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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
08hs | Problem:
In a rectangular system $xOy$ the graph of the function $f: \mathbb{R}
ightarrow \mathbb{R}$, $f(x) = x^{2}$ is drawn. The ordered triple $B, A, C$ has distinct points on the parabola, the point $D \in BC$ such that the straight line $AD$ is parallel to the axis $Oy$ and the triangles $BAD$ and $CAD$ have th... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | (4 s1 s2)^{1/3} | |
0h4s | Let $AB$ be a diameter of a circle $\omega$, and points $M$ and $C$ on $\omega$ be in different half-planes with respect to the line $AB$. Perpendiculars $MN$ and $MK$ are dropped from the point $M$ to the lines $AB$ and $AC$ respectively. Prove that the line $KN$ bisects the line segment $CM$.
(Igor Nagel) | [] | Ukraine | Ukrainian National Mathematical Olympiad | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0e4e | Problem:
Poišči vsa realna števila $x$ in $y$, za katera velja $x + y^{2} = x y + 1$ in $x y = 4 + y$. | [
"Solution:\n\nIz prve enačbe sledi $x(1-y) = 1 - y^{2}$ oziroma $(1-y)(x-1-y) = 0$.\nČe je $y = 1$, ta enačba velja, iz druge pa sledi $x = 5$.\nV primeru $y \\neq 1$ dobimo $x = 1 + y$.\nSkupaj z drugo enačbo tedaj velja $(1 + y) y = 4 + y$ oziroma $y^{2} = 4$.\nOd tod sledi $y = 2$ ali $y = -2$.\n\nEnačbi veljata... | Slovenia | 55. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | (x, y) = (5, 1), (3, 2), (-1, -2) | |
08qw | Problem:
Find all quadruples of positive integers $(p, q, a, b)$, where $p$ and $q$ are prime numbers and $a>1$, such that
$$
p^{a}=1+5 q^{b}
$$ | [
"Solution:\nFirst of all, observe that if $p, q$ are both odd, then the left hand side of the given equation is odd and the right hand side is even so there are no solutions in this case. In other words, one of these numbers has to be equal to $2$ so we can discuss the following two cases:\n\n- $p=2$\n\nIn this cas... | JBMO | JBMO | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | [(2,3,4,1), (3,2,4,4)] | |
0131 | Problem:
Let $a$ and $b$ be positive integers. Prove that if $a^{3}+b^{3}$ is the square of an integer, then $a+b$ is not a product of two different prime numbers. | [
"Solution:\nSuppose $a+b=pq$, where $p \\neq q$ are two prime numbers. We may assume that $p \\neq 3$. Since\n$$\na^{3}+b^{3}=(a+b)\\left(a^{2}-a b+b^{2}\\right)\n$$\nis a square, the number $a^{2}-a b+b^{2}=(a+b)^{2}-3 a b$ must be divisible by $p$ and $q$, whence $3 a b$ must be divisible by $p$ and $q$. But $p \... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
049k | Determine the largest possible quotient of a three-digit number and the sum of its digits. | [] | Croatia | Hrvatska 2011 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 100 | |
0ein | Problem:
Ali obstaja tako praštevilo $p$, da velja $p^{2}+p+1=n^{3}$ za neko naravno število $n$? | [
"Solution:\n\nPokazali bomo, da takšno praštevilo ne obstaja. Enačbo preoblikujemo v $p^{2}+p=n^{3}-1$ in obe strani razstavimo, da dobimo $p(p+1)=(n-1)(n^{2}+n+1)$. Če $p$ deli $n-1$, tedaj $n^{2}+n+1$ deli $p+1$. Od tod sledi $n^{2}+n+1 \\leq p+1 \\leq n$, kar pa je protislovje. Torej mora $p$ deliti $n^{2}+n+1$ ... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof only | null | |
0bse | Consider matrices $A$, $B$, $C$, $D \in \mathcal{M}_n(\mathbb{C})$, $n \ge 2$ și $k \in \mathbb{R}$ so that $AC + kBD = I_n$ and $AD = BC$. Demonstrate that $CA + kDB = I_n$ and $DA = CB$. | [
"$$\n(CA + kDB) - w(DA - CB) = I_n\n$$\nand\n$$\n(CA + kDB) + w(DA - CB) = I_n,\n$$\nwhich give the result. For $k = 0$ we get $AC = I_n$, so $CA = I_n$. From $AD = BC$ and $CA = I_n$ we obtain $ADA = B$ and $DA = CB$."
] | Romania | 67th Romanian Mathematical Olympiad | [
"Algebra > Linear Algebra > Matrices"
] | English | proof only | null | |
0e3q | Does there exist an integer $n$ such that all roots of the polynomial $p(x) = x^4 - 2011x^2 + n$ are integers? | [
"Assume that such $n$ exists. From $x^4 - 2011x^2 + n = 0$ we deduce that\n$$\nx^2 = \\frac{2011 \\pm \\sqrt{2011^2 - 4n}}{2}.\n$$\nThis has to be an integer, so $2011^2 - 4n$ has to be a perfect square. We can write $2011^2 - 4n = m^2$ for some odd positive integer $m$ or $n = \\frac{2011^2 - m^2}{4}$. So, $x^2 = ... | Slovenia | National Math Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Number Theory > Modular Arithmetic"
] | null | proof and answer | No, such an integer n does not exist. | |
03n3 | Let $a$, $b$, and $c$ be non-negative real numbers, no two of which are equal. Prove that
$$
\frac{a^2}{(b-c)^2} + \frac{b^2}{(c-a)^2} + \frac{c^2}{(a-b)^2} > 2.
$$ | [
"The left-hand side is symmetric with respect to $a$, $b$, $c$. Hence, we may assume that $a > b > c \\ge 0$. Note that replacing $(a, b, c)$ with $(a-c, b-c, 0)$ lowers the value of the left-hand side, since the numerators of each of the fractions would decrease and the denominators remain the same. Therefore, to ... | Canada | CMO 2017 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
058m | The bisector of the angle on vertex $A$ of triangle $ABC$ intersects the circumcircle of triangle $ABC$ at point $F$ ($F \neq A$). Points $D$ and $E$ are chosen on the sides $AB$ and $AC$, respectively, in such a way that the lines $DE$ and $BC$ are parallel. Let $G$ and $H$ be the points of intersection of the rays $F... | [
"Let $K$ and $L$ be the points of intersection of the line $AF$ with lines $BC$ and $DE$, respectively (Fig. 6). Then\n$$\n\\begin{aligned}\n\\angle AGD &= \\angle AGF = \\angle AGC + \\angle CGF = \\angle ABC + \\angle CAF \\\\\n&= \\angle ABK + \\angle KAB = \\angle CKA = \\angle KLD = 180^\\circ - \\angle ALD.\n... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0k9o | Problem:
A convex polygon on the plane is called wide if the projection of the polygon onto any line in the same plane is a segment with length at least $1$. Prove that a circle of radius $\frac{1}{3}$ can be placed completely inside any wide polygon.
Proposed by: Shengtong Zhang | [
"Solution:\n\nLemma. For any polygon including its boundary, there exists a largest circle contained inside it.\n\nProof. It's easy to see that for any circle inside the polygon, it can be increased in size until it is tangent to at least three sides of the polygon. Then for any three sides of the polygon, there is... | United States | HMMT February 2019 Team Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Optimizat... | null | proof only | null | |
0adh | Four points $A$, $B$, $C$ and $D$ in the plane are given such that $\overline{AB} = \overline{AC}$ and $\overline{AD} = \overline{BD}$. Let $E$ be a point from the plane $AC$, such that $A$ lies between $E$ and $C$ (see picture).
