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0bn1 | Let $(K, +, \cdot)$ be a finite field with at least four elements. Prove that the set $K^*$ can be partitioned into two nonempty subsets $A$ and $B$, such that
$$
\sum_{x \in A} x = \prod_{y \in B} y.
$$ | [
"Since the product of the elements of $K^*$ is $-1$, if $A$ and $B$ form a partition of $K^*$, then $(\\prod_{a \\in A} a) (\\prod_{b \\in B} b) = -1$, so $\\sum_{a \\in A} a = \\prod_{b \\in B} b$ if and only if\n$$\n\\left(\\sum_{a \\in A} a\\right) \\left(\\prod_{a \\in A} a\\right) = -1. \\quad (*)\n$$\nLet $|K... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Abstract Algebra > Field Theory",
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0hlm | Problem:
Determine the positive real numbers $a$ and $b$ satisfying $9 a^{2} + 16 b^{2} = 25$ such that $a \cdot b$ is maximal. What is the maximum of $a \cdot b$? Explain your answer! | [
"Solution:\n\nApplying the inequality $x^{2} + y^{2} \\geq 2 x y$ on $x = 3a$ and $y = 4b$ we get\n$$\n25 = (3a)^{2} + (4b)^{2} \\geq 2 \\cdot 3a \\cdot 4b = 24ab.\n$$\nHence $ab \\leq \\frac{25}{24}$.\n\nThe equality is attained for $x = y$ or equivalently for $3a = 4b$. In that case\n$$\n25 = (3a)^{2} + (4b)^{2} ... | United States | Berkeley Math Circle | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | null | proof and answer | Maximum ab = 25/24, attained at a = 5/(3√2) and b = 5/(4√2) (with 3a = 4b). | |
0bhx | For a non-negative integer $n$ the $n$-th iterate of a function $f: \mathbb{R} \to \mathbb{R}$ is $f^n = \underbrace{f \circ \dots \circ f}_{n \text{ times}}$, and $f^0$ is the identity function. Determine the continuous functions $f: \mathbb{R} \to \mathbb{R}$, that satisfy simultaneously the conditions:
a) The functi... | [
"We shall prove that all such functions are of the form $f(x) = -x + c$, where $c$ is a real constant. It is clear that such functions verify the given conditions.\n\nWe shall first prove that $f$ is one to one. Let $x$ and $y$ be real numbers such that $f(x) = f(y)$ and define $g_n = f^0 + \\dots + f^n$, $n \\in \... | Romania | 65th Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = -x + c for some real constant c | |
09ng | For a word $S$ written in letters $A$ and $B$, let $f(S)$ denote the maximum number of non-intersecting $ABA$ subwords in $S$. For example, $f(ABBABBA) = 0$, $f(ABABABBA) = 1$ and $f(ABABABA) = 2$.
For $n = 4k + 1$, find the sum $\sum f(S)$ where $S$ runs over all $n$ letter words written in letters $A$ and $B$.
(Nyamd... | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof and answer | ((20k - 3) * 2^(4k) + 3) / 25 | |
02tq | Problem:
a)
Verifique que $(1+\operatorname{tg} k)\left(1+\operatorname{tg}\left(45^\circ-k\right)\right)=2$.
b)
Dado que
$$
\left(1+\operatorname{tg} 1^\circ\right)\left(1+\operatorname{tg} 2^\circ\right) \cdot \ldots \cdot\left(1+\operatorname{tg} 45^\circ\right)=2^n
$$
encontre $n$. | [
"Solution:\na)\n$$\n\\begin{aligned}\n\\operatorname{tg}\\left(45^\\circ-k\\right)+1 & =\\frac{\\operatorname{sen}\\left(45^\\circ-k\\right)}{\\cos \\left(45^\\circ-k\\right)}+1 \\\\\n& =\\frac{\\operatorname{sen} 45^\\circ \\cos k-\\cos 45^\\circ \\operatorname{sen} k}{\\cos 45^\\circ \\cos k+\\operatorname{sen} 4... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | 23 | |
0ffn | Problem:
Sean $a$, $b$ y $c$ las longitudes de los lados de un triángulo. Demostrar que
$$
a^{2} b(a-b) + b^{2} c(b-c) + c^{2} a(c-a) \geq 0
$$
Determinar en qué casos se cumple la igualdad. | [] | Spain | International Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof and answer | Equality holds if and only if the triangle is equilateral (all three sides are equal). | |
0fsi | Problem:
Es seien $m$ und $n$ zwei positive ganze Zahlen, sodass $m^{2}+n^{2}-m$ durch $2 m n$ teilbar ist. Zeige, dass $m$ eine Quadratzahl ist. | [
"Solution:\n\nWir können annehmen, dass $m>1$ gilt. Für eine Primzahl $p$ und eine ganze Zahl $x$ bezeichne $\\operatorname{ord}_{p}(x)$ die grösste ganze Zahl $a$, sodass $x$ durch $p^{a}$ teilbar ist (der sogenannte $p$-Exponent von $x$ ). Sei $p$ ein Primteiler von $m$ und sei $a=\\operatorname{ord}_{p}(m)>0$ un... | Switzerland | IMO - Selektion | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
073o | Problem:
Let $A$ be a set of real numbers such that $A$ has at least four elements. Suppose $A$ has the property that $a^{2} + b c$ is a rational number for all distinct numbers $a, b, c$ in $A$. Prove that there exists a positive integer $M$ such that $a \sqrt{M}$ is a rational number for every $a$ in $A$. | [
"Solution:\nSuppose $0 \\in A$. Then $a^{2} = a^{2} + 0 \\times b$ is rational and $a b = 0^{2} + a b$ is also rational for all $a, b$ in $A$, $a \\neq 0$, $b \\neq 0$, $a \\neq b$. Hence $a = a_{1} \\sqrt{M}$ for some rational $a_{1}$ and natural number $M$. For any $b \\neq 0$, we have\n$$\nb \\sqrt{M} = \\frac{a... | India | INMO | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Intermediate Algebra > Other"
] | null | proof only | null | |
0aml | Problem:
Given $f(1-x)+(1-x) f(x)=5$ for all real number $x$, find the maximum value that is attained by $f(x)$. | [] | Philippines | 18th PMO Area Stage | [
"Algebra > Algebraic Expressions > Functional Equations"
] | null | proof and answer | 5 | |
03t0 | A sequence of real numbers $\{a_n\}$ satisfies the condition that $a_1 = \frac{1}{2}$, $a_{k+1} = -a_k + \frac{1}{2-a_k}$, $k = 1, 2, \dots$.
