2}+y^{2}+z^{2}+u^{2}= & \\left(\\frac{3}{2}+a\\right)^{2}+\\left(\\frac{3}{2}-a\\right)^{2}+\\left(\\frac{3}{2}... | Baltic Way | Baltic Way | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
07o5 | Find all integers $n$ for which
$$
f(n) = \frac{9n(n-5)^{2013} - n^2 - 7}{n^2 - 9n + 21}
$$
is an integer. | [
"Clearly $f(n)$ is an integer if $n^2 - 9n + 21 = 1$. Because $n^2 - 9n + 21 = 1$ can be rewritten as $(n - 4)(n - 5) = 0$, the solutions are $n = 4$ and $n = 5$. For the general situation, we use this insight and write $n^2 - 9n + 21 = (n - 4)(n - 5) + 1$, which suggests to use a new variable $x = n - 5$. In this\... | Ireland | Ireland | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 2, 4, 5, 7 | |
0305 | Problem:
Dizemos que um inteiro positivo é bacana se ao somarmos os quadrados de seus dígitos e repetirmos essa operação sucessivamente, obtivermos o número $1$. Por exemplo, $1900$ é bacana, pois
$$
1900 \rightarrow 82 \rightarrow 68 \rightarrow 100 \rightarrow 1
$$
a) Encontre dois números de dois dígitos consecuti... | [
"Solution:\n\na) Os números $31$ e $32$ são bacanas, pois\n$$\n31 \\rightarrow 10 \\rightarrow 1\n$$\ne\n$$\n32 \\rightarrow 13 \\rightarrow 10 \\rightarrow 1\n$$\n\nb) Considere os números\n$$\nA=\\underbrace{111 \\ldots 11}_{31 \\text{ uns}}\\underbrace{000 \\ldots 00}_{n-31 \\text{ zeros}}\n$$\ne\n$$\nB=\\underb... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | Part a: 31 and 32. Part b: For each n > 2020, take A to be a number with thirty-one ones followed by enough zeros to have n digits, and let B be A plus one; both are consecutive n-digit numbers that eventually reach one under the process. | |
0j6l | Problem:
Find the number of polynomials $p(x)$ with integer coefficients satisfying $p(x) \geq \min \{2 x^{4}-6 x^{2}+1, 4-5 x^{2}\}$ and $p(x) \leq \max \{2 x^{4}-6 x^{2}+1, 4-5 x^{2}\}$ for all $x \in \mathbb{R}$. | [
"Solution:\nWe first find the intersection points of $f(x) = 2x^{4} - 6x^{2} + 1$ and $g(x) = 4 - 5x^{2}$. If $2x^{4} - 6x^{2} + 1 = 4 - 5x^{2}$, then $2x^{4} - x^{2} - 3 = 0$, so $(2x^{2} - 3)(x^{2} + 1) = 0$, and $x = \\pm \\sqrt{\\frac{3}{2}}$. Note that this also demonstrates that $g(x) \\geq f(x)$ if and only ... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 4 | |
0223 | Problem:
Par perfeito - Dizemos que 2 números naturais formam um par perfeito quando a soma e o produto desses dois números são quadrados perfeitos. Por exemplo, $5$ e $20$ formam um par perfeito, pois $5+20=25=5^{2}$ e $5 \times 20=100=10^{2}$. Será que $122$ forma um par perfeito com outro natural? | [
"Solution:\n\nChamemos de $n$ o natural \"candidato\" a formar um par perfeito com $122$. Então, devemos ter: $122+n=A^{2}$ e $122 \\times n=B^{2}$ onde $A$ e $B$ são números naturais.\n\nComo $B^{2}=2 \\times 61 \\times n$, concluímos que $n$ tem também os fatores primos $2$ e $61$. Logo, podemos escrever $n$ como... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | Yes; 14762 | |
0kq9 | Problem:
How many ways are there to cut a $1$ by $1$ square into $8$ congruent polygonal pieces such that all of the interior angles for each piece are either $45$ or $90$ degrees? Two ways are considered distinct if they require cutting the square in different locations. In particular, rotations and reflections are c... | [
"Solution:\n\nFirst note that only triangles and quadrilaterals are possible.\nThere are $3$ possibilities:\n- $1/2$ by $1/2$ right isosceles triangles\n- $1$ by $1/8$ rectangles\n- $1/2$ by $1/4$ rectangles\n\nThe first case has $16$ possibilities (there are $2$ choices for the orientation of each quadrant).\nThe ... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Discrete Mathematics > Other"
] | null | final answer only | 54 | |
0blm | Does there exist a sequence of positive integers $a_1, a_2, a_3, \dots$ such that $a_m$ and $a_n$ are coprime if and only if the indices $m$ and $n$ are one unit apart? | [
"The answer is in the affirmative. The idea is to consider a sequence of pairwise distinct primes $p_1, p_2, p_3, \\dots$, cover the positive integers by a sequence of finite non-empty sets $I_n$ such that $I_m$ and $I_n$ are disjoint if and only if $m$ and $n$ are one unit apart, and set $a_n = \\prod_{i \\in I_n}... | Romania | THE 2015 Seventh ROMANIAN MASTER OF MATHEMATICS | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
07jd | Given is a right triangle $ABC$ with $\angle A = 90^\circ$. Let $M$ be the midpoint of $BC$ and $P$ be an arbitrary point on $AM$. The reflection of $BP$ with respect to $AB$ intersects the lines $AC$ and $AM$ at $T$ and $Q$, respectively. Let the circumcircles of $BPQ$ and $ABC$ meet for the second time at $F$. Prove ... | [
"First, note that since $AM = CM$, it follows that $\\angle MAC = \\angle MCA$, and with a bit of angle chasing, we have:\n$$\n\\begin{align*} \n\\angle QFA &= \\angle QFB - \\angle AFB = \\angle QPB - \\angle ACB = \\angle APB - \\angle MAC = \\\\ \n&= \\angle APB + \\angle BAP - 90^\\circ = 90^\\circ - \\angle AB... | Iran | Iranian 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 | |
03kw | Problem:
Let $n$ be a natural number such that $n \geq 2$. Show that
$$
\frac{1}{n+1}\left(1+\frac{1}{3}+\cdots+\frac{1}{2 n-1}\right)>\frac{1}{n}\left(\frac{1}{2}+\frac{1}{4}+\cdots+\frac{1}{2 n}\right) .
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0koi | Problem:
Compute the number of distinct pairs of the form
(first three digits of $x$, first three digits of $x^{4}$ )
over all integers $x>10^{10}$.
For example, one such pair is $(100,100)$ when $x=10^{10^{10}}$. | [
"Solution:\nGraph these points on an $x$, $y$-plane. We claim that there are integers $100=a_{0}<a_{1}<a_{2}<a_{3}<a_{4}=999$, for which the locus of these points is entirely contained in four taxicab (up/right movement by 1 unit) paths from $(a_{i}, 100)$ to $(a_{i+1}, 999)$, $i=0,1,2,3$.\n\nAs we increment $x$ ve... | United States | HMMT November | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | 4495 | |
0j0d | Problem:
How many sequences $a_{1}, a_{2}, \ldots, a_{8}$ of zeroes and ones have $a_{1} a_{2} + a_{2} a_{3} + \cdots + a_{7} a_{8} = 5$? | [
"Solution:\n\nFirst, note that we have seven terms in the left hand side, and each term can be either $0$ or $1$, so we must have five terms equal to $1$ and two terms equal to $0$. Thus, for $n \\in \\{1,2, \\ldots, 8\\}$, at least one of the $a_{n}$ must be equal to $0$. If we can find $i, j \\in \\{2,3, \\ldots,... | United States | Harvard-MIT November Tournament | [
"Discrete Mathematics > Combinatorics"
] | null | final answer only | 9 | |
0ao8 | Problem:
Sixty men working on a construction job have done $1/3$ of the work in $18$ days. The project is behind schedule and must be accomplished in the next twelve days. How many more workers need to be hired?