If $\alpha = \angle BAE$ and $\beta = \angle ADB$ and $\alpha + \beta = 200^\circ$, find ... | [
"Since $ABC$ is an isosceles triangle with base $BC$, and $\\alpha = \\angle BAE$, then $\\angle ACB = \\angle CBA = \\frac{180^\\circ - \\angle BAC}{2} = \\frac{\\alpha}{2}$.\n\nSince $ABD$ is isosceles triangle with base $AB$, then $\\angle DBA = \\angle BAD = \\frac{180^\\circ - \\beta}{2}$.\n\nFrom there, we ge... | North Macedonia | Macedonian Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10° | |
0fz9 | Problem:
Sei $n \geq 6$ eine natürliche Zahl. Betrachte eine Menge $S$ von $n$ verschiedenen reellen Zahlen. Beweise, dass es mindestens $n-1$ verschiedene zweielementige Teilmengen von $S$ gibt, sodass das arithmetische Mittel der beiden Elemente in jeder dieser Teilmengen mindestens gleich dem arithmetischen Mittel ... | [
"Solution:\n\nMit Paar meinen wir im folgenden eine zweielementige Teilmenge von $S$. Betrachte zuerst den Fall $n=6$, also $S=\\{x_{1}, x_{2}, \\ldots, x_{6}\\}$ und o.B.d.A $x_{1}<x_{2}<\\cdots<x_{6}$. Mit $M$ bezeichnen wir das arithmetische Mittel der Elemente aus $S$. Dann gilt sicher eine der folgenden Unglei... | Switzerland | IMO Selektion | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0jlx | Problem:
Find all integers $n$ for which $\frac{n^{3}+8}{n^{2}-4}$ is an integer. | [
"Solution:\n\n$0, 1, 3, 4, 6$\n\nWe have\n$$\n\\frac{n^{3}+8}{n^{2}-4} = \\frac{(n+2)\\left(n^{2}-2n+4\\right)}{(n+2)(n-2)} = \\frac{n^{2}-2n+4}{n-2}\n$$\nfor all $n \\neq -2$. Then\n$$\n\\frac{n^{2}-2n+4}{n-2} = n + \\frac{4}{n-2},\n$$\nwhich is an integer if and only if $\\frac{4}{n-2}$ is an integer. This happen... | United States | HMMT 2014 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 0, 1, 3, 4, 6 | |
0iei | Problem:
The Dingoberry Farm is a $10$ mile by $10$ mile square, broken up into $1$ mile by $1$ mile patches. Each patch is farmed either by Farmer Keith or by Farmer Ann. Whenever Ann farms a patch, she also farms all the patches due west of it and all the patches due south of it. Ann puts up a scarecrow on each of he... | [
"Solution:\nWhenever Ann farms a patch $P$, she also farms all the patches due west of $P$ and due south of $P$. So, the only way she can put a scarecrow on $P$ is if Keith farms the patch immediately north of $P$ and the patch immediately east of $P$, in which case Ann cannot farm any of the patches due north of $... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 7 | |
0frc | Dado un número entero positivo $n$, definimos $\lambda(n)$ como el número de soluciones enteras positivas de la ecuación $x^2 - y^2 = n$. Diremos que el número $n$ es “olímpico” si $\lambda(n) = 2021$. ¿Cuál es el menor entero positivo que es olímpico? ¿Y cuál es el menor entero positivo impar que es olímpico? | [
"Distinguiremos 4 casos, según $n$ sea impar o par y según $n$ sea cuadrado perfecto o no.\n\na. Sea $n = p_1^{a_1} \\cdots p_r^{a_r}$ un número impar que no es cuadrado perfecto. Si $x^2 - y^2 = (x+y)(x-y) = n$, con $x, y > 0$, entonces existen enteros positivos $a, b$, con $a > b$ y teniendo ambos la misma parida... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | smallest olímpico: 2^48 * 3^42 * 5; smallest odd olímpico: 3^46 * 5^42 * 7 | |
00vu | Let $ABCD$ be a convex quadrilateral such that $AB^2 + BC^2 = AD^2 + CD^2$. Points $X$ and $Y$ are chosen such that $XD \perp CD$, $XB \perp AB$, $YB \perp BC$ and $YD \perp AD$. Let lines $AC$ and $XY$ meet at $T$ and $M$ be the midpoint of segment $XY$. Prove that points $T, M, B, D$ lie on a circle. | [
"Let $P$ be the midpoint of $AC$. Applying the formula for the length of the median on $\\triangle ABC$ and $\\triangle ADC$, and using the fact that $AB^2 + BC^2 = AD^2 + CD^2$, we obtain $BP = DP$.\n$$\n\\text{Claim.} \\quad \\angle BXA = \\angle PBD\n$$\n*Proof.* We use directed angles.\nLet $U$ be the projectio... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0jia | Problem:
Let $m$ be an odd positive integer greater than $1$. Let $S_{m}$ be the set of all non-negative integers less than $m$ which are of the form $x+y$, where $x y-1$ is divisible by $m$. Let $f(m)$ be the number of elements of $S_{m}$.
a. Prove that $f(m n)=f(m) f(n)$ if $m, n$ are relatively prime odd integers ... | [
"Solution:\n\nFor a positive integer $n$, let $\\mathbb{Z} / n \\mathbb{Z}$ denote the set of residues modulo $n$ and $(\\mathbb{Z} / n \\mathbb{Z})^{*}$ denote the set of residues modulo $n$ that are relatively prime to $n$. Then, rephrased, $S_{m}$ is the set of residues modulo $m$ of the form $x+x^{-1}$, where $... | United States | HMMT 2013 | [
"Number Theory > Modular Arithmetic > Chinese remainder theorem",
"Number Theory > Modular Arithmetic > Inverses mod n",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | a) f(mn) = f(m) f(n) for coprime odd integers m, n > 1.
b) For an odd prime p and integer k > 0,
f(p^k) = 2 + (p^{k-1}(p − 3))/2 + (p^{k-1} − p^{(1 + (−1)^k)/2})/(p + 1). | |
0lg6 | Problem:
Prove that there exist infinitely many pairs $(m, n)$ of positive integers such that $m+n$ divides $(m!)^{n}+(n!)^{m}+1$. | [
"Solution:\nWe shall find a pair such that $m+n=p$ is prime and $n$ is even. Applying Wilson's theorem we have\n$$\nm!=(p-n)!=\\frac{(p-1)!}{(p-n+1) \\ldots (p-2)(p-1)} \\equiv \\frac{-1}{-(n-1) \\ldots (-2)(-1)} \\equiv \\frac{1}{(n-1)!} \\equiv \\frac{n}{n!} \\quad (\\bmod p)\n$$\nIt follows from Fermat's Little ... | Zhautykov Olympiad | IZhO | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0179 | For which $k$ do there exist $k$ distinct primes $p_1, p_2, \dots, p_k$ such that
$$
p_1^2 + p_2^2 + \dots + p_k^2 = 2010?
$$ | [
"We show that it is possible only if $k = 7$.\nThe 15 smallest prime squares are:\n4, 9, 25, 49, 121, 169, 289, 361, 529, 841, 961, 1369, 1681, 1849, 2209.\nSince $2209 > 2010$ we see that $k \\le 14$.\nNow we note that $p^2 \\equiv 1 \\mod 8$ if $p$ is an odd prime. We also have that $2010 \\equiv 2 \\mod 8$. If a... | Baltic Way | BALTIC WAY | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 7 | |
035i | Problem:
Let $M$ be the set of the rational numbers in the interval $(0,1)$. Does there exist a subset $A$ of $M$ such that every number from $M$ can be represented in a unique way as a sum of one or finitely many distinct numbers from $A$? | [
"Solution:\nAssume, for a contradiction, that there exists such a set.\n\nWe first prove that if $a \\in A$, then $A \\cap \\left(\\frac{a}{2}, a\\right) = \\varnothing$. To do this suppose the contrary, i.e. there exists $a'$ in $A$ and $a > a' > \\frac{a}{2}$. Then the number $a - a' < \\frac{a}{2}$ can be repres... | Bulgaria | 54. Bulgarian Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | No; there is no such subset A. | |
05t4 | Problem:
Les entiers de 1 à 2020 sont écrits au tableau. Jacques a le droit d'en effacer deux et d'écrire à la place leur différence ou leur somme, et de recommencer jusqu'à ce qu'il ne reste plus qu'un entier. Est-il possible que l'entier obtenu à la fin soit 321 ? | [
"Solution:\n\nL'énoncé présente une suite finie d'opérations et le problème demande s'il est possible de partir de la situation initiale pour arriver à une certaine situation finale. Une première idée à essayer dans ce cas est de chercher un invariant.\n\nUne seconde idée est de tester le problème avec des plus pet... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
07yo | Problem:
Determinare tutte le coppie $\{a, b\}$ di interi positivi con la seguente proprietà: comunque si colorino gli interi positivi con due colori $A$ e $B$, esistono sempre due interi positivi del colore $A$ con differenza $a$ o due interi positivi del colore $B$ con differenza $b$. | [
"Solution:\n\nLe coppie che soddisfano la condizione del testo sono quelle del tipo $a=2^{h} \\cdot (2x+1)$, $b=2^{k} \\cdot (2y+1)$ dove $h \\neq k$, ovvero le coppie tali che la massima potenza di $2$ che divide i due numeri è diversa.\n\nPer prima cosa, mostriamo che per le coppie NON di questo tipo, ovvero del ... | Italy | null | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | All pairs where a=2^h·(odd), b=2^k·(odd) with h≠k; equivalently, the largest power of two dividing a and b is different. | |
05vq | Problem:
Soit $ABCD$ un parallélogramme tel que $AC = BC$. Soit $P$ un point situé sur le prolongement du segment $[AB]$ au-delà de $B$. Soit $Q$ le point d'intersection, autre que $D$, entre le segment $[PD]$ et le cercle circonscrit à $ACD$. Soit ensuite $R$ le point d'intersection, autre que $P$, entre le segment $... | [
"Solution:\n\nTout d'abord, on constate que\n$$\n(RA, RC) = (RA, RP) = (QA, QP) = (QA, QD) = (CA, CD) = (AC, AB) = (BA, BC)\n$$\nce qui signifie que les points $A$, $B$, $C$ et $R$ sont cocycliques.\n\nDe même, si l'on note $X$ le point d'intersection des droites $(AQ)$ et $(CD)$, on constate que\n$$\n(QR, QX) = (Q... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0fv7 | Problem:
Seien $a, b, c$ positive reelle Zahlen mit $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1$. Beweise die Ungleichung
$$
\sqrt{a b+c}+\sqrt{b c+a}+\sqrt{c a+b} \geq \sqrt{a b c}+\sqrt{a}+\sqrt{b}+\sqrt{c}
$$ | [
"Solution:\n\nDie Nebenbedingung ist äquivalent zu $a b c=a b+b c+c a$. Mit C.S. folgt\n$$\n\\begin{aligned}\n\\sqrt{a b+c} & =\\sqrt{\\frac{a b c+c^{2}}{c}}=\\sqrt{\\frac{a b+b c+c a+c^{2}}{c}}=\\frac{\\sqrt{(a+c)(b+c)}}{\\sqrt{c}} \\\\\n& \\geq \\frac{\\sqrt{a b}+c}{\\sqrt{c}}=\\frac{1}{c} \\sqrt{a b c}+\\sqrt{c}... | Switzerland | IMO Selektion | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
02cn | Problem:
Número ímpar - Se $n$ é um número inteiro qualquer, qual das seguintes opções é um número ímpar?