Prove the following inequality:
$$
\left( \frac{n}{2(a_1 + a_2 + \cdots + a_n)} - 1 \right)^n \le \left[ \frac{a_1 + a_2 + \cdots + a_n}{n} \right]^n \left( \frac{1}{a_1} - 1 \ri... | [
"**Proof** First, we use induction to prove that $0 < a_n \\le \\frac{1}{2}$, $n = 1, 2, \\dots$.\nWhen $n = 1$, it is obvious.\nNow suppose it is true for $n$ ($n \\ge 1$), i.e. $0 < a_n \\le \\frac{1}{2}$.\nLet $f(x) = -x + \\frac{1}{2-x}$ for $x \\in [0, \\frac{1}{2}]$. Then $f(x)$ is a decreasing function. So\n... | China | China Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof only | null | |
0eja | Problem:
Kvadrat s stranico dolžine $2$ je razdeljen na $4$ trikotnike, od teh sta $2$ trikotnika enakokraka (glej sliko). Ploščina enega od enakokrakih trikotnikov je dvakrat tolikšna, kot je ploščina drugega enakokrakega trikotnika. Koliko je ploščina osenčenega trikotnika?
(A) $\frac{2}{... | [
"Solution:\n\nNarišemo diagonalo kvadrata in označimo z $A, B, C, D, E$ in $F$ nekatere točke na kvadratu (glej sliko).\n\n\n\nOznačimo z $x=|B F|$. Ker je ploščina trikotnika $A C E$ dvakrat tolikšna kot ploščina trikotnika $A B C$, je $|E F|=2|B F|=2 x$. Trikotnik $B F C$ je pravokoten in... | Slovenia | 65. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | MCQ | A | |
0a3u | We call an integer $n \ge 3$ polypythagorean if there are $n$ distinct positive integers that you can put around a circle such that the sum of the squares of each pair of neighbouring numbers is a square. Thus, $3$ is a polypythagorean integer because for example for the triple $(44, 117, 240)$, we have $44^2+117^2 = 1... | [
"We prove by induction that all integers greater than or equal to $2$ are polypythagorean. Extend the definition of polypythagorean to $n = 2$ in the logical way. For the induction basis, take $(3, 4)$ for $n = 2$ and $(44, 117, 240)$ from the example for $n = 3$.\n\nFor the induction step, let $n \\ge 4$ and assum... | Netherlands | IMO Team Selection Test 3 | [
"Number Theory > Diophantine Equations > Pythagorean triples",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | all integers greater than or equal to 3 | |
0krn | Problem:
Parallel lines $\ell_{1}$, $\ell_{2}$, $\ell_{3}$, $\ell_{4}$ are evenly spaced in the plane, in that order. Square $ABCD$ has the property that $A$ lies on $\ell_{1}$ and $C$ lies on $\ell_{4}$. Let $P$ be a uniformly random point in the interior of $ABCD$ and let $Q$ be a uniformly random point on the perim... | [
"Solution:\n\nThe first thing to note is that the area of $ABCD$ does not matter in this problem, so for the sake of convenience, introduce coordinates so that $A = (0, 0)$, $B = (1, 0)$, and $C = (0, 1)$.\n\nSuppose $A$ and $B$ lie on the same side of $\\ell_{2}$. Then, by symmetry, $C$ and $D$ lie on the same sid... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals"
] | null | final answer only | 6100 | |
02zk | Problem:
No quadrado $A B C D$, os pontos $M$ e $N$ são interiores aos lados $B C$ e $C D$ de modo que $\angle M A N=45^{\circ}$. Seja $O$ o ponto de interseção do círculo que passa por $C, M$ e $N$ com o segmento $A C$.
a) Verifique que $O M=O N$.