(a) $60$
(b) $180$
(c) $120$
(d) $240$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | c | |
0a8v | Problem:
Find all positive integers $k$ such that the product of the digits of $k$, in the decimal system, equals
$$
\frac{25}{8} k-211
$$ | [
"Solution:\nLet\n$$\na = \\sum_{k=0}^{n} a_{k} 10^{k}, \\quad 0 \\leq a_{k} \\leq 9, \\text{ for } 0 \\leq k \\leq n-1, 1 \\leq a_{n} \\leq 9\n$$\nSet\n$$\nf(a) = \\prod_{k=0}^{n} a_{k}\n$$\nSince\n$$\nf(a) = \\frac{25}{8} a - 211 \\geq 0\n$$\n$a \\geq \\frac{8}{25} \\cdot 211 = \\frac{1688}{25} > 66$. Also, $f(a)$... | Nordic Mathematical Olympiad | Nordic Mathematical Contest, NMC 19 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 72 and 88 | |
0bq2 | We will call a number *good* if it is a positive integer with at least two digits and by removing one of its digits we get a number which is equal to the sum of its initial digits (for instance, $109$ is good: remove $9$ to get $10 = 1 + 0 + 9$).
a) Find the smallest good number.
b) Find how many numbers are good. | [] | Romania | 67th NMO Shortlisted Problems | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | a) 10; b) infinitely many | |
09x3 | Problem:
Gegeven zijn twee positieve gehele getallen $k$ en $n$ met $k \leq n \leq 2k-1$. Julian heeft een grote stapel rechthoekige $k \times 1$-tegels. Merlijn noemt een positief geheel getal $m$ en ontvangt van Julian $m$ tegels om op een $n \times n$-bord te plaatsen. Op elke tegel schrijft Julian eerst of het een... | [
"Solution:\n\nWe bewijzen dat de grootste $m$ die Merlijn kan noemen, gelijk is aan $\\min(n, 3(n-k)+1)$. Eerst bewijzen we dat $m \\leq \\min(n, 3(n-k)+1)$.\n\nAls Merlijn $n+1$ tegels bestelt, kan Julian ze allemaal als horizontaal bestempelen. Aangezien $n \\leq 2k-1$ is het niet mogelijk om meer dan één horizon... | Netherlands | IMO-selectietoets III | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | min(n, 3(n-k)+1) | |
0e5x | Problem:
Dokaži, da ne obstaja injektivna funkcija $f: \mathbb{R} \rightarrow \mathbb{R}$, za katero bi veljalo
$$
f(f(x)+y)=f(x+y)+f(2012) \quad \text{ za vse } x, y \in \mathbb{R}
$$ | [
"Solution:\n\nČe v enačbo vstavimo $y=-x$, dobimo $f(f(x)-x)=f(0)+f(2012)$. Ker je na desni strani konstanta, $f$ pa je injektivna funkcija, mora biti tudi $f(x)-x$ konstanta. Torej je $f(x)=x+c$ za neko realno število $c$. Če to vstavimo v začetno enačbo, dobimo $x+y+2c=x+y+2c+2012$, kar nam da protislovje $0=2012... | Slovenia | 56. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0lay | Given a circle $(O)$ with diameter $AB$ on the plane. A point $P$ moves on the tangent at $B$ to $(O)$. The line $PA$ intersects $(O)$ in the second point $C$. Let $D$ be the point symmetric to $C$ with respect to $O$. The line $PD$ intersects $(O)$ in the second point $E$.
1/ Show that the lines $AE$, $BC$ and $PO$ p... | [
"1/ Let $F$ be the intersection of lines $AE$ and $BP$.\nWe have $\\overline{ACE} = 90^\\circ + \\overline{BCE} = 90^\\circ + \\overline{FAB} = \\overline{EFP}$. Consequently $\\overline{EFP} + \\overline{ECP} = 180^\\circ$.\n\nHence $CEFP$ is a cyclic quadrilateral. Consequently $\\overline{CFP} = \\overline{CEP} ... | Vietnam | Vijetnam 2011 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalitie... | English | proof and answer | The three lines AE, BC, and PO are concurrent at M. The area of triangle ABC is maximized when the moving point is at distance √2·R from the tangency point along the tangent (two symmetric positions), and the maximal area is R^2/√2. | |
0d0x | Let $ABC$ be a triangle right-angled at $A$. A circle passing through $B$ and $C$ intersects the sides $AB$ and $AC$ at $M$, respectively $N$. Prove that if $BM \cdot CN \cdot BC = MN^3$, then the symmetric point of $A$ with respect to the midpoint of the segment $MN$ belongs to $BC$. | [
"Let $BC = a$, $CA = b$, $AB = c$, $AM = x$.\n\n\n\nTriangles $AMN$ and $ACB$ are similar, so\n$$\n\\frac{x}{b} = \\frac{AN}{c} = \\frac{MN}{a}.\n$$\nWe obtain\n$$\nAN = \\frac{cx}{b}, \\quad MN = \\frac{ax}{b}. \\qquad (1)\n$$\nThe relation $BM \\cdot CN \\cdot BC = MN^3$ is equivalent to\... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0a4n | Problem:
In triangle $ABC$, points $D$ and $E$ lie on the interior of segments $AB$ and $AC$, respectively, such that $AD = 1$, $DB = 2$, $BC = 4$, $CE = 2$ and $EA = 3$. Let $DE$ intersect $BC$ at $F$. Determine the length of $CF$. | [
"Solution:\n\nFirst notice that the sidelengths of $\\triangle ABC$ are $3$, $4$ and $5$. By Pythagoras this implies that triangle $ABC$ is right-angled at $B$. Now we can put the diagram on coordinate axes such that $B = (0,0)$, $A = (0,3)$ and $C = (4,0)$. Furthermore we get $D = (0,2)$ and since $E$ divides $CA$... | New Zealand | NZMO Round One | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 | |
07do | Prove that for each positive integer $m$, one can find $m$ consecutive positive integers like $n$ such that the following expression is not a perfect power
$$
(1^3 + 2018^3)(2^3 + 2018^3)\cdots(n^3 + 2018^3)
$$ | [
"Let $p$ be a prime number of the form $3k+2$, such that $p > \\max(m, 2018)$. (Such $p$ exists because there are infinitely many prime numbers of the form $3k+2$.) We shall prove that $n = p - 2019 + i$, $1 \\le i \\le m$ satisfies the problem's conditions.\n\nIt suffices to prove that\n$$\nv_p((1^3 + 2018^3)(2^3 ... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Other"
] | null | proof only | null | |
0jrz | Problem:
Let $a$ and $b$ be integers (not necessarily positive). Prove that $a^{3} + 5b^{3} \neq 2016$. | [
"Solution:\nSince cubes are $0$ or $\\pm 1$ modulo $9$, by inspection we see that we must have $a^{3} \\equiv b^{3} \\equiv 0 \\pmod{3}$ for this to be possible. Thus $a$, $b$ are divisible by $3$. But then we get $3^{3} \\mid 2016$, which is a contradiction.\n\nOne can also solve the problem in the same manner by ... | United States | HMMT February 2016 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0bdj | Let $n$ be an integer greater than $1$ and let $S$ be the set of $n$-element subsets of the set $\{1, 2, \dots, 2n\}$. Determine
$$
\max_{S \in \mathcal{S}} \min_{x, y \in S, x \neq y} [x, y],
$$
where $[x, y]$ denotes the least common multiple of the integers $x$ and $y$. | [
"The required value is $6(\\lfloor n/2 \\rfloor + 1)$, unless $n=4$ in which case it is $24$. Let $S$ be a member of $\\mathcal{S}$. We first show that\n$$\n\\min_{x, y \\in S, x \\neq y} [x, y] \\le 6(\\lfloor n/2 \\rfloor + 1), \\quad (*)\n$$\nunless $n = 4$. To this end, for each $x$ in $S$, choose a positive in... | Romania | 64th NMO Selection Tests for the Balkan and International Mathematical Olympiads | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 6(floor(n/2) + 1), except when n = 4 in which case it is 24 | |
0ac5 | The difference of two complementary angles $\alpha$ and $\beta$ is $20^\circ 52'$. Determine $\alpha$ and $\beta$. | [
"For the angles $\\alpha$ and $\\beta$ we have $\\alpha + \\beta = 90^\\circ$ and $\\alpha - \\beta = 20^\\circ 52'$. Hence $\\beta = (90^\\circ - 20^\\circ 52') : 2 = 34^\\circ 34'$ and $\\alpha = 90^\\circ - 34^\\circ 34' = 55^\\circ 26'.$"
] | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | α = 55°26′, β = 34°34′ | |
001v | Se tiene un tablero rectangular de $5 \times 50$, dividido en casillas de $1 \times 1$, y fichas de dominó de $1 \times 2$, cada una con dos números escritos: un $1$ y un $-1$, uno en cada mitad. Cada ficha de dominó cubre exactamente dos casillas vecinas del tablero. Gabriel debe cubrir el tablero con estas fichas, si... | [] | Argentina | XX OLIMPIADA MATEMÁTICA ARGENTINA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | español | proof and answer | 5×50: impossible; 5×100: possible | |
08p0 | Problem:
A positive integer is called a repunit, if it is written only by ones. The repunit with $n$ digits will be denoted by $\underbrace{11 \ldots 1}_{n}$. Prove that:
a) the repunit $\underbrace{11 \ldots 1}_{n}$ is divisible by 37 if and only if $n$ is divisible by 3 ;
b) there exists a positive integer $k$ such... | [
"Solution:\n\na) Let $n = 3m + r$, where $m$ and $r$ are non-negative integers and $r < 3$.\nDenote by $\\underbrace{00 \\ldots 0}_{p}$ a recording with $p$ zeroes and $\\underbrace{abcabc \\ldots abc}_{p}$ a recording with $p$ times $abc$. We have:\n$$\n\\underbrace{11 \\ldots 1}_{n} = \\underbrace{11 \\ldots 1}_{... | JBMO | Junior Balkan Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | a) divisible by thirty-seven if and only if n is a multiple of three; b) k = 5 | |
03u0 | Suppose points $O$ and $I$ are the circumcenter and incenter of $\triangle ABC$ respectively, and the inscribed circle of $\triangle ABC$ is tangent to the sides $BC$, $CA$, $AB$ at points $D$, $E$, $F$ respectively. Lines $FD$ and $CA$ intercept at point $P$, while lines $DE$ and $AB$ intercept at point $Q$. And point... | [
"We first consider $\\triangle ABC$ and segment $PFD$. By Menelaus theorem we have\n$$\n\\frac{CP}{PA} \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = 1.\n$$\nThen\n$$\nPA = CP \\cdot \\frac{AF}{FB} \\cdot \\frac{BD}{DC} = (PA + b) \\frac{p-a}{p-c}.\n$$\n(We define $a = BC$, $b = CA$, $c = AB$, $p = \\frac{1}{2}(a+b+... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, ... | English | proof only | null | |
05ma | Problem:
Déterminer tous les entiers naturels $a$ pour lesquels il existe des nombres premiers $p$, $q$, $r$, pas forcément distincts, tels que
$$
a = \frac{p+q}{r} + \frac{q+r}{p} + \frac{r+p}{q}
$$ | [
"Solution:\n\nTout d'abord, si $p = q = r$, alors on obtient $a = 6$, qui est bien une solution.\n\nMontrons qu'il s'agit de la seule solution, et supposons qu'il existe une solution $a = \\frac{p+q}{r} + \\frac{q+r}{p} + \\frac{r+p}{q}$ avec $p, q, r$ premiers et non tous égaux.\n\n- Si deux des nombres $p, q, r$ ... | France | Olympiades Françaises de Mathématiques | [
"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 | 6 | |
0isv | At a certain mathematical conference, every pair of mathematicians are either friends or strangers. At mealtime, every participant eats in one of two large dining rooms. Each mathematician insists upon eating in a room which contains an even number of his or her friends. Prove that the number of ways that the mathemati... | [
"Let $n$ be the number of participants at the conference. We proceed by induction on $n$.\n\nIf $n = 1$, then we have one participant who can eat in either room; that gives us total of $2 = 2^1$ options.\nLet $n \\ge 2$. The case in which some participant, $P$, has no friends is trivial. In this case, $P$ can eat i... | United States | USAMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Vectors"
] | null | proof only | null | |
02pb | Problem:
Dez pontos são marcados ao redor de uma circunferência, como ilustra a figura.