(a) $n^{2}-n+2$
(b) $n^{2}+n+2$
(c) $n^{2}+n+5$
(d) $n^{2}+5$
(e) $n^{3}+5$ | [
"Solution:\n\nLembremos que:\n- $n$ e $n^{2}$ têm a mesma paridade: $(\\text{par})^{2}=$ par e (ímpar $)^{2}=$ ímpar;\n- a soma ou diferença de números de mesma paridade é um número par: (par $\\pm$ par = par e ímpar $\\pm$ ímpar = par).\n\nSolução 1: Observemos que $n^{2}+n$ e $n^{2}-n$ são soma e diferença de doi... | Brazil | null | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | MCQ | c | |
00f7 | Suppose there are 997 points given in a plane. If every two points are joined by a line segment with its midpoint coloured in red, show that there are at least 1991 red points in the plane. Can you find a special case with exactly 1991 red points? | [
"Embed the points in the cartesian plane such that no two points have the same $y$-coordinate. Let $P_{1}, P_{2}, \\ldots, P_{997}$ be the points and $y_{1}<y_{2}<\\ldots<y_{997}$ be their respective $y$-coordinates. Then the $y$-coordinate of the midpoint of $P_{i} P_{i+1}$, $i=1,2, \\ldots, 996$ is $\\frac{y_{i}+... | Asia Pacific Mathematics Olympiad (APMO) | APMO 1991 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | At least 1991 red points; equality is achieved, for example, by placing the 997 points at P_i = (0, 2i), which yields exactly 1991 distinct red midpoints. | |
03hv | Problem:
Let $AB$ be a diameter of a circle, $C$ be any fixed point between $A$ and $B$ on this diameter, and $Q$ be a variable point on the circumference of the circle. Let $P$ be the point on the line determined by $Q$ and $C$ for which $\frac{AC}{CB} = \frac{QC}{CP}$. Describe, with proof, the locus of the point $P... | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Circles"
] | null | proof only | null | |
0kyc | What is the value of $101 \cdot 9,901 - 99 \cdot 10,101$?
(A) 2 (B) 20 (C) 21 (D) 200 (E) 2020 | [
"**Answer (A):** Write the difference as\n$$\n(100 + 1) \\cdot (9900 + 1) - 99 \\cdot (10,000 + 100 + 1).\n$$\nApplying the distributive property gives\n$$\n(990,000 + 9,900 + 100 + 1) - (990,000 + 9,900 + 99) = 100 + 1 - 99 = 2.\n$$\n\nLet $x = 100$. Then the minuend (the first quantity in the subtraction operatio... | United States | AMC 10 A | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | A | |
0ix9 | Problem:
The following grid represents a mountain range; the number in each cell represents the height of the mountain located there. Moving from a mountain of height $a$ to a mountain of height $b$ takes $(b-a)^2$ time. Suppose that you start on the mountain of height $1$ and that you can move up, down, left, or righ... | [
"Solution:\n\nAnswer: $212$\n\nConsider the diagonals of the board running up and to the right - so the first diagonal is the square $1$, the second diagonal is the squares $2$ and $3$, and so on. The $i$th ascent is the largest step taken from a square in the $i$th diagonal to a square in the $i+1$st. Since you mu... | United States | 2nd Annual Harvard-MIT November Tournament | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof and answer | 212 | |
0l4j | Problem:
C/1 Sugar Station sells 44 different kinds of candies, packaged one to a box. Each box is priced at a positive integer number of cents, and it costs $1.51$ to buy one of every kind. (There is no discount based on the number of candies in a purchase.) Unfortunately, Anna only has $0.75$.
a) Show that Anna can... | [
"Solution:\n\nFor part (a), pick boxes containing 22 different candies chosen at random. Let their total cost be $m$ cents. If $m \\leq 75$, Anna can buy these candies. Otherwise, $m \\geq 76$. In this case, the other 22 candies have a total cost of $151-m \\leq 75$ cents, so Anna can buy those candies instead.\n\n... | United States | 25th Bay Area Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof only | null | |
0d10 | Find the greatest real number $a$ such that for every positive real numbers $x, y, z$ we have
$$
\frac{x+1}{y} + \frac{2y+1}{z} + \frac{3z+1}{x} > a.
$$ | [
"For any positive real numbers $x, y, z$ we have\n$$\n\\begin{aligned}\n\\frac{x+1}{y} + \\frac{2y+1}{z} + \\frac{3z+1}{x} &= \\frac{x}{y} + \\frac{2y}{z} + \\frac{3z}{x} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} \\\\\n&\\geq 3\\sqrt{6} + \\frac{1}{x} + \\frac{1}{y} + \\frac{1}{z} > 3\\sqrt{6}.\n\\end{aligned}\n... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof and answer | 3√6 | |
0gv4 | Let $ABC$ be a scalene triangle, $I$ be its incenter and $O$ be its circumcenter. The line $IO$ intersects the lines $BC$, $CA$, $AB$ at points $D$, $E$, $F$, respectively. Let $A_1$ be the intersection of $BE$ and $CF$. The points $B_1$ and $C_1$ are defined similarly. The incircle of $ABC$ is tangent to sides $BC$, $... | [
"Let $M$ be the Miquel point of the quadrilateral defined by the lines $AB$, $AC$, $BC$, $IO$. We will prove that $M$ lies on all three circles. Since the statement is symmetric, we will only show that $M$ lies on the circle with diameter $AA_2$.\n\nDefine $S = A_1 \\cap BC$, or equivalently, as the point satisfyin... | Turkey | Team Selection Test for IMO 2024 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometr... | English | proof only | null | |
0czy | Let $p \geq 3$ be a prime. For $j=1,2, \ldots, p-1$, let $r_{j}$ be the remainder when the integer $\frac{j^{p-1}-1}{p}$ is divided by $p$. Prove that
$$
r_{1}+2 r_{2}+\ldots+(p-1) r_{p-1} \equiv \frac{p+1}{2}(\bmod p)
$$ | [
"For $j=1,2, \\ldots, p-1$, we have\n$$\n\\frac{j^{p-1}-1}{p}=a_{j} p+r_{j}\n$$\nfor some integer $a_{j}$. It follows\n$$\n\\frac{j^{p}-j}{p}=j a_{j} p+j r_{j},\n$$\nhence\n$$\n\\frac{j^{p}-j+(p-j)^{p}-(p-j)}{p}=j a_{j} p+j r_{j}+(p-j) a_{p-j} p+(p-j) r_{p-j}.\n$$\nWe obtain\n$$\n\\frac{j^{p}+(p-j)^{p}}{p}=j a_{j} ... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | English | proof only | null | |
0ewl | Problem:
We place labeled points on a circle as follows. At step 1, take two points at opposite ends of a diameter and label them both $1$. At step $n > 1$, place a point at the midpoint of each arc created at step $n - 1$ and label it with the sum of the labels at the two adjacent points. What is the total sum of the... | [
"Solution:\n\nAnswer: $2 \\cdot 3^{n - 1}$.\n\nTrue for $n = 1$. The new points added at step $n + 1$ have twice the sum of the points after step $n$, because each old point contributes to two new points. Hence the total after step $n + 1$ is three times the total after step $n$."