b) Verifique que $O$ é o centro da circunfe... | [
"Solution:\n\na) Como $\\angle M C O=\\angle O C N=45^{\\circ}$, segue que os comprimentos das cordas $M O$ e $N O$ são iguais a um mesmo valor $L$.\n\nb) Seja $R$ o comprimento do raio da circunferência que passa por $A, M$ e $N$. Pela Lei dos Senos,\n$$\n\\frac{M N}{\\operatorname{sen}(\\angle M A N)}=2 R \\Right... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0ew0 | Problem:
a. Given a quadruple $(a, b, c, d)$ of positive reals, transform to the new quadruple $(ab, bc, cd, da)$. Repeat arbitrarily many times. Prove that you can never return to the original quadruple unless $a = b = c = d = 1$.
b. Given $n$ a power of $2$, and an $n$-tuple $(a_1, a_2, \ldots, a_n)$ transform to a... | [
"Solution:\n\na.\nLet $Q_0$ be the original quadruple $(a, b, c, d)$ and $Q_n$ the quadruple after $n$ transformations. If $abcd > 1$, then the products form a strictly increasing sequence, so return is impossible. Similarly if $abcd < 1$. So we must have $abcd = 1$. Let the largest of the four values of a quadrupl... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Algebra ... | null | proof only | null | |
0dz8 | In the triangle $ABC$ the lengths of the sides are given: $|AB| = 15$ cm, $|BC| = 14$ cm and $|CA| = 13$ cm. Let $D$ be the foot of the altitude from $A$ and let $E$ be a point on this altitude such that $\angle BAD = \angle DEC$. Denote the intersection of lines $AB$ and $CE$ by $F$. Find $|EF|$. | [
"First, let us find the lengths of the segments $AD$ and $CD$. Write $|AD| = v$ and $|CD| = x$. By Pythagoras's theorem $v^2 = |AC|^2 - x^2 = |AB|^2 - (|BC| - x)^2$, so $13^2 - x^2 = 15^2 - (14 - x)^2$ or, equivalently, $13^2 = 15^2 - 14^2 + 2 \\cdot 14 \\cdot x$. We see that $x = 5$ and $v = 12$.\n\nTriangles $EDC... | Slovenia | Slovenija 2008 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof and answer | 10/3 | |
0afp | Секоја точка од рамнината е обоена во една од две бои, сина или црвена. Да се докаже дека во таа рамнина постои рамностран триаголник чии темиња се обоени во една иста боја. | [
"Понатаму ќе покажеме дека постои отсечка чии крајни точки и средишна точка се обоени во иста боја.\nНека $АВ$ е отсечка чии крајни точки се обоени на пример во сина боја (таква отсечка постои според претходното). Нека $D$ и $E$ (од различни страни на $А$ и $В$) се точки такви што $\\overline{AD} = \\overline{AB} =... | North Macedonia | Регионален натпревар по математика за средно образование | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Macedonian, English | proof only | null | |
003x | Carlos escribe en el pizarrón, de izquierda a derecha, los 20 números enteros del $1$ al $20$ en algún orden. Luego, debajo del primer número escrito, escribe el producto del primero con el segundo; debajo del segundo número escribe el producto del segundo con el tercero y así sucesivamente, debajo del penúltimo escrib... | [] | Argentina | XV Olimpiada Matemática Rioplatense | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | Español | proof and answer | 1024 | |
02mq | Problem:
Soma de quadrados - Encontre três números, numa progressão aritmética de razão $2$, tais que a soma de seus quadrados seja um número formado de quatro algarismos iguais. | [] | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 41, 43, 45 | |
0ai4 | Jeck and Lisa are playing a game on an $m \times n$ board, with $m, n > 2$. Lisa starts by putting a knight onto the board. Then in turn Jeck and Lisa put a new piece onto the board according to the following rules:
1. Jeck puts a queen on an empty square that is two squares horizontally and one square vertically, or a... | [
"*Lisa's winning strategy*\nSuppose the game is played on an $m \\times n$ board with $m$ and $n$ both odd. Then Lisa puts her knight in a corner and partitions the remaining squares of the board into \"dominoes\". In each turn Jeck has to put a queen in one of these dominoes and Lisa puts a knight on the other squ... | North Macedonia | European Mathematical Cup | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | English | proof and answer | Lisa has a winning strategy if and only if both dimensions are odd; otherwise Jeck wins. | |
038i | Two circles $k_1$ and $k_2$ with centers $O_1$ and $O_2$, respectively, are externally tangent at point $P$. A circle $k_3$ is externally tangent to $k_1$ at $Q$ and to $k_2$ at $R$. The lines $PQ$ and $PR$ meet $k_3$ at points $A$ and $B$, respectively. If $AO_2$ meets $BO_1$ at a point $S$ prove that $SP \perp O_1O_2... | [] | Bulgaria | Second selection test for IMO 2007, Vietnam | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | English | proof only | null | |