(a) Quantas cordas podem ser formadas ligando dois quaisquer destes pontos? (Uma corda é um segmento de reta ligando dois pontos sobre uma circunferência.)
(b) Quantos triângulos podem ser formados ligando... | [
"Solution:\n(a) De cada ponto saem 9 cordas e temos 10 pontos. Mas cada corda é contada duas vezes (uma corda $AB$ é contada por sair de $A$ e por sair de $B$), assim temos $9 \\times 10 / 2 = 45$ cordas.\n\n(b) Cada corda é lado de 8 triângulos (basta escolher um ponto que não seja extremidade da corda escolhida) ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) 45; b) 120 | |
048f | Prove that the product of any two elements of the set
$$
\{m \mid m = a^2 - 5b^2,\ a, b \in \mathbb{N}\}
$$
belongs to that set as well. | [] | Croatia | CroatianCompetitions2011 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Algebraic Number Theory > Quadratic fields"
] | null | proof only | null | |
08kj | Problem:
Let $ABCD$ be a parallelogram, $P$ a point on $CD$, and $Q$ a point on $AB$. Let also $M = AP \cap DQ$, $N = BP \cap CQ$, $K = MN \cap AD$, and $L = MN \cap BC$. Show that $BL = DK$. | [
"Solution:\n\nLet $O$ be the intersection of the diagonals. Let $P_1$ be on $AB$ such that $PP_1 \\parallel AD$, and let $Q_1$ be on $CD$ such that $QQ_1 \\parallel AD$. Let $\\sigma$ be the central symmetry with center $O$. Let $P' = \\sigma(P)$, $Q' = \\sigma(Q)$, $P_1' = \\sigma(P_1)$.\n\nLet $M_1 = AQ_1 \\cap D... | JBMO | OJBM | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
03b6 | Do there exist positive integers $n$ and $k$, $1 \le k \le n-2$, such that
$$
\binom{n}{k}^2 + \binom{n}{k+1}^2 = \binom{n}{k+2}^4 ?
$$ | [
"By applying the formula $\\binom{a}{b} = \\frac{a!}{b!(a-b)!}$ we obtain the equation\n$$\n1 + \\frac{(n-k)^2}{(k+1)^2} = \\frac{(n-k)^2(n-k-1)^2}{(k+1)^2(k+2)^2} \\binom{n}{k+2}^2.\n$$\nHence $(k+2)^2 [(k+1)^2 + (n-k)^2] = (n-k)^2(n-k-1)^2 \\binom{n}{k+2}^2$, which implies that $(k+1)^2 + (n-k)^2$ is a perfect sq... | Bulgaria | Bulgaria | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0hdv | Point $P$ is chosen on the smaller arc $BC$ of the circumscribed circle of an acute-angled triangle $ABC$. Points $R$ and $S$ on sides $AB$ and $AC$ respectively are chosen such that $CPRS$ is a parallelogram. Point $T$ on arc $AC$ of the circumscribed circle of $\triangle ABC$ is such that $BT \parallel CP$. Prove tha... | [
"Let line $PR$ intersect the circumscribed circle of $\\triangle ABC$ at point $K$ (Fig. 27).\n\nThen, $AKPC$ and $BPCT$ are isosceles trapezoids, hence, $\\angle RSA = \\angle PCA = 180^\\circ - \\angle RKA$, so $AKRS$ is inscribed, which makes it an isosceles trapezoid.\n\nThen, $\\angle RKS = \\angle RAS = \\ang... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0776 | Problem:
Let $m, n$ be distinct positive integers. Prove that
$$
\operatorname{gcd}(m, n)+\operatorname{gcd}(m+1, n+1)+\operatorname{gcd}(m+2, n+2) \leq 2|m-n|+1
$$
Further, determine when equality holds. | [
"Solution:\nObserve that\n$$\n\\operatorname{gcd}(m+j, n+j)=\\operatorname{gcd}(m+j,|m-n|)\n$$\nfor $j=0,1,2$. Hence we can find positive integers $a, b, c$ such that\n$$\n\\operatorname{gcd}(m, n)=\\frac{|m-n|}{a}, \\quad \\operatorname{gcd}(m+1, n+1)=\\frac{|m-n|}{b}, \\quad \\operatorname{gcd}(m+2, n+2)=\\frac{|... | India | Indian National Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds if and only if (m, n) are consecutive integers or they are consecutive even integers, i.e., (m, n) = (k, k+1) or (2k, 2k+2), up to swapping m and n, for some positive integer k. | |
04h6 | Tamara has written an even positive integer on a board. After that, she wrote twelve numbers consecutively, so that every number is by $5$ greater than the square of the previously written number. Determine all possible last digits of the last written number.
(Italy 2012) | [] | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Modular Arithmetic",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | 0 and 6 | |
0gvd | For an acute triangle $ABC$: its circumcircle $\omega$ with center $O$, the circumcircle $\omega_1$ of the triangle $AOC$ and the diameter $OQ$ of $\omega_1$ were drawn. The points $M$ and $N$ were taken on the lines $AQ$ and $AC$ respectively in such a way that the quadrilateral $AMBN$ is a parallelogram. Prove that t... | [
"Let the line $BQ$ meet the circle $\\omega_1$ again at the point $T$. We will show that the points $M$, $T$, $N$ are collinear. Since $OQ$ is the diameter of the circle $\\omega_1$, $MQ$ and $CQ$ are tangents to the circle $\\omega$. Let $\\angle ABC = \\beta$. Then $\\angle CAQ = \\angle ACQ = \\beta$ (as angles ... | Ukraine | Ukrainian Mathematical Olympiad, Final Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point ... | null | proof only | null | |
0bsx | The numbers $1$, $2$, $3$, $\ldots$, $16$ are written in the squares of a $4 \times 4$ table, one in each square, and we add the numbers in each column. If one of the sums is larger than the other three, we denote it $S$.
a) Give an example with $S = 40$.
b) Which is the smallest possible value of $S$? | [
"a) An example is given in figure 1.\n\n<table><tr><td>1</td><td>2</td><td>3</td><td><b>10</b></td></tr><tr><td>8</td><td>7</td><td>6</td><td><b>5</b></td></tr><tr><td>9</td><td>4</td><td>11</td><td><b>12</b></td></tr><tr><td>16</td><td>15</td><td>14</td><td><b>13</b></td></tr></table>\n\nFigure 1\n\nb) The sum of ... | Romania | 67th Romanian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | a) 40; b) 35 | |
02z2 | Problem:
Em um tabuleiro $4 \times 4$, deve-se colocar os números de 1 a 16 nas casas, sem repetir, de forma que a soma dos números de cada linha, coluna e diagonal seja a mesma. Chamamos essa soma de Soma Mágica.