] | Soviet Union | 3rd ASU | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2 * 3^(n - 1) | |
0996 | Бүх $x, y \in \mathbb{R}$-ийн хувьд
$$
f([x]y) = f(x)[f(y)], \qquad (1)
$$
байх бүх $f: \mathbb{R} \to \mathbb{R}$ функцийг ол $([z]$-нь $z$-ээс үл хэтрэх хамгийн их бүхэл тоо). | [
"(1)-д $x = 0$ гэвэл\n$$\nf(0) = f(0)[f(y)] \\qquad (2)\n$$\nболно.\n\na) $f(0) \\neq 0$ гэе. (2)-оос $\\forall y \\in \\mathbb{R}, [f(y)] = 1$. Иймд (1) нь $f([x]y) = f(x)$ болно, энд $y = 0$ гэвэл $f(x) = f(0) = C \\neq 0$. $[f(y)] = 1 = [c]$-ээс $1 \\le c < 2$.\n\nb) $f(0) = 0$ гэе. Дараахь 2 дэд тохиол байна.\n... | Mongolia | International Mathematical Olympiad 51 | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | Mongolian | proof and answer | All constant functions f(x) = c for all real x, where either c = 0 or 1 ≤ c < 2. | |
0boq | Given a positive integer $n$, prove that there are only finitely many sequences of $n$ consecutive positive integers such that $n!$ can be constructed from these integers only using the elementary operations of addition, subtraction, multiplication, and division, the integers being used exactly once each. | [
"A straightforward induction on $n$ shows that the outcome of each such construction is a number of the form\n$$\n\\frac{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} a_{\\alpha_1, \\dots, \\alpha_n} x_1^{\\alpha_1} \\cdots x_n^{\\alpha_n}}{\\sum_{\\alpha_1, \\dots, \\alpha_n \\in \\{0, 1\\}} b_{\\alpha_1, \... | Romania | THE 2015 Seventh ROMANIAN MASTER OF MATHEMATICS | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
02xk | Problem:
A logomarca de uma empresa deve ser criada sobrepondo-se um triângulo equilátero e um quadrado, conforme a figura. Se a medida do lado do triângulo é $12~\mathrm{cm}$ e o lado do triângulo intercepta o lado do quadrado em seu ponto médio, qual a diferença entre a área sobreposta (escura) e a soma das áreas sem... | [
"Solution:\nVamos chamar a medida do lado do quadrado de $2n$, então quatro triângulos retângulos que não foram sobrepostos (todos congruentes) têm catetos medindo $n$ e $(6-n)$ e o triângulo equilátero não sobreposto tem lado medindo $(4n-12)$.\n\nComo os ângulos do triângulo equilátero me... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 1026 - 576 sqrt(3) cm^2 | |
062x | Problem:
Gegeben sei ein konvexes Fünfeck $A B C D E$ mit den Eigenschaften $B C \| A E$ und $\overline{A B}=\overline{A E}$. Weiter sei $F$ ein Punkt auf der Strecke $A E$, so dass $\overline{A B}=\overline{B C}+\overline{A F}$ sowie $\Varangle C B A=\Varangle F D C$ erfüllt ist. Schließlich sei $M$ der Mittelpunkt d... | [
"Solution:\n\nAus den Bedingungen folgt $\\overline{F E}=\\overline{A E}-\\overline{A F}=\\overline{A B}-(\\overline{A B}-\\overline{B C})=\\overline{B C}$.\nDaher ist $B C E F$ ein Parallelogramm, dessen beide Diagonalen $C F$ und $B E$ einander in $M$ halbieren.\n\nWegen der Punktsymmetrie an $M$ gilt $\\Varangle... | Germany | Auswahlwettbewerb zur Internationalen Mathematik-Olympiade | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous >... | null | proof only | null | |
0km7 | Problem:
Find all positive integers $N, n$ such that $N^{2}$ is 1 away from $n(N+n)$. | [
"Solution:\nThe solutions are $(N, n)=\\left(F_{i+1}, F_{i}\\right)$.\n\nIf $N>n$, $N^{2}-n(N+n)=N(N-n)-n^{2}$, so $(N, n)$ works if and only if $(n, N-n)$ works.\n\nIf $N \\leq n$, $n(N+n)-N^{2} \\geq n^{2} \\geq 1$, so the only solution is $(N, n)=(1,1)$.\n\nThus, all solutions eventually become $(1,1)$ after rep... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"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 | (N, n) = (F_{i+1}, F_i) for all integers i ≥ 1 | |
06rl | Let $ABC$ be an acute triangle with circumcircle $\omega$. Let $t$ be a tangent line to $\omega$. Let $t_{a}, t_{b}$, and $t_{c}$ be the lines obtained by reflecting $t$ in the lines $BC, CA$, and $AB$, respectively. Show that the circumcircle of the triangle determined by the lines $t_{a}, t_{b}$, and $t_{c}$ is tange... | [
"To avoid a large case distinction, we will use the notion of oriented angles. Namely, for two lines $\\ell$ and $m$, we denote by $\\angle(\\ell, m)$ the angle by which one may rotate $\\ell$ anticlockwise to obtain a line parallel to $m$. Thus, all oriented angles are considered modulo $180^{\\circ}$.\n\n![](atta... | IMO | 52nd International Mathematical Olympiad 2011 Shortlist | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Advanced Configurations > Miquel point",
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | null | proof only | null | |
072h | Problem:
Find all functions $f: \mathbf{R} \rightarrow \mathbf{R}$ such that
$$
f\left(x^{2}+y f(z)\right)=x f(x)+z f(y)
$$
for all $x, y, z$ in $\mathbf{R}$. (Here $\mathbf{R}$ denotes the set of all real numbers.) | [
"Solution:\nTaking $x=y=0$ in (1), we get $z f(0)=f(0)$ for all $z \\in \\mathbf{R}$. Hence we obtain $f(0)=0$.\n\nTaking $y=0$ in (1), we get\n$$\nf\\left(x^{2}\\right)=x f(x)\n$$\nSimilarly $x=0$ in (1) gives\n$$\nf(y f(z))=z f(y)\n$$\nPutting $y=1$ in (3), we get\n$$\nf(f(z))=z f(1) \\quad \\forall z \\in \\math... | India | INMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
0aof | Problem:
Let $3x$, $4y$, $5z$ form a geometric sequence while $\frac{1}{x}$, $\frac{1}{y}$, $\frac{1}{z}$ form an arithmetic sequence. Find the value of $\frac{x}{z} + \frac{z}{x}$. | [
"Solution:\nLet $3x$, $4y$, $5z$ be in geometric progression. Then there exists a common ratio $r$ such that:\n\n$$\n4y = 3x \\cdot r, \\quad 5z = 4y \\cdot r\n$$\n\nFrom the first equation:\n$$\n4y = 3x r \\implies y = \\frac{3x r}{4}\n$$\nFrom the second equation:\n$$\n5z = 4y r \\implies z = \\frac{4y r}{5}\n$$\... | Philippines | AREA STAGE | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 34/15 | |
0k7z | Problem:
For how many positive integers $a$ does the polynomial
$$
x^{2}-a x+a
$$
have an integer root? | [
"Solution:\nLet $r, s$ be the roots of $x^{2}-a x+a=0$. By Vieta's, we have $r+s = a$ and $r s = a$. Note that if one root is an integer, then both roots must be integers, as they sum to an integer $a$.\n\nThen,\n$$\nr s - (r + s) + 1 = a - a + 1 = 1 \\implies (r - 1)(s - 1) = 1\n$$\nBecause we require $r, s$ to be... | United States | HMMT November 2019 | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 1 | |
03cc | Consider a table $19 \times 2015$. *Block* is the figure consisting of a square $10 \times 10$ and a single cell pasted to the right of the most upper-right cell of the square. Rotation of a block is not allowed. Find the number of ways in which maximum number of blocks can be positioned on the table. (The blocks do no... | [] | Bulgaria | First Team Selection Test for 56th IMO | [
"Discrete Mathematics > Combinatorics > Generating functions",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | English | proof and answer | C(250, 199) | |
0a9u | Problem:
Given an equilateral triangle, find all points inside the triangle such that the distance from the point to one of the sides is equal to the geometric mean of the distances from the point to the other two sides of the triangle.
[The geometric mean of two numbers $x$ and $y$ equals $\sqrt{x y}$.] | [
"Solution:\nLet $P$ be a point inside $\\triangle ABC$. Denote its orthogonal projections on $AB$, $BC$, $CA$ by $X$, $Y$, $Z$, respectively. We have $\\angle XPZ = \\angle YPX = 120^\\circ$.\n\nAssume that $PX^2 = PY \\cdot PZ$. Together with $\\angle XPZ = \\angle YPX = 120^\\circ$, this gives $\\triangle XPZ \\s... | Nordic Mathematical Olympiad | The 28th Nordic Mathematical Contest | [
"Geometry > Plane Geometry > Triangles",
"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 and answer | The locus is the union of three circular arcs inside the triangle. Each arc is the part of the circle passing through a pair of vertices and the triangle’s center, and consists of points P with angle at P between those two vertices equal to one hundred twenty degrees. | |
0d0n | Let $a$, $b$, $c$ be rational numbers such that
$$
\frac{1}{a+bc} + \frac{1}{b+ac} = \frac{1}{a+b}.