0hvj | Problem:
Find all perfect squares that can be written as the sum of two powers of $2$. | [
"Solution:\nWe claim that the squares which work are $4^{k+1}$ or $9 \\cdot 4^{k}$ for $k \\in \\mathbb{Z}_{\\geq 0}$. These work because $4^{k+1} = 2^{2k+1} + 2^{2k+1}$ and $9 \\cdot 4^{k} = 2^{2k+3} + 2^{2k}$.\n\nWrite $n^{2} = 2^{a} + 2^{b}$ for $n, a, b \\in \\mathbb{Z}_{\\geq 0}$ and $a \\geq b$. If $a = b$, $... | United States | Berkeley Math Circle: Monthly Contest 2 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | All such squares are 4^{k+1} or 9*4^{k} for k >= 0. | |
0g2b | Problem:
Seien $a$, $b$, $c$, $d$ und $e$ positive reelle Zahlen. Bestimme den grössten Wert, den folgender Ausdruck annehmen kann:
$$
\frac{a b+b c+c d+d e}{2 a^{2}+b^{2}+2 c^{2}+d^{2}+2 e^{2}}
$$ | [
"Solution:\n\nL'idée est clairement d'appliquer AM-GM. On doit donc décomposer $b^{2}=x b^{2}+y b^{2}$ et probablement $2 c^{2}=c^{2}+c^{2}$. Les nombres $x, y$ doivent donc satisfaire $x+y=1$ et $2 x=y$. On obtient $x=1/3$ et $y=2/3$. Par AM-GM, on obtient\n$$\n\\begin{aligned}\n2 a^{2}+1/3 b^{2} & \\geq 2 \\sqrt{... | Switzerland | SMO - Finalrunde | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt(3/8) | |
0i04 | Problem:
A combination lock has a 3 number combination, with each number an integer between $0$ and $39$ inclusive. Call the numbers $n_{1}$, $n_{2}$, and $n_{3}$. If you know that $n_{1}$ and $n_{3}$ leave the same remainder when divided by $4$, and $n_{2}$ and $n_{1}+2$ leave the same remainder when divided by $4$, h... | [
"Solution:\nThere are $40$ choices for the last number, and for each of these we have $10$ choices for each of the first two numbers, thus giving us a total of $4000$ possible combinations. It is interesting to note that these restrictions are actually true for Master locks."
] | United States | Harvard-MIT Math Tournament | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 4000 | |
0g0z | Problem:
Soit $ABC$ un triangle aigu et soit $H$ son orthocentre. Soit $G$ l'intersection de la parallèle à $AB$ passant par $H$ avec la parallèle à $AH$ passant par $B$. Soit $I$ le point sur la droite $GH$ tel que $AC$ coupe le segment $HI$ en son milieu. Soit $J$ la deuxième intersection de $AC$ avec le cercle circ... | [
"Solution:\n\nSoient $A_{1}, B_{1}$ et $C_{1}$ les pieds des hauteurs passant respectivement par $A, B$ et $C$ et soit $P$ le point d'intersection de $AC$ avec $GI$. Nous montrons d'abord que les deux quadrilatères $CHBG$ et $BA_{1}B_{1}A$ sont des quadrilatères inscrits.\n\nEn effet, puisque $H$ est l'orthocentre ... | Switzerland | SMO - Finalrunde | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
06ty | Let $I$ be the incentre of a non-equilateral triangle $ABC$, $I_{A}$ be the $A$-excentre, $I_{A}'$ be the reflection of $I_{A}$ in $BC$, and $l_{A}$ be the reflection of line $A I_{A}'$ in $A I$. Define points $I_{B}, I_{B}'$ and line $l_{B}$ analogously. Let $P$ be the intersection point of $l_{A}$ and $l_{B}$.
a. Pr... | [
"a. Let $A'$ be the reflection of $A$ in $BC$ and let $M$ be the second intersection of line $A I$ and the circumcircle $\\Gamma$ of triangle $ABC$. As triangles $ABA'$ and $AOC$ are isosceles with $\\angle ABA' = 2\\angle ABC = \\angle AOC$, they are similar to each other. Also, triangles $ABI_{A}$ and $AIC$ are s... | IMO | IMO 2016 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilate... | English | proof only | null | |
0c2v | If $a, b \in [0, +\infty)$, $c \in \mathbb{R}$ and $\lfloor x \rfloor + \lfloor x + a \rfloor + \lfloor x + b \rfloor = \lfloor c x \rfloor$, for every $x \in \mathbb{R}$, prove that $\{a, b\} = \{\frac{1}{3}, \frac{2}{3}\}$ and $c = 3$. | [] | Romania | Shortlisted problems for the 2018 Romanian NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | {a, b} = {1/3, 2/3}, c = 3 | |
0ivq | Problem:
Circle $\Omega$ has radius $5$. Points $A$ and $B$ lie on $\Omega$ such that chord $AB$ has length $6$. A unit circle $\omega$ is tangent to chord $AB$ at point $T$. Given that $\omega$ is also internally tangent to $\Omega$, find $AT \cdot BT$. | [
"Solution:\n\nLet $M$ be the midpoint of chord $AB$ and let $O$ be the center of $\\Omega$. Since $AM = BM = 3$, Pythagoras on triangle $AMO$ gives $OM = 4$.\n\nNow let $\\omega$ be centered at $P$ and say that $\\omega$ and $\\Omega$ are tangent at $Q$. Because the diameter of $\\omega$ exceeds $1$, points $P$ and... | United States | 12th Annual Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 2 | |
0j37 | Problem:
Let $a$, $b$, $c$, $x$, $y$, and $z$ be complex numbers such that
$$
a = \frac{b+c}{x-2}, \quad b = \frac{c+a}{y-2}, \quad c = \frac{a+b}{z-2}.