a) Qual a Soma Mágica deste tabuleiro?
b) Se a soma das casas marcadas com $X$ no tabuleiro abaixo é 34... | [
"Solution:\n\na) Como são 4 linhas (assim como 4 colunas), a Soma Mágica vale:\n$$\n\\frac{1+2+3+\\ldots+16}{4}=34\n$$\n\nb) Se somarmos as duas diagonais, teremos exatamente a soma das casas marcadas com $X$ e com $Y$. Assim, a soma das casas marcadas com $Y$ é $2 \\cdot 34 - 34 = 34$.\n\nc) Temos:\n$$\n\\begin{al... | Brazil | Brazilian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | a) 34; b) 34; c) 5 | |
016b | Find all positive integers $x$, $y$ and $n$ such that
$$
x^n - y^n = 2010.
$$ | [
"We first notice that $x > y$ and $x \\equiv y \\pmod 2$. The prime factor decomposition of $2010$ is $2010 = 2 \\cdot 3 \\cdot 5 \\cdot 67$. For even $n$, $x^n \\equiv y^n \\pmod 4$. This is not possible, so $n$ is odd. If $n = 1$, the equations has as solutions all numbers $x$ and $y$ such that $x = 2010 + y$.\n\... | Baltic Way | Baltic Way SHL | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems"
] | null | proof and answer | All solutions are with n = 1 and x = y + 2010 (with y any positive integer). No solutions exist for n ≥ 2. | |
0k7f | Problem:
A $5 \times 5$ grid of unit squares is partitioned into $5$ pairwise incongruent rectangles with sides lying on the gridlines. Find the maximum possible value of the product of their areas. | [
"Solution:\n\nThe greatest possible value for the product is $3 \\cdot 4 \\cdot 4 \\cdot 6 \\cdot 8 = 2304$, achieved when the rectangles are $3 \\times 1$, $1 \\times 4$, $2 \\times 2$, $2 \\times 3$, $4 \\times 2$. To see that this is possible, orient these rectangles so that the first number is the horizontal di... | United States | HMMT February 2019 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2304 | |
07f5 | For every positive integer $k > 1$ prove that there exists a real number $x$ such that for every positive integer $n < 1398$:
$$
\{x^n\} < \{x^{n-1}\} \iff k \mid n.
$$ | [
"Take a sufficiently large $m$ ($m > 2^{3000}$) and put $x = m + \\frac{1}{k-1}$. Note that\n$$\n\\left\\{ \\left( m + \\frac{1}{k-1} \\right)^n \\right\\} = \\sum_{k_i > n} \\binom{n}{i} \\frac{m^{n-i}}{m^{(k-1)i}}\n$$\nBecause\n$$\n\\sum_{k_i > n} \\binom{n}{i} \\frac{m^{n-i}}{m^{(k-1)i}} < \\frac{1}{m} \\sum_{i=... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and pr... | English | proof only | null | |
0jex | Problem:
The number $989 \cdot 1001 \cdot 1007 + 320$ can be written as the product of three distinct primes $p, q, r$ with $p < q < r$. Find $(p, q, r)$. | [
"Solution:\n\nAnswer: $(991, 997, 1009)$\n\nLet $f(x) = x(x - 12)(x + 6) + 320 = x^{3} - 6x^{2} - 72x + 320$, so that $f(1001) = 989 \\cdot 1001 \\cdot 1007 + 320$.\n\nBut $f(4) = 4(-8)(10) + 320 = 0$, so $f(x) = (x - 4)(x^{2} - 2x - 80) = (x - 4)(x - 10)(x + 8)$.\n\nThus $f(1001) = 991 \\cdot 997 \\cdot 1009$, as ... | United States | HMMT November 2013 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | final answer only | (991, 997, 1009) | |
07kb | Prove that a triangle $ABC$ is right-angled if and only if
$$
sin^2 A + sin^2 B + sin^2 C = 2.
$$ | [
"Suppose one of $\\angle A, \\angle B, \\angle C$ is a right angle. Say $\\angle A$ is. Then $\\angle B + \\angle C = \\pi/2$ and so\n$$\nsin^2 A + sin^2 B + sin^2 C = 1 + sin^2 B + sin^2(\\pi/2 - B) = 1 + sin^2 B + cos^2 B = 2.\n$$\n\nSuppose the identity holds. By the Sine Rule,\n$$\n\\frac{\\sin A}{a} = \\frac{\... | Ireland | Irish Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof only | null | |
07bq | Find all of the solutions of the following equation in natural numbers:
$$n^{n^n} = m^m.$$ | [
"We start with a lemma.\n**Lemma 1.** Let $n$ be a positive integer and $p, q$ some positive rational numbers. If $n^p = q$, then $q$ is itself an integer.\n*Proof.* Suppose $p = \\frac{a}{b}$ and $q = \\frac{c}{d}$ where $a, b, c, d \\in \\mathbb{N}$. We have\n$$\nn^p = q \\Rightarrow n^{\\frac{a}{b}} = \\frac{c}{... | Iran | Iranian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | m = n = 1 | |
0cwu | Diagonals of a convex quadrilateral *ABCD* intersect at *E*. The four points of tangency of the circles (*ABE*) and (*CDE*) with their external common tangents lie on a circle $\omega$. Analogously, the four points of tangency of the circles (*ADE*) and (*BCE*) with their external common tangents lie on a circle $\gamm... | [
"Let us denote the centers of the circumscribed circles of triangles *ABE*, *BCE*, *CDE*, *ADE* by $O_{AB}$, $O_{BC}$, $O_{CD}$, $O_{AD}$ respectively. Let $T_1, T_2$ be the points of tangency of one of the common tangents with the circumscribed circles of triangles *ABE* and *CDE*, respectively; denote by *O* and ... | Russia | LI Всероссийская математическая олимпиада школьников | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | Russian | proof only | null | |
0cs9 | Given are 111 distinct positive integers not exceeding 500. May it happen that for each of these numbers, its last digit coincides with the last digit of the sum of all other numbers? | [
"Suppose this is possible. Denote the given numbers by $a_1, a_2, \\dots, a_{111}$ and let their sum be $S$. By the condition, for each index $k$, the numbers $a_k$ and $S - a_k$ have the same last digit. Hence, their difference $S - 2a_k$ is divisible by $10$. Therefore, for any $k$, the number $2a_k$ ends with th... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Pigeonhole principle"
] | null | proof and answer | No | |
08ys | 2021 integers $a_1, a_2, \dots, a_{2021}$ satisfy
$$
a_{n+5} + a_n > a_{n+2} + a_{n+3}
$$
for all integers $n = 1, \dots, 2016$. Determine the minimum value of the difference between maximum and minimum of $a_1, a_2, \dots, a_{2021}$. | [
"85008 is the minimum value of the difference.\n\nFirst we show the difference between maximum and minimum is greater than or equal to 85008. Let $a_1, a_2, \\dots, a_{2021}$ be 2021 integers satisfying the condition. Let $n$ be an integer satisfying $1 \\le n \\le 2016$ then the condition shows $a_{n+5} - a_{n+3} ... | Japan | Japan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 85008 | |
01s8 | Prove that
$$
|(a-b)(b-c)(c-d)(d-a)| \le \frac{abcd}{4},
$$
if real numbers $a, b, c, d$ belong to the segment $[1; 2]$. | [
"It is evident that\n$$\n|(a-b)(b-c)(c-d)(d-a)| \\le \\frac{abcd}{4} \\iff\n$$\n$$\n\\frac{(a-b)^2}{ab} \\cdot \\frac{(b-c)^2}{bc} \\cdot \\frac{(c-d)^2}{cd} \\cdot \\frac{(d-a)^2}{da} \\le \\frac{1}{16}. \\qquad (*)\n$$\n\nNote that\n$$\n\\frac{(a-b)^2}{ab} \\le \\frac{1}{2}. \\qquad (1)\n$$\nIndeed, we have\n$$\n... | Belarus | FINAL ROUND | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
01c5 | Let $S$ be a finite set of positive rational numbers. Let $x_1, x_2, x_3, \dots$ be a sequence such that $x_1 = 0$ and for every positive integer $n$ there exists some $q_n \in S$ such that $x_{n+1} = \sqrt{x_n + q_n}$. Suppose that all the numbers $x_1, x_2, x_3, \dots$ are rational. Show that there are only finitely ... | [
"Let $x_n = \\frac{y_n}{z_n}$ where $y_n, z_n \\ge 0$ are coprime integers, and similarly let $q_n = \\frac{a_n}{b_n}$, where $a_n, b_n \\ge 0$ are coprime integers. Then\n$$\nx_{n+1}^2 = \\frac{y_{n+1}^2}{z_{n+1}^2} = x_n + q_n = \\frac{b_n y_n + a_n z_n}{b_n z_n}.\n$$\nLet $d_n = \\gcd(b_n y_n + a_n z_n, b_n z_n)... | Baltic Way | Baltic Way | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | proof only | null | |
096z | Problem:
În interiorul triunghiului isoscel $ABC$ ($AC = BC$) cu $m(\angle C) = 80^{\circ}$ este situat punctul $P$ astfel încât $m(\angle PAB) = 30^{\circ}$ și $m(\angle PBA) = 10^{\circ}$. Determinați măsura în grade a unghiului $CPB$. | [
"Solution:\n\nÎn $\\triangle ABC$ $m(\\angle A) = m(\\angle B) = \\frac{180^{\\circ} - 80^{\\circ}}{2} = 50^{\\circ}$, deci $m(\\angle PBD) = 50^{\\circ} - 10^{\\circ} = 40^{\\circ}$.\n\n$\\angle BPD$ - exterior $\\triangle ABP \\Rightarrow m(\\angle BPD) = 30^{\\circ} + 10^{\\circ} = 40^{\\circ}$, prin urmare $\\t... | Moldova | Olimpiada Republicană la Matematică | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 70° | |
0ht6 | Problem:
Every two members of a certain society are either friends or enemies. Suppose that there are $n$ members of the society, that there are exactly $q$ pairs of friends, and that in every set of three persons there are two who are enemies to each other. Prove that there is at least one member of the society among... | [
"Solution:\n\nDenote by $S$ the set of all members of the society, by $A$ the set of all pairs of friends, and by $N$ the set of all pairs of enemies. For every $x \\in S$, denote by $f(x)$ the number of friends of $x$ and by $F(x)$ the number of pairs of friends among enemies of $x$. It is easy to prove:\n$$\n\\be... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof only | null | |
0c2n | For a fixed integer $n \ge 2$, determine complex numbers $z$ that satisfy the relations:
$$
a)\ z^n + z^{n-1} + \dots + z^2 + |z| = n;
$$
$$
b)\ |z|^{n-1} + |z|^{n-2} + \dots + |z|^2 + z = n z^n.