$$
Prove that $\sqrt{\frac{c-3}{c+1}}$ is rational. | [
"The given relation is equivalent to\n$$\n(b + ac + a + bc)(a + b) = ab + c(a^2 + b^2) + abc^2,\n$$\nso therefore\n$$\n(a+b)^2c + (a+b)^2 = ab(c^2+1) + c(a^2+b^2).\n$$\nIt follows\n$$\n\\begin{aligned}\n(a+b)^2 &= ab(c^2+1) + c[a^2 + b^2 - (a+b)^2] \\\\\n&= ab(c^2+1) - 2abc = ab(c-1)^2.\n\\end{aligned} \\quad (1)\n... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Other",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof only | null | |
03ox | Assume $P_1, P_2, \dots, P_n$ ($n \ge 2$) is an arbitrary permutation of $1, 2, \dots, n$. Prove that
$$
\frac{1}{P_1 + P_2} + \frac{1}{P_2 + P_3} + \dots + \frac{1}{P_{n-2} + P_{n-1}} + \frac{1}{P_{n-1} + P_n} > \frac{n-1}{n+2}.
$$ | [
"**Proof** By Cauchy's inequality, we can get\n$$\n[(P_1 + P_2) + (P_2 + P_3) + \\dots + (P_{n-1} + P_n)] \\cdot \\left(\\frac{1}{P_1 + P_2} + \\frac{1}{P_2 + P_3} + \\dots + \\frac{1}{P_{n-1} + P_n}\\right) \\ge (n-1)^2.\n$$\nTherefore\n$$\n\\begin{aligned}\n& \\frac{1}{P_1 + P_2} + \\frac{1}{P_2 + P_3} + \\dots +... | China | China Girls' Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
04g4 | Each of the numbers $x_1, x_2, \dots, x_{2014}$ is $1$, $0$, or $-1$. What is the minimal possible value of the sum of products of all the pairs of those numbers, i.e. the sum of all $x_i x_j$ for $1 \le i < j \le 2014$? (USSR 1965) | [
"First note that the double sum of all the products $x_i x_j$ for $1 \\le i < j \\le 2014$ equals\n$$\n(x_1 + \\cdots + x_{2014})^2 - (x_1^2 + \\cdots + x_{2014}^2).\n$$\nDenote $A = (x_1 + \\cdots + x_{2014})^2$ and $B = x_1^2 + \\cdots + x_{2014}^2$.\nWe want to minimize $A$ and maximize $B$ at the same time.\nOb... | Croatia | Mathematica competitions in Croatia | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | -1007 | |
0hq9 | Problem:
Let $n$ be a positive integer. Prove that there exist distinct positive integers $x$, $y$, $z$ such that
$$
x^{n-1} + y^n = z^{n+1}.
$$ | [
"Solution:\nOne solution is\n$$\nx = 2^{n^2} 3^{n+1}, \\quad y = 2^{n^2 - n} 3^n, \\quad z = 2^{n^2 - 2n + 2} 3^{n-1}.\n$$"
] | United States | Berkeley Math Circle | [
"Number Theory > Diophantine Equations",
"Algebra > Intermediate Algebra > Exponential functions"
] | null | proof only | null | |
0cj0 | Determine all sets $M$ having at least two elements, all of which are prime natural numbers, and with the property that for any two distinct elements chosen from $M$, their difference is equal to 1 or to a prime number belonging to the set $M$. | [] | Romania | 75th NMO | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | {2, 3} and {2, 3, 5} | |
03gj | Problem:
Let $c$ be the length of the hypotenuse of a right angle triangle whose other two sides have lengths $a$ and $b$. Prove that $a + b \leq \sqrt{2} c$. When does the equality hold? | [] | Canada | Canadian Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM / Power Mean"
] | null | proof and answer | Equality holds precisely when the two legs are equal (the triangle is isosceles right). | |
0aay | There are two empty pots on disposal. The first pot can contain exactly 3 liters of liquid and the other exactly 5 liters. Is it possible with these pots to measure exactly 4 liters of liquid? | [
"First we fill the pot that can contain exactly 3 liters of liquid and we pour its content into the 5-liter pot. Then in the 5-liter pot there is room for 2 liters. Then we fill again the 3-liter pot and we fill with it the 5-liter pot. Now, in the 3-liter pot, we have exactly one liter left. Then we pour out the c... | North Macedonia | Macedonian Mathematical Competitions | [
"Math Word Problems"
] | null | proof and answer | Yes | |
012v | Problem:
Let $ABCD$ be a square. Let $M$ be an inner point on side $BC$ and $N$ be an inner point on side $CD$ with $\angle MAN = 45^{\circ}$. Prove that the circumcentre of $AMN$ lies on $AC$. | [
"Solution:\n\nDraw a circle $\\omega$ through $M$, $C$, $N$; let it intersect $AC$ at $O$. We claim that $O$ is the circumcentre of $AMN$.\n\nClearly $\\angle MON = 180^{\\circ} - \\angle MCN = 90^{\\circ}$. If the radius of $\\omega$ is $R$, then $OM = 2R \\sin 45^{\\circ} = R \\sqrt{2}$; similarly $ON = R \\sqrt{... | 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 > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / C... | null | proof only | null | |
0co1 | Nine skiers participated in a race. They started one by one, and each skier passed the distance with a constant speed (which could be different for different skiers). Determine if it could happen that each skier participated in an overtaking exactly four times. (In each overtaking, exactly two skiers participated: the ... | [
"It could not happen.\n\nSuppose it is possible. Since the speeds are constant, any two skiers met at most once. Then the skier who started first could not overtake anyone; therefore, he was overtaken by four skiers and finished fifth. On the other hand, the skier who started last could not be overtaken by anyone, ... | Russia | Regional round | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English; Russian | proof and answer | It could not happen. | |
02no | Let $n$ be an integer and $n_1$ be one of its divisors. Let $A$ be a $n \times n$ symmetric matrix defined by $a_{i,i} = 4$, $a_{i,i+1} = a_{i+1,i} = -1$ for all $i$ such that $1 \le i \le n-1$ and $i+1$ is not a multiple of $n_1$, $a_{i,i+n_1} = a_{i+n_1,i} = -1$ and $a_{i,j} = 0$ otherwise. | [
"See problem 3, grades 10–12."
] | Brazil | Brazilian Math Olympiad | [
"Algebra > Linear Algebra > Matrices",
"Algebra > Linear Algebra > Determinants"
] | null | proof only | null | |
0cv0 | Determine if there exists a triangle whose side lengths $x, y, z$ satisfy $x^3 + y^3 + z^3 = (x+y)(y+z)(z+x)$.
Существует ли треугольник, длины сторон которого $x, y, z$ удовлетворяют равенству $x^3 + y^3 + z^3 = (x+y)(y+z)(z+x)$? | [
"No, such a triangle does not exist.\n\nLet us consider the expression:\n$$(x + y)(x + z)(y + z) = x^2(y + z) + y^2(x + z) + z^2(x + y) + 2xyz$$\nBy the triangle inequality, $x, y, z$ are positive and the sum of any two is greater than the third. Therefore,\n$$x^2(y + z) + y^2(x + z) + z^2(x + y) + 2xyz > x^3 + y^3... | Russia | XLIII Russian mathematical olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English; Russian | proof and answer | No | |
01gb | Find all real $x, y, z$ so that
$$
\begin{aligned}
& x^2 y + y^2 z + z^2 = 0 \\
& z^3 + z^2 y + z y^3 + x^2 y = \frac{1}{4}(x^4 + y^4)
\end{aligned}
$$ | [
"Answer: $x = y = z = 0$.\n\n$y = 0 \\implies z^2 = 0 \\implies z = 0 \\implies \\frac{1}{4}x^4 = 0 \\implies x = 0$. $x = y = z = 0$ is a solution, so assume that $y \\neq 0$. Then $z = 0 \\implies x^2y = 0 \\implies x = 0 \\implies \\frac{1}{4}y^4 = 0$, which is a contradiction. Hence $z \\neq 0$. Now we solve th... | Baltic Way | Baltic Way 2020 | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x = y = z = 0 | |
0hw4 | Problem:
In the interior of a triangle $ABC$ with area $1$, points $D$, $E$, and $F$ are chosen such that $D$ is the midpoint of $AE$, $E$ is the midpoint of $BF$, and $F$ is the midpoint of $CD$. Find the area of triangle $DEF$. | [
"Solution:\n\nLet $x$ be the area of $\\triangle DEF$. Comparing triangles $ABE$ and $DEF$, we find that base $AE$ is twice base $DE$ but, since $E$ bisects $BF$, the heights to these bases are equal. Thus $\\triangle ABE$ has area $2x$. Symmetrically, triangles $BCF$ and $CAD$ have area $2x$. Since these four tria... | United States | Berkeley Math Circle Monthly Contest 1 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 1/7 | |
0b1y | Problem:
How many permutations of the string "000011112222" contain the substring "2020"? | [
"Solution:\n\nRemoving the string \"2020\", there are two $0$'s, four $1$'s, and two $2$'s remaining. There are $\\frac{8!}{2!4!2!} = 420$ ways to arrange these digits, multiplied to $9$ possible placements for the string \"2020\", for a product of $3780$.\n\nHowever, by PIE, we still need to subtract the number of... | Philippines | 22nd Philippine Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | proof and answer | 3575 | |