$$
If $xy + yz + zx = 67$ and $x + y + z = 2010$, find the value of $xyz$. | [
"Solution:\nManipulate the equations to get a common denominator: $a = \\frac{b+c}{x-2} \\Longrightarrow x-2 = \\frac{b+c}{a} \\Longrightarrow x-1 = \\frac{a+b+c}{a} \\Longrightarrow \\frac{1}{x-1} = \\frac{a}{a+b+c}$; similarly, $\\frac{1}{y-1} = \\frac{b}{a+b+c}$ and $\\frac{1}{z-1} = \\frac{c}{a+b+c}$. Thus\n$$\... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | -5892 | |
02f6 | $ABCD$ is a convex quadrilateral with $\angle BAC = 30^\circ$, $\angle CAD = 20^\circ$, $\angle ABD = 50^\circ$, $\angle DBC = 30^\circ$. If the diagonals intersect at $P$, show that $PC = PD$. | [
"We have $\\angle ADB = 80^\\circ$, $\\angle ACB = 70^\\circ$. Applying the sine rule repeatedly we have $PD = \\frac{\\sin 20^\\circ}{\\sin 80^\\circ} PA$, $PC = \\frac{\\sin 30^\\circ}{\\sin 70^\\circ} PB = \\frac{\\sin 30^\\circ}{\\sin 70^\\circ} \\frac{\\sin 30^\\circ}{\\sin 50^\\circ} PA$. So we have to show t... | Brazil | XV OBM | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0d5r | Let $p$, $q$ be two different odd prime numbers and $n$ an integer such that $p q$ divides $n^{p q} + 1$. Prove that if $p^{3} q^{3}$ divides $n^{p q} + 1$ then either $p^{2}$ divides $n + 1$ or $q^{2}$ divides $n + 1$. | [
"Because $p q$ divides $n^{p q} + 1$, neither $p$ nor $q$ divides $n$.\nAssume $p < q$. We have from Fermat's little theorem\n$$\n0 \\equiv n^{p q} + 1 \\equiv n^{q} + 1 \\quad (\\bmod p).\n$$\nTherefore $n^{2 q} \\equiv 1 (\\bmod p)$. But $n^{p-1} \\equiv 1 (\\bmod p)$ and $\\gcd(p-1, q) = 1$, since $p-1 < q$ and ... | Saudi Arabia | SAMC 2015 | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Modular Arithmetic > Polynomials mod p",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English, Arabic | proof only | null | |
071c | Problem:
A game is played on a $2001 \times 2001$ board as follows. The first player's piece is the policeman, the second player's piece is the robber. Each piece can move one square south, one square east or one square northwest. In addition, the policeman (but not the robber) can move from the bottom right to the to... | [
"Solution:\n\nColor the squares with three colors as follows:\n\n```\n0 1 2 0 1 2 0 \\ldots 2\n1 2 0 1 2 0 1 \\cdots 0\n2 0 1 2 0 1 2 \\cdots 1\n```\n\nThe middle square is color $2$ (moving $999+1$ squares $E$ from the top left increases the color by $1$, then moving $999+1~S$ increases it by another $1$) and the ... | Ibero-American Mathematical Olympiad | Iberoamerican Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0fis | Problem:
Sean los polinomios:
$$
\begin{aligned}
& P(x)=x^{4}+a x^{3}+b x^{2}+c x+1 \\
& Q(x)=x^{4}+c x^{3}+b x^{2}+a x+1
\end{aligned}
$$
Halla las condiciones que deben cumplir los parámetros reales $a, b$ y $c,(a \neq c)$, para que $P(x)$ y $Q(x)$ tengan dos raíces comunes, y resuelve en ese caso las ecuaciones $P(x... | [
"Solution:\nLas raíces comunes a ambos polinomios serán raíces de la diferencia\n$$\nP(x)-Q(x)=(a-c) x^{3}+(c-a) x\n$$\nResolvemos la ecuación $P(x)-Q(x)=0$, sacando primero $x$ factor común\n$$\nx\\left((a-c) x^{2}+(c-a)\\right)=0\n$$\nLas tres raíces son $0$, $1$ y $-1$, y entre ellas tienen que estar las raíces ... | Spain | Olimpiada Matemática Española | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Conditions: b = -2 and a = -c (with the given assumption a ≠ c). Then P(x) has roots 1, -1, (-a ± sqrt(a^2 + 4)) / 2, and Q(x) has roots 1, -1, (a ± sqrt(a^2 + 4)) / 2. | |
0fnu | Dados los números $r$, $q$ y $n$, tales que
$$
\frac{1}{r+qn} + \frac{1}{q+rn} = \frac{1}{r+q}
$$
probar que
$$
\sqrt{\frac{n-3}{n+1}}
$$
es un número racional. | [] | Spain | L Olimpiada Matemática Española | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Spanish | proof only | null | |
055c | Do there exist two positive powers of $5$ such that the number obtained by writing one after the other is also a power of $5$? | [
"Suppose that $5^x \\cdot 10^n + 5^y = 5^z$, where $5^y$ has $n$ digits. Then $5^{x+n} \\cdot 2^n = 5^y \\cdot (5^{z-y} - 1)$, whence $2^n = 5^{z-y} - 1$.\n\nCase $n = 1$ does not work.\n\nFor case $n = 2$ we get $z - y = 1$. Since $5^y$ has $2$ digits, the only possibility is $y = 2$ and $z = 3$, whence $x = 0$, w... | Estonia | IMO Team Selection Contest I | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