$$ | [] | Romania | 2018 Romanian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Prealgebra / Basic Algebra > Other"
] | null | proof and answer | a) z = 1.
b) z = 0 or z is a nonnegative real number r satisfying r + r^2 + ... + r^{n-1} = n r^n (equivalently, for r ≠ 1, r^{n-1}(n + 1 − n r) = 1). | |
03lp | Problem:
Let $ABC$ be an acute-angled triangle. Inscribe a rectangle $DEFG$ in this triangle so that $D$ is on $AB$, $E$ is on $AC$ and both $F$ and $G$ are on $BC$. Describe the locus of (i.e., the curve occupied by) the intersections of the diagonals of all possible rectangles $DEFG$. | [
"Solution:\n\nThe locus is the line segment joining the midpoint $M$ of $BC$ to the midpoint $K$ of the altitude $AH$. Note that a segment $DE$ with $D$ on $AB$ and $E$ on $AC$ determines an inscribed rectangle; the midpoint $F$ of $DE$ lies on the median $AM$, while the midpoint of the perpendicular from $F$ to $B... | Canada | 38th Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | null | proof only | null | |
0d2a | Given an integer $n \geq 2$, determine the number of ordered $n$-tuples of integers $(a_{1}, a_{2}, \ldots, a_{n})$ such that
a) $a_{1} + a_{2} + \cdots + a_{n} \geq n^{2}$; and
b) $a_{1}^{2} + a_{2}^{2} + \cdots + a_{n}^{2} \leq n^{3} + 1$. | [
"First solution. Notice that we have\n$$\n\\begin{aligned}\n\\left(a_{1}-n\\right)^{2}+\\cdots+\\left(a_{n}-n\\right)^{2} & =\\left(a_{1}^{2}+\\cdots+a_{n}^{2}\\right)-2 n\\left(a_{1}+\\cdots+a_{n}\\right)+n^{3} \\\\\n& \\leq n^{3}+1-2 n^{3}+n^{3}=1 .\n\\end{aligned}\n$$\nTherefore, there are two cases:\n\nThe firs... | Saudi Arabia | Selection tests for the International Mathematical Olympiad 2013 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | English | proof and answer | 1 | |
0gts | Let $O_1O_2O_3$ be an acute angled triangle. Let $\omega_1, \omega_2, \omega_3$ be the circles with centres $O_1, O_2, O_3$ respectively such that any two of them are tangent to each other. Circumcircle of $O_1O_2O_3$ intersects with $\omega_1$ at $A_1$ and $B_1$, with $\omega_2$ at $A_2$ and $B_2$, with $\omega_3$ at ... | [
"Let the circumcircle of the triangle $O_1O_2O_3$ be $\\Gamma$. Let the tangency point of $\\omega_1$ and $\\omega_2$ be $D$, $\\omega_1$ and $\\omega_3$ be $E$, $\\omega_2$ and $\\omega_3$ be $F$. Let the incenter of $O_1O_2O_3$ be $I$, then it's easy to see that $ID, IE, IF$ are perpendicular to the respective si... | Turkey | Team Selection Test for EGMO 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
0e59 | Let $E$ be the midpoint of the side $AB$ in the quadrilateral $ABCD$ and let $F$ be a point on the diagonal $AC$, such that the line $BF$ is perpendicular to the diagonal $AC$. Find the ratio of the sides of the rectangle $ABCD$, if the segment $EF$ is perpendicular to the diagonal $BD$. | [
"Write $\\angle BAC = \\alpha$. Since $ABF$ is a right triangle and $E$ is the midpoint of the hypotenuse, it is also the circumcentre of the triangle $ABF$ and $|AE| = |BE| = |FE|$. So, $\\angle AFE = \\angle EAF = \\alpha$ and $\\angle EFB = \\frac{\\pi}{2} - \\angle AFE = \\frac{\\pi}{2} - \\alpha$, which implie... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof and answer | sqrt(3) | |
0kpw | Problem:
Let $\triangle ABC$ be an isosceles right triangle with $AB = AC = 10$. Let $M$ be the midpoint of $BC$ and $N$ the midpoint of $BM$. Let $AN$ hit the circumcircle of $\triangle ABC$ again at $T$. Compute the area of $\triangle TBC$. | [
"Solution:\nNote that since quadrilateral $BAC T$ is cyclic, we have\n$$\n\\angle BTA = \\angle BCA = 45^\\circ = \\angle CBA = \\angle CTA\n$$\nHence, $TA$ bisects $\\angle BTC$, and $\\angle BTC = 90^\\circ$. By the angle bisector theorem, we then have\n$$\n\\frac{BT}{TC} = \\frac{BN}{NC} = \\frac{1}{3}.\n$$\nBy ... | United States | HMMT November 2022 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | final answer only | 30 | |
029q | Problem:
Na tabela ao lado, com 6 colunas e diversas linhas, estão escritos os números $1,2,3,4, \ldots$ Qual é a posição do número $1000$?