002z | Se desea colorear cada entero positivo con un color utilizando la mayor cantidad posible de colores de manera que se verifique la siguiente condición: Si, en notación decimal, el número $B$ se puede obtener a partir del número $A$, suprimiéndole a $A$ dos dígitos iguales consecutivos (aa) o suprimiéndole a $A$ cuatro d... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Español | proof and answer | 1024 | |
0kva | Problem:
The points $A=\left(4, \frac{1}{4}\right)$ and $B=\left(-5,-\frac{1}{5}\right)$ lie on the hyperbola $x y=1$. The circle with diameter $A B$ intersects this hyperbola again at points $X$ and $Y$. Compute $X Y$. | [
"Solution:\n\n\n\nLet $A=(a, 1/a)$, $B=(b, 1/b)$, and $X=(x, 1/x)$. Since $X$ lies on the circle with diameter $\\overline{A B}$, we have $\\angle A X B = 90^{\\circ}$. Thus, $\\overline{A X}$ and $\\overline{B X}$ are perpendicular, and so the product of their slopes must be $-1$. We deduc... | United States | HMMT November 2023 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | sqrt(401/5) | |
0aid | Let $k_1$, $k_2$ and $k_3$ be three circles with centers $O_1$, $O_2$ and $O_3$ respectively, such that none of the centers lies inside any of the two other circles. The circles $k_1$ and $k_2$ intersect in $A$ and $P$, $k_1$ and $k_3$ intersect in $C$ and $P$ and $k_2$ and $k_3$ intersect in $B$ and $P$. Let $X$ be a ... | [
"We will first show that the points $Y$, $B$ and $Z$ are collinear. Since the quadrilateral $BYAP$ is inscribed we have $\\angle PBY = \\angle PAX$. Since the quadrilateral $AXCP$ is inscribed we have $\\angle PAX = \\angle PCZ$. Since the quadrilateral $CPBZ$ is inscribed we obtain $\\angle PBZ + \\angle PCZ = 180... | North Macedonia | Macedonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | Yes; the maximum of four times the area is attainable when the constructed triangle has the center triangle as its medial triangle (specifically, when the auxiliary parallel through the intersection point is used so the constructed points coincide with those from that configuration). | |
0iub | Problem:
Let $f(x) = 2x^{3} - 2x$. For what positive values of $a$ do there exist distinct $b, c, d$ such that $(a, f(a))$, $(b, f(b))$, $(c, f(c))$, $(d, f(d))$ is a rectangle? | [
"Solution:\nSay we have four points $(a, f(a)), (b, f(b)), (c, f(c)), (d, f(d))$ on the curve which form a rectangle. If we interpolate a cubic through these points, that cubic will be symmetric around the center of the rectangle. But the unique cubic through the four points is $f(x)$, and $f(x)$ has only one point... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof and answer | a in [√3/3, 1] | |
03c3 | In a quadrilateral $ABCD$ sides $AB$ and $CD$ are not parallel. The midpoints of $AD$ and $BC$ are denoted by $M$ and $N$, respectively. The line $MN$ intersects the diagonals $AC$ and $BD$ at points $K$ and $L$, respectively. Prove that the circumcircles of triangles $AKM$ and $BNL$ intersect on the line $AB$. | [] | Bulgaria | First Team Selection Test for 56th IMO | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
06ut | Queenie and Horst play a game on a $20 \times 20$ chessboard. In the beginning the board is empty. In every turn, Horst places a black knight on an empty square in such a way that his new knight does not attack any previous knights. Then Queenie places a white queen on an empty square. The game gets finished when someb... | [
"We show two strategies, one for Horst to place at least $100$ knights, and another strategy for Queenie that prevents Horst from putting more than $100$ knights on the board.\n\nA strategy for Horst: Put knights only on black squares, until all black squares get occupied.\n\nColour the squares of the board black a... | IMO | IMO Shortlisted Problems | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 100 | |
00pn | Let $M$ be the point of intersection of the diagonals of a cyclic quadrilateral $ABCD$. Let $I_1$ and $I_2$ be the incenters of triangles $AMD$ and $BMC$, respectively, and let $L$ be the point of intersection of the lines $DI_1$ and $CI_2$. The foot of the perpendicular from the midpoint $T$ of $I_1I_2$ to $CL$ is $N$... | [
"The point $L$ is the midpoint of the arc $AB$ and, as $I_1$, $M$, $I_2$ are collinear, we have $\\angle LI_2I_1 = \\angle I_2CM + \\angle I_2MC = \\angle I_1DM + \\angle I_1MD = \\angle LI_1I_2$. Therefore the triangle $LI_1I_2$ is isosceles and $LT \\perp I_1I_2$.\n\nLet $Z$ be the midpoint of $NI_2$. Then $FZ \\... | Balkan Mathematical Olympiad | Balkan 2012 shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
03q4 | Let $x_1, x_2, \dots, x_5$ be nonnegative real numbers with $\sum_{i=1}^5 \frac{1}{1+x_i} = 1$.
Prove that $\sum_{i=1}^5 \frac{x_i}{4+x_i^2} \le 1$. (posed by Li Shenghong) | [
"Let $y_i = \\frac{1}{1+x_i}$, $i=1, 2, \\dots, 5$, then $x_i = \\frac{1-y_i}{y_i}$, $i=1, 2, \\dots, 5$ and $\\sum_{i=1}^5 y_i = 1$.\nWe have\n$$\n\\begin{align*}\n\\sum_{i=1}^{5} \\frac{x_i}{4+x_i^2} &\\le 1 \\Leftrightarrow \\sum_{i=1}^{5} \\frac{-y_i^2+y_i}{5y_i^2-2y_i+1} \\le 1 \\\\\n&\\Leftrightarrow \\sum_{i... | China | China Western Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
066j | Let $\triangle ABC$ be an acute angled triangle with $AB < AC$ and $O$ be the center of its circumcircle $\omega$. Let $D$ be a point on the segment $BC$ such that $\angle BAD = \angle CAO$. Let $E$ be the second point of intersection of $\omega$ and the line $AD$. If $M, N$ and $P$ are the midpoints of the line segmen... | [] | Greece | Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordin... | English | proof only | null | |
05c3 | Circles $\omega_1$ and $\omega_2$ touch at point $K$. The line through the centres of the circles intersects the circle $\omega_1$ once more at point $A$. A line through the point $A$ intersects the circle $\omega_1$ once more at point $B$ and the circle $\omega_2$ at points $C$ and $D$, where the points $A, B, C, D$ l... | [
"Let the radii of $\\omega_1$ and $\\omega_2$ be $r_1$ and $r_2$ respectively. Let $E$ be the second intersection of $AK$ and $\\omega_2$ (Fig. 40). As $AK$ and $KE$ are diameters of $\\omega_1$ and $\\omega_2$ respectively, the angles $ABK$ and $KDE$ must be right angles. In triangle $AKC$, the segment $KB$ is bot... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof and answer | 1/3 | |
059o | On the first line of a notebook Juku writes the number $43$. On every following line he writes the number $x^2 - 66x + 1122$, where $x$ is the number on the previous line. Find the number that Juku will write on the $2021$st line. | [
"Let $x_i$ be the number written on the $i$th line, then for all $i = 1, 2, \\dots$ we have $x_{i+1} = x_i^2 - 66x_i + 1122$. Notice that this is equivalent to $x_{i+1} - 33 = x_i^2 - 66x_i + 1089 = (x_i - 33)^2$. Denoting $a_n = x_n - 33$, we acquire $a_{i+1} = a_i^2$ for all $i = 1, 2, \\dots$, which means that $... | Estonia | Estonian Math Competitions | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 10^{2^{2020}} + 33 | |
0bq6 | Problem:
a) Demonstraţi că $\frac{1}{\sqrt{n+1}} < 2(\sqrt{n+1} - \sqrt{n}) < \frac{1}{\sqrt{n}}$ pentru orice $n \in \mathbf{N}^{*}$.
b) Demonstraţi că $\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \cdots + \frac{1}{\sqrt{n}} < 2 \sqrt{n-1}$, pentru orice $n \in \mathbf{N}, n \geq 2$. | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Algebra > Algebraic Expressions > Sequences and Series > Telescoping series",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0i41 | Problem:
An $(l, a)$-design of a set is a collection of subsets of that set such that each subset contains exactly $l$ elements and that no two of the subsets share more than $a$ elements. How many $(2,1)$-designs are there of a set containing 8 elements? | [
"Solution:\n\nThere are $\\binom{8}{2} = 28$ 2-element subsets. Any two distinct such subsets have at most 1 common element; hence, for each subset, we can decide independently whether or not it belongs to the design, and we thus obtain $2^{28}$ designs."
] | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 2^28 | |
06vx | Version 1. Let $n$ be a positive integer, and set $N=2^{n}$. Determine the smallest real number $a_{n}$ such that, for all real $x$,
$$
\sqrt[N]{\frac{x^{2 N}+1}{2}} \leqslant a_{n}(x-1)^{2}+x .
$$
Version 2. For every positive integer $N$, determine the smallest real number $b_{N}$ such that, for all real $x$,
$$
\sq... | [
"Solution 1 (for Version 1). First of all, assume that $a_{n}<N / 2$ satisfies the condition. Take $x=1+t$ for $t>0$, we should have\n$$\n\\frac{(1+t)^{2 N}+1}{2} \\leqslant\\left(1+t+a_{n} t^{2}\\right)^{N}\n$$\nExpanding the brackets we get\n$$\n\\begin{equation*}\n\\left(1+t+a_{n} t^{2}\\right)^{N}-\\frac{(1+t)^... | IMO | IMO 2020 Shortlisted Problems | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Chebyshev polynomials",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Math... | null | proof and answer | Version 1: a_n = 2^{n-1}. Version 2: b_N = N/2. | |
011u | Problem:
There are $2n$ cards. On each card some real number $x$, $1 \leqslant x \leqslant 2$, is written (there can be different numbers on different cards). Prove that the cards can be divided into two heaps with sums $s_1$ and $s_2$ so that
$$
\frac{n}{n+1} \leqslant \frac{s_1}{s_2} \leqslant 1.
$$ | [
"Solution:\n\nLet the numbers be $x_1 \\leqslant x_2 \\leqslant \\ldots \\leqslant x_{2n-1} \\leqslant x_{2n}$. We will show that the choice $s_1 = x_1 + x_3 + x_5 + \\cdots + x_{2n-1}$ and $s_2 = x_2 + x_4 + \\cdots + x_{2n}$ solves the problem. Indeed, the inequality $\\frac{s_1}{s_2} \\leqslant 1$ is obvious and... | Baltic Way | Baltic Way | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0df9 | Let $BB'$, $CC'$ be the altitudes of an acute-angled triangle $ABC$. Two circles passing through $A$ and $C'$ are tangent to $BC$ at points $P$ and $Q$. Prove that $A$, $B'$, $P$, $Q$ are concyclic. | [
"Since $BP^2 = BQ^2 = BA \\cdot BC'$ and the quadrilaterals $AC'A'C$, $AB'A'B$ are cyclic ($AA'$ is the altitude) we have\n$$\nCP \\cdot CQ = CB^2 - BP^2 = CB^2 - BA \\cdot BC' = BC^2 - BC \\cdot BA' = BC \\cdot CA' = CA \\cdot CB'\n$$\nClearly this is equivalent to the required assertion. $\\square$"
] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0kz4 | A group of 100 students from different countries meet at a mathematics competition. Each student speaks the same number of languages, and, for every pair of students $A$ and $B$, student $A$ speaks some language that student $B$ does not speak, and student $B$ speaks some language that student $A$ does not speak. What ... | [
"Suppose the languages spoken are labeled $L_1, L_2, L_3, \\dots, L_n$. Note that the collection of all subsets of $\\{L_1, L_2, L_3, \\dots, L_n\\}$ of size $r$ will satisfy the conditions in the problem for any $r$ in the range $1 \\le r < n$. For a given $n$, there are $\\binom{n}{r}$ subsets, and $\\binom{n}{r}... | United States | AMC 10 B | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients"
] | null | MCQ | A | |
0edi | Draw 4 distinct lines in the plane and let $n$ denote the number of intersections (if more than one line passes through the same point it still only counts as one intersection). Which of the following is the set of all possible values of $n$?
(A) $\{0, 2, 3, 4, 5, 6\}$
(B) $\{0, 1, 3, 4, 5, 6\}$
(C) $\{0, 1, 3, 4, 6\}$... | [
"As shown in the figures, four distinct lines can intersect in $0$, $1$, $3$, $4$, $5$ or $6$ points.\n\n\nAssume that there are exactly two intersections. At each of these either two or three lines meet. Suppose that at one of the intersections, denote it by $A$, three of the lines meet. O... | Slovenia | Slovenija 2016 | [
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | MCQ | B | |
0afs | Збирот од должините на страните на правоаголникот е 40 см. Едната страна на правоаголникот е 4 пати подолга од другата.
а) Да се определат должините на страните на правоаголникот.
б) Дали може, со 3 паралелни прави, дадениот правоаголник да се подели на 4 еднакви квадрати? Нацртајте!
в) За кол... | [
"а) $x + x + 4x + 4x = 40$, $10x = 40$, $x = 4$ см.\nЕдната страна е долга 16 см, а другата 4 см.\n\nб) Може, страната на секој квадрат е долга по 4 см.\n\nв) Збирот на страните на квадратот е $4 + 4 + 4 + 4 = 16$ см. $40 - 16 = 24$ см."
] | North Macedonia | Регионален натпревар по математика за основно образование | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | a) Side lengths are 4 cm and 16 cm. b) Yes; three parallel lines can divide it into four equal squares of side 4 cm. c) The difference in perimeters is 24 cm. | |
00qu | We have $3366$ film critics who sent their preferences for the best actor and best actress for the Oscars. It turns out that for every integer $n \in \{1, 2, \dots, 100\}$ there is an actor or an actress who has been voted exactly $n$ times. Show that there are two critics which voted in exactly the same manner. | [
"Call the vote of each critic, i.e. his choice for the pair of an actor and an actress, as a double-vote, and call as a single-vote each one of the two choices he makes, i.e. the one for an actor and the other one for an actress. In this terminology, a double-vote corresponds to two single-votes.\n\nFor each $n = 3... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
06zx | Problem:
We say that two non-negative integers are related if their sum uses only the digits $0$ and $1$. For example, $22$ and $79$ are related. Let $A$ and $B$ be two infinite sets of non-negative integers such that:
(1) if $a \in A$ and $b \in B$, then $a$ and $b$ are related,
(2) if $c$ is related to every membe... | [
"Solution:\n\nSuppose there is a member of $A$ with last digit $d$. Then every member of $B$ must have one of two possible last digits. Suppose there are members of $B$ with both possibilities. Then every member of $A$ must have last digit $d$. So either every member of $A$ has the same last digit or every member o... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06fo | There are $n$ points on the plane, no three of which are collinear. Each pair of points is joined by a red, yellow or green line. For any three points, the sides of the triangle they form consist of exactly two colours. Show that $n < 13$. | [
"It suffices to show that the case $n = 13$ is impossible since we can remove extra points. For the $j$th point, let $r_j, y_j, g_j$ be the numbers of lines having this point as an endpoint which are in red, yellow, and green respectively.\n\nFor any $\\triangle XYZ$, WLOG assume $XY$ and $XZ$ are red. Then this tr... | Hong Kong | CHKMO | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0b5f | Let $x_1, x_2, \dots, x_n$ and $y_1, y_2, \dots, y_n$ be positive real numbers so that
$$
x_1 + x_2 + \dots + x_n \geq x_1 y_1 + x_2 y_2 + \dots + x_n y_n.
$$
Show that for any non-negative integer $p$ the following inequality holds
$$
\frac{x_1}{y_1^p} + \frac{x_2}{y_2^p} + \dots + \frac{x_n}{y_n^p} \geq x_1 + x_2 + \... | [
"Assume by contradiction that\n$$\n\\frac{x_1}{y_1^p} + \\frac{x_2}{y_2^p} + \\dots + \\frac{x_n}{y_n^p} < x_1 + x_2 + \\dots + x_n.\n$$\nOn the other hand, by multiplying the given relation by $p \\in \\mathbb{N}$, we have\n$$\np(x_1 + x_2 + \\dots + x_n) \\geq p(x_1 y_1 + x_2 y_2 + \\dots + x_n y_n).\n$$\nAdding ... | Romania | Local Mathematical Competitions | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | English | proof only | null | |
0f7j | Problem:
$AB$ is a chord of the circle center $O$. $P$ is a point outside the circle and $C$ is a point on the chord. The angle bisector of $APC$ is perpendicular to $AB$ and a distance $d$ from $O$. Show that $BC = 2d$. | [] | Soviet Union | 21st ASU | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0ddx | Let $ABC$ be an acute, non-isosceles triangle inscribed in $(O)$ and $BB'$, $CC'$ are altitudes. Denote $E$, $F$ as the intersections of $BB'$, $CC'$ with $(O)$ and $D$, $P$, $Q$ are projections of $A$ on $BC$, $CE$, $BF$. Prove that the perpendicular bisector of $PQ$ bisects two segments $AO$, $BC$. | [] | Saudi Arabia | Saudi Arabian Mathematical Competitions | [
"Geometry > Plane Geometry > Advanced Configurations > Simson line",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0iw6 | Problem:
Consider an isosceles triangle $T$ with base $10$ and height $12$. Define a sequence $\omega_{1}, \omega_{2}, \ldots$ of circles such that $\omega_{1}$ is the incircle of $T$ and $\omega_{i+1}$ is tangent to $\omega_{i}$ and both legs of the isosceles triangle for $i > 1$.