0h5u | Positive integers $a$, $p$ satisfy: $p = 2^a - 1$. Find all $a$ such that $\frac{1}{2}(p^2+1)$ is a square of an integer. | [
"For $a=1$ we have $p=1$ and $\\frac{1}{2}(p^2+1)=1$ satisfies the problem.\n\nFor $a=2$ we have $p=3$ and $\\frac{1}{2}(p^2+1)=5$ doesn't satisfy the problem.\n\nAssume now $a \\ge 3$. Let $\\frac{1}{2}(p^2+1) = p_1^2$, then $p^2 - 2p_1^2 = -1$. Hence: $2^{2a} - 2^{a+1} + 1 - 2p_1^2 = -1$ or $2^{2a-1} - 2^a = p_1^... | Ukraine | 55rd Ukrainian National Mathematical Olympiad - Fourth Round | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | a = 1 and a = 3 | |
03kg | Problem:
Let $a$, $b$, and $c$ be positive real numbers. Prove that
$$
a^{a} b^{b} c^{c} \geq (a b c)^{\frac{a+b+c}{3}}
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Equations and Inequalities > Jensen / smoothing"
] | null | proof only | null | |
08yk | Let $I$ be the in-center of a triangle $ABC$ satisfying $AB > AC$, and let $D, E$ be points on the sides $AB, AC$, respectively, dividing the sides into two segments with $1:8$ ratio. If the triangle $DIE$ becomes the regular triangle with side length $1$, what is the length of $AB$? Here by $XY$ we denote the length o... | [
"$$\n\\frac{81 + 9\\sqrt{13}}{16}\n$$\nLet $P, Q$ be the foot of the perpendicular line drawn from $I$ to sides $AB, AC$, respectively. Then, from $IP = IQ$ and $ID = IE$ it follows that the right triangles $IDP$ and $IEQ$ are congruent. If we suppose that the points $P$ and $Q$ are located in the same side of the ... | Japan | 2019 Japan Mathematical Olympiad First Stage | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | (81 + 9*sqrt(13))/16 | |
0ldr | Given an acute, scalene triangle $ABC$, $D$ is a point on side $BC$. Let $E$, $F$ be the points on $AB$, $AC$ such that $\angle DEB = \angle DFC$. Lines $DF$, $DE$ intersect $AB$, $AC$ at points $M$, $N$, respectively. Denote $(I_1)$, $(I_2)$ by the circumcircles of triangles $DEM$, $DFN$. The circle $(J_1)$ touches $(... | [
"a. Note that $\\angle DEB = \\angle DFC$ then $\\angle DEA = \\angle DFA$, which implies $MNEF$ is a cyclic quadrilateral.\n\nWe have $\\angle DI_2F = 2\\angle DNF = 2\\angle EMF$ and\n$$\n\\angle I_2DF = 90^\\circ - \\frac{1}{2} \\angle DI_2F\n$$\nso $I_2D \\perp ME$. We also have $J_1K \\perp ME$ and it follows ... | Vietnam | VMO | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
08kf | Problem:
Let $ABCDEF$ be a regular hexagon. The points $M$ and $N$ are internal points of the sides $DE$ and $DC$ respectively, such that $\angle AMN = 90^{\circ}$ and $AN = \sqrt{2} \cdot CM$. Find the measure of the angle $\angle BAM$. | [
"Solution:\nSince $AC \\perp CD$ and $AM \\perp MN$ the quadrilateral $AMNC$ is inscribed. So, we have\n$$\n\\angle MAN = \\angle MCN\n$$\nLet $P$ be the projection of the point $M$ on the line $CD$. The triangles $AMN$ and $CPM$ are similar implying\n$$\n\\frac{AM}{CP} = \\frac{MN}{PM} = \\frac{AN}{CM} = \\sqrt{2}... | JBMO | OJBM | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 75° | |
0alj | Problem:
Marc and Jon together have $66$ marbles although Marc has twice as many marbles as Jon. Incidentally, Jon found a bag of marbles which enabled him to have three times as many marbles as Mark. How many marbles were in the bag that Jon found? | [] | Philippines | 18th PMO Area Stage | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 110 | |
03xg | Suppose that $m$ and $k$ are non-negative integers, and $p = 2^{2^m} + 1$ is a prime number. Prove that
a. $2^{2^{m+1}} p^k \equiv 1 \pmod{p^{k+1}}$;
b. $2^{m+1} p^k$ is the smallest positive integer $n$ satisfying the congruence equation $2^n \equiv 1 \pmod{p^{k+1}}$. | [
"We want to prove that $2^{2^{m+1}} p^k = p^{k+1} t_k + 1$ for some integer $t_k$ not divisible by $p$. We proceed by induction on $k$.\n\nWhen $k = 0$, it follows from $2^{2^m} = p-1$ that $2^{2^m} = (p-1)^2 = p(p-2)+1$, in this case, $t_0 = p-2$.\n\nFor inductive step, suppose that $2^{2^{m+1}} p^k = p^{k+1} t_k ... | China | China Western Mathematical Olympiad | [
"Number Theory > Residues and Primitive Roots > Multiplicative order",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 2^{m+1} p^k | |
0iri | Problem:
Triangle $ABC$ obeys $AB = 2AC$ and $\angle BAC = 120^\circ$. Points $P$ and $Q$ lie on segment $BC$ such that
$$