| 1 | 2 | 3 | 4 | 5 | 6 |
| :---: | :---: | :---: | :---: | :---: | :---: |
| 7 | 8 | 9 | 10 | 11 | 12 |
| 13 | 14 | $\cdots$ | | | |
| | | | | | |
| | | | | | |
| ... | [
"Solution:\nComo a tabela tem 6 colunas, em cada linha escrevemos 6 números consecutivos. Dividindo $1000$ por $6$ obtemos\n$$\n1000 = 6 \\times 166 + 4\n$$\n\n| $1^{\\text{a}}$ linha | 1 | 2 | 3 | 4 | 5 | 6 |\n| :---: | :---: | :---: | :---: | :---: | :---: | :---: |\n| $2^{\\text{a}}$ linha | 7 | 8 | 9 | 10 | 11 ... | Brazil | Nível 2 | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | row 167, column 4 | |
07yu | Problem:
Qual è il massimo numero intero positivo che ha lo stesso numero di cifre in base 10 e in base 16? (Le risposte sono espresse in base 10)
(A) 1024
(B) 99'999
(C) 999"999
(D) 1“600”000
(E) Nessuna delle precedenti | [] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | MCQ | B | |
0f6c | Problem:
The sequence $\{a_1, a_2, a_3, \ldots\}$ satisfies $a_{4n + 1} = 1$, $a_{4n + 3} = 0$, $a_{2n} = a_n$. Show that it is not periodic. | [] | Soviet Union | 19th ASU | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0cc0 | A store sold 235 robots over the course of twelve months. Each month, 16, 20, or 25 robots were sold. Find the number of months in which exactly 20 robots were sold. | [
"If exactly 16 robots were sold each month, the number of the robots sold would have been $12 \\cdot 16 = 192$. The difference of 43 robots comes from the months where 20 robots were sold (4 more each month) and from the months where 25 robots were sold (9 more each month). Denote by $a$ and $b$ the number of month... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - DISTRICT ROUND | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 4 | |
09pw | Problem:
Zij $n \geq 10$ een geheel getal. We schrijven $n$ in het tientallig stelsel. Zij $S(n)$ de som van de cijfers van $n$. Een stomp van $n$ is een positief geheel getal dat verkregen is door een aantal (minstens één, maar niet alle) cijfers van $n$ aan het rechteruiteinde weg te halen. Bijvoorbeeld: $23$ is een... | [
"Solution:\n\nVan rechts naar links geven we de cijfers van $n$ aan met $a_{0}, a_{1}, \\ldots, a_{k}$. Er geldt dus\n$$\nn = a_{0} + 10 a_{1} + \\cdots + 10^{k} a_{k}.\n$$\nEen stomp van $n$ bestaat van rechts naar links uit de cijfers $a_{i}, a_{i+1}, \\ldots, a_{k}$ waarbij $1 \\leq i \\leq k$. Zo'n stomp is dan... | Netherlands | Dutch TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
09gq | Let $ABC$ be a triangle. Take the points $D$ and $E$ outside and $F$ inside of the triangle. If $ADB$, $BEC$ and $CFA$ are all similar to each other and isosceles triangles with bases $AB$, $BC$ and $CA$, respectively then prove that $DBEF$ is a parallelogram. | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0dyp | On every square of an $n \times n$ chessboard we write one of the numbers $1$ or $-1$. Let $a_k$ be the product of all numbers in the $k^{\text{th}}$ row and $b_l$ be the product of all numbers in the $l^{\text{th}}$ column. Assuming $n = 2007$, can we choose the numbers in such a way that the sum
$$
a_1 + a_2 + \dots ... | [
"If $n = 2008$ fill the first row of the board with $-1$ and all other squares with $1$. Then $a_k = 1$ for all $k$ and $b_l = -1$ for all $l$, so\n$$\na_1 + a_2 + \\dots + a_{2008} + b_1 + b_2 + \\dots + b_{2008} = 1 + \\dots + 1 + (-1) + \\dots + (-1) = 0.\n$$\n\nNow, let $n = 2007$. We will show that we cannot c... | Slovenia | Slovenija 2008 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | For n = 2008: Yes. For n = 2007: No. | |
0chp | Let $n \ge 3$ be a positive integer, set $M = \{1, 2, \dots, n\}$ and let $k > 0$ be a real number. Let's associate each non-empty subset of $M$ with a point in the plane, such that any two distinct subsets correspond to different points. If the absolute value of the difference between the arithmetic means of the eleme... | [
"To show that $\\frac{1}{2}$ is the required minimum, notice that:\n\n* $1 \\le m_A \\le n$, for every nonempty subset $A \\subset M$, (1);\n* any two one-element subsets are connected with a sequence of subsets, (2).\n\nIndeed, for $k < p$, consider the sequence $\\{k\\}, \\{k, k+1\\}, \\{k+1\\}, \\{k+1, k+2\\}, \... | Romania | 74th NMO Selection Tests for JBMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof and answer | 1/2 | |
02fx | Two players play a game as follows. There are $n > 1$ rounds and $d \ge 1$ is fixed. In the first round A picks a positive integer $m_1$, then B picks a positive integer $n_1 \ne m_1$. In round $k$ (for $k = 2, \dots, n$), A picks an integer $m_k$ such that $m_{k-1} < m_k \le m_{k-1} + d$. Then B picks an integer $n_k$... | [
"$B$ has a winning strategy. Let $s = N!$ for $N$ sufficiently large (so $N!$ is divisible by all $m_i$'s). $B$ can compute $N$ after $A$ makes his first choice, because he knows that the biggest number that $A$ can choose in any round is at most $m_1 + (n-1)d$, so $N$ may be any number bigger than $m_1 + (n-1)d$. ... | Brazil | XX OBM | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof and answer | B | |
021a | Problem:
Let $n$ be a positive integer. In a coordinate grid, a path from $(0,0)$ to $(2 n, 2 n)$ consists of $4 n$ consecutive unit steps $(1,0)$ or $(0,1)$. Prove that the number of paths that divide the square with vertices $(0,0)$, $(2 n, 0)$, $(2 n, 2 n)$, $(0,2 n)$ into two regions with even areas is
$$
\frac{\bi... | [
"Solution:\nLet $X$ denote the set of paths for which $A$ and $B$ have even area and let $Y$ denote the set of paths for which $A$ and $B$ both have odd area. Because $A$ and $B$ together form a square of area $4 n^{2}$, which is even, $|X|+|Y|$ equals the total number of paths from $(0,0)$ to $(2 n, 2 n)$, which i... | Benelux Mathematical Olympiad | 16th Benelux Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | proof only | null | |
071t | Problem:
Let $ABC$ be a triangle and $D$ be the mid-point of side $BC$. Suppose $\angle DAB = \angle BCA$ and $\angle DAC = 15^{\circ}$. Show that $\angle ADC$ is obtuse. Further, if $O$ is the circumcentre of $ADC$, prove that triangle $AOD$ is equilateral. | [
"Solution:\n\nLet $\\alpha$ denote the equal angles $\\angle BAD = \\angle DCA$. Using sine rule in triangles $DAB$ and $DAC$, we get\n$$\n\\frac{AD}{\\sin B} = \\frac{BD}{\\sin \\alpha}, \\quad \\frac{CD}{\\sin 15^{\\circ}} = \\frac{AD}{\\sin \\alpha}\n$$\nEliminating $\\alpha$ (using $BD ... | India | INMO | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circl... | null | proof only | null | |
00q0 | Let $c(O, R)$ be a circle, $AB$ a diameter and $C$ an arbitrary point different than $A$ and $B$ such that $AOC > 90^\circ$. On the radius $OC$ we consider point $K$ and the circle $c_1(K, KC)$. The extension of the segment $KB$ meets the circle ($c$) at point $E$. From $E$ we consider the tangents $ES$ and $ET$ to the... | [
"Let the lines $BE$ and $ST$ intersect at point $L$. It is enough to prove that $AC$ passes through $L$, that is, the points $A, L, C$ are collinear (Figure 1).\n\n\n\nWe observe that the circles $c(O, R)$ and $c_1(K, KC)$ are homothetic with respect to homothety $H(C, m)$, that is, homothe... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Brocard point, symmedians",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
05qo | Problem:
Soient $p$ et $n$ des entiers strictement positifs, avec $p$ premier, et $n \geqslant p$. On suppose que $1+n p$ est un carré parfait. Montrer que $n+1$ est une somme de $p$ carrés parfaits non nuls (non nécessairement distincts). | [
"Solution:\n\nSoit $k$ tel que $1+n p = k^{2}$, alors $n p = (k-1)(k+1)$.\n\nPremier cas : $p$ divise $k-1$. Alors on peut écrire $k-1 = p \\ell$, donc $k = p \\ell + 1$. En reportant dans l'égalité, on obtient que\n$$\nn+1 = p \\ell^{2} + 2 \\ell + 1 = (p-1) \\ell^{2} + (\\ell+1)^{2}.\n$$\n\nDeuxième cas : $p$ div... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0eas | Which digit of the 7-digit number $2345678$ should we delete to get a 6-digit number divisible by $9$?
(A) $8$ (B) $7$ (C) $6$ (D) $5$ (E) $4$ | [
"A positive integer is divisible by $9$ if and only if the sum of its digits is divisible by $9$. The sum of the digits of $2345678$ is equal to $35 = 3 \\cdot 9 + 8$, so we should delete the digit $8$. We obtain the 6-digit number $234567$. The sum of its digits is equal to $27 = 3 \\cdot 9$, so it must be divisib... | Slovenia | National Math Olympiad in Slovenia | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | A | |
0jox | Problem:
Find the largest real number $k$ such that there exists a sequence of positive reals $\{a_{i}\}$ for which $\sum_{n=1}^{\infty} a_{n}$ converges but $\sum_{n=1}^{\infty} \frac{\sqrt{a_{n}}}{n^{k}}$ does not. | [
"Solution:\nFor $k > \\frac{1}{2}$, I claim that the second sequence must converge. The proof is as follows: by the Cauchy-Schwarz inequality,\n$$\n\\left(\\sum_{n \\geq 1} \\frac{\\sqrt{a_{n}}}{n^{k}}\\right)^{2} \\leq \\left(\\sum_{n \\geq 1} a_{n}\\right)\\left(\\sum_{n \\geq 1} \\frac{1}{n^{2k}}\\right)\n$$\nSi... | United States | HMMT November 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | 1/2 | |
07er | Let $ABC$ be an acute-angled triangle. The altitudes $BE$, $CF$ meet at $H$. $O$ is the circumcenter of triangle $ABC$. $M$ is the midpoint of $BC$. $P$ is the point on $EF$ such that $HP \perp HO$. $Q$ is the point on $HA$ such that $PQ \perp HM$. Prove that $QA = 3QH$. | [
"Let $(O)$ be the circumcircle of triangle $ABC$. $AK$ is the diameter of $(O)$. If $D$, $S$, $T$ are the reflection points of $H$ through $BC$, $CA$, $AB$, then $D$, $S$, $T$ are on $(O)$. It's easily seen that $HCKB$ is a parallelogram so $M$ lies on $HK$. Let $KH$ intersect $(O)$ again at $G$. Lines $HP$ interse... | Iran | Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous ... | English | proof only | null | |
0fvp | Problem:
Sei $ABC$ ein Dreieck und $D$ ein Punkt in dessen Inneren. Sei $E$ ein von $D$ verschiedener Punkt auf der Geraden $AD$. Seien $\omega_{1}$ und $\omega_{2}$ die Umkreise der Dreiecke $BDE$ bzw. $CDE$. $\omega_{1}$ und $\omega_{2}$ schneiden die Seite $BC$ in den inneren Punkten $F$ bzw. $G$. Der Schnittpunkt ... | [
"Solution:\n\nDie Schnittpunkte von $\\omega_{1}$ und $\\omega_{2}$ mit den Seiten $AB$ bzw. $AC$ seien $P$ bzw. $Q$. Weil $A$ auf der Potenzlinie der beiden Kreise liegt, gilt\n$$\nAP \\cdot AB = AE \\cdot AD = AQ \\cdot AC\n$$\nSomit ist $BCQP$ ein Sehnenviereck und es gilt $\\angle ABC = \\angle AQP$. Um zu zeig... | Switzerland | IMO Selektion | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0jas | Determine, with proof, whether or not there exist integers $a, b, c > 2010$ satisfying the equation
$$
a^3 + 2b^3 + 4c^3 = 6abc + 1.