Find the total area contained in al... | [
"Solution:\n\nAnswer: $\\frac{180 \\pi}{13}$\n\nUsing the notation from the previous solution, the area contained in the $i$th circle is equal to $\\pi r_{i}^{2}$. Since the radii form a geometric sequence, the areas do as well. Specifically, the areas form a sequence with initial term $\\pi \\cdot \\frac{100}{9}$ ... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Algebra > Algebraic Expressions > Sequences and Series > Sums ... | null | proof and answer | 180*pi/13 | |
037q | Problem:
Let $ABC$ be a non-equilateral triangle and let $M$ and $N$ be interior points of it such that $\Varangle BAM = \Varangle CAN$, $\Varangle ABM = \Varangle CBN$ and
$$
AM \cdot AN \cdot BC = BM \cdot BN \cdot CA = CM \cdot CN \cdot AB = k
$$
Prove that:
a) $3k = AB \cdot BC \cdot CA$;
b) the midpoint of the ... | [
"Solution:\n\nThe angle equality implies that $M$ and $N$ are isogonal conjugate points in $\\triangle ABC$. Therefore $\\Varangle BCM = \\Varangle ACN$. Denote by small letters the affixes of the corresponding points in the complex plane. We have that\n$$\n\\arg \\frac{b-a}{m-a} = \\arg \\frac{n-a}{c-a}\n$$\nand\n... | Bulgaria | Team selection test for 47. IMO | [
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Eule... | null | proof and answer | 3k = AB · BC · CA; the midpoint of MN is the centroid of triangle ABC. | |
0lfq | Problem:
In a scalene triangle $A B C$, $I$ is the incenter and $C N$ is the bisector of angle $C$. The line $C N$ meets the circumcircle of $A B C$ again at $M$. The line $\ell$ is parallel to $A B$ and touches the incircle of $A B C$. The point $R$ on $\ell$ is such that $C I \perp I R$. The circumcircle of $M N R$ ... | [
"Solution:\n\nIn this solution we make use of directed angles. A directed angle $\\angle(n, m)$ between lines $n$ and $m$ is the angle of counterclockwise rotation transforming $n$ into a line parallel to $m$.\n\nLet $d$ be the tangent to the circumcircle of $\\triangle A B C$ containing $N$ and different from $A B... | Zhautykov Olympiad | XVI International Zhautykov Olympiad in Mathematics | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0dfj | Given an equilateral triangle $ABC$. Points $D$, $E$, $F$ lie on sides $BC$, $CA$, $AB$, respectively, and satisfy $AF = BD$ and $DF = EF \neq DE$. Prove that $\angle CDE = 90^\circ$. | [] | Saudi Arabia | Saudi Arabian IMO Booklet | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
023o | Problem:
Retângulo quase quadrado - Um terreno retangular é quase quadrado: sua largura e seu comprimento são números inteiros de metros que diferem exatamente de 1 metro. A área do terreno, em metros quadrados, é um número de 4 algarismos, sendo o das unidades de milhar e o das centenas iguais, e o mesmo ocorre com o... | [
"Solution:\n\nA área é um número da forma $a a b b$, onde $a$ e $b$ representam algarismos; agora lembre que\n$$\na a b b = 1100 a + 11 b = 11(100 a + b)\n$$\nSeja $x$ a largura do terreno, logo\n$$\nx(x+1) = 11(100 a + b) \\quad (I)\n$$\ne deduzimos que $x$ ou $x+1$ é um múltiplo de 11. Procurar múltiplos de 11 qu... | Brazil | null | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 33 by 34, 66 by 67, 99 by 100 | |
01m9 | Each student of a group is friends with at least a half of the other students of this group. At the beginning, all students are partitioned into pairs according to the seats they occupy in the classroom. Per move any two students may swap their pairs.
Is it possible that after a finite number of such moves each pair co... | [] | Belarus | 61st Belarusian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | Yes | |
0iud | Problem:
There are $2008$ distinct points on a circle. If you connect two of these points to form a line and then connect another two points (distinct from the first two) to form another line, what is the probability that the two lines intersect inside the circle? | [
"Solution:\n\nGiven four of these points, there are $3$ ways in which to connect two of them and then connect the other two, and of these possibilities exactly one will intersect inside the circle. Thus $1 / 3$ of all the ways to connect two lines and then connect two others have an intersection point inside the ci... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Combinations"
] | null | proof and answer | 1/3 | |
0fds | Problem:
Demuestra que en un triángulo se verifica: si $r$ es una recta que pasa por su baricentro y no pasa por ningún vértice, la suma de las distancias a dicha recta de los vértices que quedan en un mismo semiplano es igual a la distancia del tercer vértice a dicha recta. | [
"Solution:\n\nEl triángulo $G M M'$ es semejante a $G A A'$ con razón de semejanza 2 (pues $A G = 2 G M$). Por tanto, $A A' = 2 M M'$.\n\nPor otro lado, $M M'$ es la paralela media del trapecio $B B' C' C$, de donde $M M' = \\left(B B' + C C'\\right) / 2$.\n\nEn consecuencia: $A A' = 2 M M'... | Spain | XLVII Olimpiada Matemática Española Primera Fase | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
08r2 | Five distinct points $A$, $M$, $B$, $C$ and $D$ are on a circle $O$ in this order with $MA = MB$.
Let the lines $AC$ and $MD$ intersect at $P$, the lines $BD$ and $MC$
at $Q$. Let the line $PQ$ meet the circle $O$ at $X$ and $Y$. Prove that
$MX = MY$. | [
"Since $\\angle ACM = \\angle BDM$ (because of $MA = MB$), $\\angle PCQ = \\angle PDQ$, so four points $C$, $D$, $P$ and $Q$ are concyclic. So $\\angle PQD = \\angle PCD = \\angle ACD = \\angle ABD$. Therefore $AB$ and $PQ$, hence $AB$ and $XY$ are parallel. Since $M$ is the middle point of the arc $AB$, it is also... | Japan | The 16th Japanese Mathematical Olympiad - The Final Round | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
056y | On a horizontal line, one colors $2k$ points red and, to the right of them, $2k$ points blue. On every move, one chooses two points of different color, such that there is exactly one colored point between them, and interchanges the colors of the chosen points. How many different configurations can one obtain using thes... | [
"Enumerate the colored points by positive integers from the left to the right. Every move can influence two points with the same parity, whereby the total number of red or blue points with this parity does not change. Thus in each configuration that can be achieved there are $k$ red and $k$ blue points with each pa... | Estonia | Final Round of National Olympiad | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | ((2k choose k))^2 | |
020l | Problem:
a. Determine the minimal value of
$$
\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}-2018\right)+\left(y+\frac{1}{x}\right)\left(y+\frac{1}{x}-2018\right)
$$
where $x$ and $y$ vary over the positive reals.
b. Determine the minimal value of
$$
\left(x+\frac{1}{y}\right)\left(x+\frac{1}{y}+2018\right)+\left(y+\f... | [
"Solution:\n\nSolution 1. By the inequality between arithmetic and quadratic means,\n$$\n\\left(x+\\frac{1}{y}\\right)^{2}+\\left(y+\\frac{1}{x}\\right)^{2} \\geqslant \\frac{1}{2}\\left(x+\\frac{1}{y}+y+\\frac{1}{x}\\right)^{2}\n$$\nwith equality if and only if $x+1 / y=y+1 / x$, which holds if $x=y$. It follows t... | Benelux Mathematical Olympiad | 10th Benelux Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a: -2036162, b: 8080 | |
04ij | The sum of squares of all solutions of the equation $x^4 + a x^2 + b = 0$ is $32$, and the product of all solutions of that equation is $4$. Determine $a$ and $b$. (Tamara Srnec) | [
"Let the roots of $x^4 + a x^2 + b = 0$ be $x_1$, $x_2$, $x_3$, $x_4$.\n\nLet us factor the quartic as follows:\nLet $y = x^2$, so the equation becomes $y^2 + a y + b = 0$.\nLet the roots of this quadratic be $y_1$ and $y_2$.\nThen the roots of the quartic are $x_1 = \\sqrt{y_1}$, $x_2 = -\\sqrt{y_1}$, $x_3 = \\sqr... | Croatia | Croatia Mathematical Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | proof and answer | a = -16, b = 4 | |
0ic3 | A $2004 \times 2004$ array of points is drawn. Find the largest integer $n$ such that it is possible to draw a convex $n$-sided polygon whose vertices lie on the points of the array. | [
"For a vector $v = (x, y)$, define $\\|v\\| = |x| + |y|$, the so-called taxicab distance (or taxicab norm). Embed the array of points in the plane such that they correspond to the lattice points in $\\{(x, y) : 1 \\le x, y \\le 2004\\}$.\n\nConsider a convex $n$-gon drawn in our square array, and imagine that we wa... | United States | USA IMO | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | 561 | |
0ic8 | Problem:
Suppose $f$ is a function that assigns to each real number $x$ a value $f(x)$, and suppose the equation
$$
f\left(x_{1}+x_{2}+x_{3}+x_{4}+x_{5}\right)=f\left(x_{1}\right)+f\left(x_{2}\right)+f\left(x_{3}\right)+f\left(x_{4}\right)+f\left(x_{5}\right)-8
$$
holds for all real numbers $x_{1}, x_{2}, x_{3}, x_{4},... | [
"Solution:\nPlug in $x_{1}=x_{2}=x_{3}=x_{4}=x_{5}=0$. Then the equation reads $f(0)=5 f(0)-8$, so $4 f(0)=8$, so $f(0)=2$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 2 |
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