\begin{aligned}
AB^2 + BC \cdot CP &= BC^2 \\
3AC^2 + 2BC \cdot CQ &= BC^2
\end{aligned}
$$
Find $\angle PAQ$ in degrees. | [
"Solution:\n\nFind $\\angle PAQ$ in degrees.\n\nAnswer: $40^\\circ$\n\nWe have $AB^2 = BC(BC - CP) = BC \\cdot BP$, so triangle $ABC$ is similar to triangle $PBA$.\n\nAlso, $AB^2 = BC(BC - 2CQ) + AC^2 = (BC - CQ)^2 - CQ^2 + AC^2$, which rewrites as $AB^2 + CQ^2 = BQ^2 + AC^2$.\n\nWe deduce that $Q$ is the foot of t... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 40° | |
0jd4 | Problem:
Two cars are driving directly towards each other such that one is twice as fast as the other. The distance between their starting points is $4$ miles. When the two cars meet, how many miles is the faster car from its starting point? | [
"Solution:\n\nNote that the faster car traveled twice the distance of the slower car, and together, the two cars traveled the total distance between the starting points, which is $4$ miles. Let the distance that the faster car traveled be $x$. Then,\n\n$$\nx + \\frac{x}{2} = 4 \\Longrightarrow x = \\frac{8}{3}.\n$$... | United States | HMMT November 2013 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 8/3 | |
05q2 | Problem:
a) Prouver qu'il existe des entiers $a$, $b$, $c$ tels que $(a, b, c) \neq (0,0,0)$ et $|a|,|b|,|c|<10^{6}$ pour lesquels
$$
|a+b \sqrt{2}+c \sqrt{3}|<10^{-11}
$$
b) Soit $a$, $b$, $c$ des entiers tels que $(a, b, c) \neq (0,0,0)$ et $|a|,|b|,|c|<10^{6}$. Prouver que
$$
|a+b \sqrt{2}+c \sqrt{3}|>10^{-21}
$$ | [
"Solution:\n\na) Soit $E$ l'ensemble des $10^{18}$ nombres de la forme $a+b \\sqrt{2}+c \\sqrt{3}$, avec $a, b, c$ entiers naturels et $a, b, c<10^{6}$. On pose $d=(1+\\sqrt{2}+\\sqrt{3}) 10^{6}$.\nPour tout $x \\in E$, on a $0 \\leqslant x<d$. De plus $0<d<(1+2+3) \\times 10^{6}<10^{7}-10^{-11}$. On divise alors l... | France | Olympiades Françaises de Mathématiques - Envoi 2 (Algèbre) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0d8m | Let $ABC$ be an acute triangle with $AT$, $AS$ respectively the internal and external angle bisectors of $ABC$ and $T, S \in BC$. On the circle with diameter $TS$, take an arbitrary point $P$ that lies inside the triangle $ABC$. Denote $D, E, F, I$ as the incenter of triangle $PBC$, $PCA$, $PAB$, $ABC$. Prove that four... | [
"First, we note that the circle of diameter $TS$ is the Apollonius circle of triangle $ABC$ then\n$$\n\\frac{BP}{CP} = \\frac{BA}{CA} = \\frac{BT}{CT}\n$$\nor\n$$\n\\frac{BP}{BA} = \\frac{CP}{CA},\n$$\nwhich implies that the bisector of angle $B$ in triangle $ABP$ and the bisector of angle $C$ in triangle $ACP$ pas... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Circles > Circle of Apollonius",
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | English | proof only | null | |
0j7c | Problem:
An ordering of a set of $n$ elements is a bijective map between the set and $\{1,2, \ldots, n\}$. Call an ordering $\rho$ of the 10 unordered pairs of distinct integers from the set $\{1,2,3,4,5\}$ admissible if, for any $1 \leq a<b<c \leq 5$, either $p(\{a, b\})<p(\{a, c\})<p(\{b, c\})$ or $p(\{b, c\})<p(\{a... | [
"Solution:\n\nAnswer: 768\nThis problem is a special case of the higher Bruhat order, a class of combinatorial object widely studied for its connection to an assortment of mathematical areas such as algebraic geometry, algebraic combinatorics, and computational geometry.\nAn admissible order in our problem—the high... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | final answer only | 768 | |
0iol | Problem:
Compute the number of sequences of numbers $a_{1}, a_{2}, \ldots, a_{10}$ such that
$$
\begin{aligned}
& \text{ I. } a_{i}=0 \text{ or } 1 \text{ for all } i \\
& \text{ II. } a_{i} \cdot a_{i+1}=0 \text{ for } i=1,2, \ldots, 9 \\
& \text{ III. } a_{i} \cdot a_{i+2}=0 \text{ for } i=1,2, \ldots, 8
\end{align... | [
"Solution:\n\nAnswer: 60. Call such a sequence \"good,\" and let $A_{n}$ be the number of good sequences of length $n$. Let $a_{1}, a_{2}, \\ldots, a_{n}$ be a good sequence. If $a_{1}=0$, then $a_{1}, a_{2}, \\ldots, a_{n}$ is a good sequence if and only if $a_{2}, \\ldots, a_{n}$ is a good sequence, so there are ... | United States | 10th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 60 | |