$$ | [
"We claim there do exist such integers $a, b, c$.\n\nNote that $(a_1, b_1, c_1) = (1, 1, 1)$ satisfy the given equation. For $n > 1$, define $(a_{n+1}, b_{n+1}, c_{n+1})$ by\n$$\n(a_{n+1}, b_{n+1}, c_{n+1}) = (a_n + 2c_n + 2b_n, b_n + a_n + 2c_n, c_n + b_n + a_n).\n$$\nIt is not hard to verify algebraically that\n$... | United States | TST | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | Yes, such integers exist (in fact, infinitely many). | |
0i31 | Problem:
Find the volume of the tetrahedron with vertices $(5,8,10)$, $(10,10,17)$, $(4,45,46)$, $(2,5,4)$. | [
"Solution:\nEach vertex $(x, y, z)$ obeys $x + y = z + 3$, so all the vertices are coplanar and the volume of the tetrahedron is $0$."
] | United States | Harvard-MIT Math Tournament | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | final answer only | 0 | |
09dy | Let $a$, $b$, $c$, $d$ be positive real numbers with $a + b + c + d = 4$. Prove the inequality
$$
\frac{(a + \sqrt{b})^2}{\sqrt{a^2 - ab + b^2}} + \frac{(b + \sqrt{c})^2}{\sqrt{b^2 - bc + c^2}} + \frac{(c + \sqrt{d})^2}{\sqrt{c^2 - cd + d^2}} + \frac{(d + \sqrt{a})^2}{\sqrt{d^2 - da + a^2}} \le 16.
$$ | [
"$$(a + \\sqrt{b})^2 = a^2 + 2a\\sqrt{b} + b \\le a^2 + a(b + 1) + b = (a + b)(a + 1).$$\nBy Cauchy's mean theorem $\\sqrt{a^2 - ab + b^2} \\ge \\sqrt{a^2 - \\frac{a^2 + b^2}{2} + b^2} = \\sqrt{\\frac{a^2 + b^2}{2}} \\ge \\frac{a + b}{2}$ from where we get $\\frac{(a + \\sqrt{b})^2}{\\sqrt{a^2 - ab + b^2}} \\ge \\f... | Mongolia | Mongolian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0069 | Sea $ABC$ un triángulo isósceles de base $AB$. Una semicircunferencia $\Gamma$ con centro en el segmento $AB$ es tangente a los lados iguales $AC$ y $BC$. Se considera una recta tangente a $\Gamma$ que corta los segmentos $AC$ y $BC$ en $D$ y $E$, respectivamente. Las rectas perpendiculares a $AC$ y $BC$ trazadas respe... | [] | Argentina | XIX Olimpiada de Matemática de Países del Cono Sur | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | Spanish | proof only | null | |
0exw | Problem:
A rectangle $ABCD$ is drawn on squared paper with its vertices at lattice points and its sides lying along the gridlines. $AD = k \; AB$ with $k$ an integer. Prove that the number of shortest paths from $A$ to $C$ starting out along $AD$ is $k$ times the number starting out along $AB$. | [
"Solution:\n\nLet $ABCD$ have $n$ lattice points along the side $AB$. Then it has $kn$ lattice points along the side $AD$. Let $X$ be the first lattice point along $AB$ after leaving $A$. A shortest path from $X$ to $C$ must involve a total of $kn + n - 1$ moves between lattice points, $n - 1$ in the direction $AB$... | Soviet Union | 6th ASU | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0573 | a. Given a convex quadrilateral $ABCD$ with $AD < AB$ and $CD < CB$, is the internal angle at $B$ always less than the internal angle at $D$?
b. The same question for a non-convex quadrilateral. | [
"a.\nConsider the triangles $ADB$ and $CDB$ (Fig. 21). The claim $AD < AB$ implies $\\angle ABD < \\angle ADB$ because the longer side is opposite to the larger angle. Similarly, $CD < CB$ implies $\\angle CBD < \\angle CDB$. As $ABCD$ is convex, $\\angle ABD + \\angle CBD = \\angle ABC$ and $\\angle ADB + \\angle ... | Estonia | Final Round of National Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | a: Yes. b: No. | |
02pt | Problem:
Com os algarismos $a$, $b$ e $c$ construímos o número de três algarismos $abc$ e os números de dois algarismos $ab$, $bc$ e $ca$. Ache todos os possíveis valores de $a$, $b$ e $c$ tais que $$\frac{abc + a + b + c}{ab + bc + ca}$$ seja um número inteiro.
Sugestão: Mostre que o denominador é sempre divisível p... | [
"Solution:\n\nObservemos que\n$$\nab + bc + ca = (10a + b) + (10b + c) + (10c + a) = 11(a + b + c)\n$$\nde forma que o denominador da fração é divisível por $11$. Como a fração é um inteiro, o numerador\n$$\nabc + a + b + c = (100a + 10b + c) + a + b + c = 101a + 11b + 2c\n$$\ntambém é divisível por $11$. Como\n$$\... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (a,b,c) = (5,1,6), (9,1,2), (6,4,5), (3,7,8), (5,7,6), (7,7,5), (9,7,2) | |
0931 | Problem:
Determine the smallest possible real constant $C$ such that the inequality
$$
\left|x^{3}+y^{3}+z^{3}+1\right| \leqslant C\left|x^{5}+y^{5}+z^{5}+1\right|
$$
holds for all real numbers $x, y, z$ satisfying $x+y+z=-1$. | [
"Solution:\nThe key for our solution is the replacement of $1$ by $-(x+y+z)^{3}$ and $-(x+y+z)^{5}$ on the LHS and RHS, respectively, of the inequality under consideration. Thus we have to deal with the equivalent inequality\n$$\n\\left|x^{3}+y^{3}+z^{3}-(x+y+z)^{3}\\right| \\leqslant C \\cdot\\left|x^{5}+y^{5}+z^{... | Middle European Mathematical Olympiad (MEMO) | Middle European Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 9/10 | |
0evg | Let $n$ and $k$ be integers satisfying $n \ge 2$ and $k \ge \frac{5}{2}n - 1$. Prove that in every choice of $k$ distinct points among all integer points $(x, y)$ with $1 \le x, y \le n$, there exists a circle going through at least four distinct chosen points. | [
"Let $a_i$ be the number of chosen points on the line $y = i$. If $a_i \\ge 2$ and $x_1 < x_2 < \\dots < x_{a_i}$ are the x-coordinates of the chosen points on the line $y = i$, then because\n$$\nx_1 + x_2 < x_1 + x_3 < x_2 + x_3 < x_2 + x_4 < x_3 + x_4 < \\dots < x_{a_1-1} + x_{a_1}\n$$\nwe deduce that the number ... | South Korea | Korean Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | English | proof only | null | |
09fk | Let $x$, $y$, $z$ be positive real numbers satisfying $x + y + z + xyz = 4$. Show that
$$
\frac{x}{\sqrt{2y+3z}} + \frac{y}{\sqrt{2z+3x}} + \frac{z}{\sqrt{2x+3y}} \ge \frac{1}{\sqrt{5}}(x+y+z).
$$ | [
"First note that the function $x \\mapsto x^{-\\frac{1}{2}}$ is convex on $(0, \\infty)$. Thus by Jensen's inequality, we have\n$$\n\\begin{aligned}\n& \\frac{x}{x+y+z} (2y+3z)^{-\\frac{1}{2}} + \\frac{y}{x+y+z} (2z+3x)^{-\\frac{1}{2}} + \\frac{z}{x+y+z} (2x+3y)^{-\\frac{1}{2}} \\\\\n\\ge & \\left( \\frac{x(2y+3z) ... | Mongolia | 51st Mongolian National Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0jy4 | Problem:
Michael writes down all the integers between $1$ and $N$ inclusive on a piece of paper and discovers that exactly $40\%$ of them have leftmost digit $1$. Given that $N > 2017$, find the smallest possible value of $N$. | [
"Solution:\n\nLet $d$ be the number of digits of $N$. Suppose that $N$ does not itself have leftmost digit $1$. Then the number of integers $1, 2, \\ldots, N$ which have leftmost digit $1$ is\n$$\n1 + 10 + 10^{2} + \\ldots + 10^{d-1} = \\frac{10^{d} - 1}{9}\n$$\nso we must have $\\frac{10^{d} - 1}{9} = \\frac{2N}{5... | United States | HMMT November 2017 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 1481480 | |
07lb | Let $n = (p^2 + 2)^2 - 9(p^2 - 7)$ where $p$ is a prime number. What is the smallest value of the sum of the digits of $n$ and for what prime numbers $p$ is this value attained? | [
"When $p = 2$, $n = 63$. When $p = 3$, $n = 103$. When $p = 5$, $n = 567$.\n\n$$\n\\begin{aligned}\n(p^2 + 2)^2 - 9(p^2 - 7) &= p^4 - 5p^2 + 4 + 63 \\\\\n&= (p^2 - 1)(p^2 - 4) + 63 \\\\\n&= (p - 1)(p + 1)(p - 2)(p + 2) + 63 \\\\\n&= (p - 2)(p - 1)(p + 1)(p + 2) + 63\n\\end{aligned}\n$$\n\nWhen $p \\neq 3$, $(p-2)(p... | Ireland | Irska | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | 4; attained at p = 3 | |
0bdt | In a division of two positive integers, the dividend and the divisor are directly proportional with the remainder and the quotient. The remainder and the quotient are relatively prime. Prove that the dividend is a perfect square. | [
"Let the dividend be $a$, the divisor be $b$, the quotient be $q$, and the remainder be $r$, so that\n$$\na = bq + r,\n$$\nwith $0 \\leq r < b$.\n\nWe are told that $a$ and $b$ are directly proportional to $r$ and $q$, respectively. That is, there exists a constant $k > 0$ such that\n$$\na = k r, \\quad b = k q.\n$... | Romania | Shortlisted Problems for the 64th NMO | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
02cf | Problem:
Em uma circunferência foram marcados $15$ pontos brancos e $1$ ponto preto. Consideremos todos os possíveis polígonos (convexos) com seus vértices nestes pontos.
Vamos separá-los em dois tipos:
- Tipo 1: os que possuem somente vértices brancos.
- Tipo 2: os que possuem o ponto preto como um dos vértices.
Exist... | [
"Solution:\nObserve que para cada polígono do tipo $1$ podemos construir um polígono do tipo $2$ adicionando o ponto preto.\nPor outro lado, se temos um polígono do tipo $2$ e retirarmos o ponto preto, a única forma de não gerar um polígono é se sobrarem exatamente dois pontos brancos.\nPortanto, existem mais políg... | Brazil | Desafios | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | Type 2 has 105 more polygons. | |
0bgv | Problem:
Fie $n \geq 3$ un număr întreg şi fie un cerc pe care marcăm $n+1$ puncte echidistante. Considerăm toate numerotările acestor puncte cu numerele $0,1, \ldots, n$ astfel încât fiecare număr este folosit exact o dată; două astfel de numerotări se consideră identice dacă printr-o rotaţie a cercului coincid. O nu... | [] | Romania | Romania Olympiad | [
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
09df | Бат орцныхоо оршин суугчдаас авсан асуулгаар тэлгээрийн 25-нь шатар тоглодог, 30 нь гадаад явж үзсэн, 28-нь онгоцоор нисч үзсэн гэсэн хариулт авав. Мөн онгоцоор нисч байсан хүмүүсээс 18 нь шатар тоглодог, 17 нь гадаад явж үзсэн байв. Шатар тоглодог, гадаад явж байсан 16 оршин суугчийн 15 нь онгоцоор нисч байсан ба орцн... | [
"Шатар тоглодог хүмүүсийн олонлогийг $A_1$, онгоцоор нислэг хүмүүсийн олонлогийг $A_2$, гадаад явсан хүмүүсийн олонлогийг $A_3$ гэж тэмдэглэвэл:\n\n$|A_1| = 25$\n\n$|A_2| = 28$\n\n$|A_3| = 30$\n\n$|A_1 \\cap A_2| = 18$\n\n$|A_2 \\cap A_3| = 17$\n\n$|A_3 \\cap A_1| = 16$\n\n$|A_1 \\cap A_2 \\cap A_3| = 15$\n\nНэгтгэ... | Mongolia | ММО-48 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | Mongolian | proof and answer | No | |
00mw | A (convex) trapezoid $ABCD$ shall be called *good* if it is inscribed, has parallel sides $AB$ and $CD$, and $CD$ is shorter than $AB$. For a good trapezoid, we fix the following notations.
* The line parallel to $AD$ through $B$ intersects the line $CD$ in $S$.
* The tangents through $S$ to the circumcircle of the tra... | [
"**Answer.** The angles $\\angle BSE$ and $\\angle FSC$ are equal if and only if $\\angle BAD = 60^\\circ$ or $AB = AD$.\n\nWe denote the circumcircle of the trapezoid by $u$, the second intersection point of the line $SB$ with $u$ by $T$ and the centre of $u$ by $M$, see Figure 4. As the trapezoid is inscribed, it... | Austria | Austria2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocente... | English | proof and answer | The angles ∠BSE and ∠FSC are equal if and only if ∠BAD = 60° or AB = AD. | |
00ej | Magalí's calculator has a special button ⋆ that works as follows. Every time she presses ⋆, the calculator multiplies the number on the screen by itself, then adds 6, and finally shows the result on the screen. For example, if the number on the screen is $11$ and Magalí presses ⋆, the number that will appear on the scr... | [
"The answer is 3. First, notice that the last digit of Magalí's initial number, name it $n_1$, determines the last digit of the following numbers. This is explained in the following table, where we name $n_2 = n_1^2 + 6$, $n_3 = n_2^2 + 6$, $n_4 = n_3^2 + 6$ and $n_5 = n_4^2 + 6$:\n\n| $n_1$ | $n_2$ | $n_3$ | $n_4$... | Argentina | Rioplatense Mathematical Olympiad | [
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 3 | |
085b | Problem:
Due circonferenze con lo stesso raggio si intersecano in $X$ e $Y$. Sia $P$ un punto su un arco $XY$ di una circonferenza interno all'altra. Sapendo che il segmento $XY$ è lungo $3$ e che l'angolo $X\widehat{P}Y$ misura $120^\circ$, qual è l'area dell'intersezione tra i due cerchi?
(A) $2\left(\pi-\frac{1}{4... | [
"Solution:\n\nLa risposta è (E). L'intersezione delle due circonferenze è divisa da $XY$ in due figure congruenti, delimitate dagli archi $XY$, quindi per ottenere il risultato basta calcolare l'area di una di queste due figure e moltiplicare per due. Sia $O$ il centro della circonferenza su cui giace $P$; per le p... | Italy | Progetto Olimpiadi di Matematica - GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | MCQ | E | |
019u | For which positive integers $k$ can the integers $1, 2, 3, \dots, (2k)^2$ be arranged as a $2k \times 2k$ table in such a way that all row sums and column sums were of the same parity, opposite to that of $k$? | [
"**Answer:** for all $k \\ge 2$.\n\nSolution:\nSuch an arrangement is impossible for $k = 1$. In order to make all row sums and column sums even, both odd numbers should occur in the same row and also in the same column, which is impossible. In the rest, let $0$ and $1$ denote any even and odd number, respectively.... | Baltic Way | Baltic Way 2013 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | k >= 2 | |
0696 | Let $\triangle AB\Gamma$ be an acute angled triangle with circumcircle $c(O, R)$. From the midpoint $\Delta$ of the side $B\Gamma$ we draw a line perpendicular to $AB$ which meets $AB$ at $E$. If the line $AO$ intersects the line $\varepsilon$ at $Z$, prove that the points $A$, $Z$, $\Delta$, $\Gamma$ are cyclic. | [
"The external angle $\\mathrm{E}\\hat{Z}A$ of the quadrilateral $AZ\\Delta\\Gamma$ belongs to the orthogonal triangle $AEZ$, with the acute angle $\\mathrm{E}\\hat{A}Z = \\omega$ equal to the angle $A\\hat{B}O$, since $OA = OB$. Hence $\\mathrm{E}\\hat{Z}A = 90^\\circ - \\omega$\n\nLet the extension of the radius $... | Greece | SELECTION EXAMINATION | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0eur | Prove that there are no positive integers $x, y, z$ such that
$$
x^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 = 0.
$$ | [
"We will prove this statement by contradiction. Assume that there are positive integers $x, y, z$ satisfying the equation\n$$\nx^2 y^4 - x^4 y^2 + 4x^2 y^2 z^2 + x^2 z^4 - y^2 z^4 = 0. \\quad (1)\n$$\nIt is easy to check that $x \\neq y$. If there are solutions of the equation (1), we have a solution of the equatio... | South Korea | 24th Korean Mathematical Olympiad Final Round | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof only | null |
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