0903 | Consider an equilateral triangle $ABC$ with each side measuring $10$. A circle passing through $A$ touches the side $BC$ (excluding the endpoints) at point $X$, and intersects sides $AB$ and $AC$ at points $D$ and $E$ respectively, both of which are different from point $A$. Given that $BX > CX$ and $AD + AE = 13$, fin... | [
"$5 + \\sqrt{10}$\n\nLet $x = BX$. Since $CX = 10 - x$ and $BX > CX$, we have $5 < x < 10$. By the power of a point theorem,\n$$\nx^2 + (10-x)^2 = BX^2 + CX^2 = BD \\cdot BA + CE \\cdot CA = 10(BD + CE)\n$$\nholds. Also, from the condition $AD + AE = 13$, we find\n$$\nBD + CE = (10 - AD) + (10 - AE) = 20 - (AD + AE... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | 5 + sqrt(10) | |
0b6n | A triangle has side lengths $a$, $b$, $c$ and perimeter $3$. Prove that
a) it is not possible that all the altitudes have integer length;
b) if $\sqrt{a+b-c} + \sqrt{b+c-a} + \sqrt{c+a-b} = 3$, then the triangle is equilateral. | [] | Romania | Shortlisted Problems for the Romanian NMO | [
"Geometry > Plane Geometry > Triangles",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0lat | $n$ countries send delegates to an international Festival, each country send $k$ delegates ($n$ and $k$ are integers satisfying $n > k > 1$). The organizing committee of the Festival divides $nk$ delegates into $n$ discussion groups, each group consists of $k$ delegates. Show that one can choose $n$ delegates, in such ... | [] | Vietnam | Vietnamese Team Selection for IMO | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem"
] | null | proof only | null | |
0d51 | Prove that for any integer $n \geq 2$, there exists a unique finite sequence $x_{0}, x_{1}, \ldots, x_{n}$ of real numbers which satisfies $x_{0}=x_{n}=0$ and $x_{i+1}-8 x_{i}^{3}- 4 x_{i}+3 x_{i-1}+1=0$ for all $i=1,2, \ldots, n-1$. Prove moreover that $\left|x_{i}\right| \leq \frac{1}{2}$ for all $i=1,2, \ldots, n-1$... | [
"Let $P_{1}(X)=X$, $P_{2}(X)=8 X^{3}+4 X-1$ and define by induction $P_{k+1}(X)$ by $P_{k+1}(X)=8 P_{k}(X)^{3}+4 P_{k}(X)-3 P_{k-1}(X)-1$ for all integer $k \\geq 2$. Clearly, $P_{k}(X)$ is a polynomial of odd degree for all $k \\geq 1$.\n\nLet $n \\geq 2$ be an integer and $a$ a real zero of the polynomial $P_{n}(... | Saudi Arabia | SAMC 2015 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Polynomials > Intermediate Value Theorem",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English, Arabic | proof only | null | |
01kv | Let $a$, $b$, $c$ denote integers. Prove that $49$ divides $10a^2 + 23ab + 12b^2$, if $7$ divides $3a^2 + 2ab - 2b^2$. | [
"Since $3a^2 + 2ab - 2b^2$ is divisible by $7$, we see that\n$$10a^2 + 23ab + 12b^2 = (3a^2 + 2ab - 2b^2) + 7(a^2 + 3ab + 2b^2)$$\nis divisible by $7$.\n\nWe have\n$$10a^2 + 23ab + 12b^2 = 10a^2 + 15ab + 8ab + 12b^2 = 5a(2a + 3b) + 4b(2a + 3b) = (2a + 3b)(5a + 4b)$$\nwhich is divisible by $7$.\n\nSince $7$ is a pri... | Belarus | Belarusian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0gpr | Find the greatest value of the expression
$$
|(x-y)(y-z)(z-x)|
$$
for all real numbers $x$, $y$, $z$ satisfying $x + y + z = 0$ and $x^2 + y^2 + z^2 = 6$. | [
"** **Without the loss of generality we assume that $x \\ge y \\ge z$. Let $x-y = a$ and $y-z = b$. We have to find the maximum value of $ab(a+b)$. Since $x+y+z=0$ and $x^2+y^2+z^2=6$, we get $a^2+b^2+(a+b)^2 = (x-y)^2+(y-z)^2+(x-z)^2 = 3(x^2+y^2+z^2)-(x+y+z)^2 = 18$. Thus, $a^2+ab+b^2 = 9$. By AM-GM inequality,\n$... | Turkey | 18th Junior Turkish Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof and answer | 6*sqrt(3) | |
0izn | Problem:
How many different collections of 9 letters are there? A letter can appear multiple times in a collection. Two collections are equal if each letter appears the same number of times in both collections. | [
"Solution:\n\nWe put these collections in bijection with binary strings of length $34$ containing $9$ zeroes and $25$ ones. Take any such string—the $9$ zeroes will correspond to the $9$ letters in the collection. If there are $n$ ones before a zero, then that zero corresponds to the $(n+1)$st letter of the alphabe... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 34 choose 9 | |
08ui | Suppose a quadrilateral $ABCD$ is inscribed in a circle of radius $1$, and its diagonals intersect with the angle of $60^\circ$. Let $P$ be the point of intersection of the diagonals. Suppose that it is known that $AP = \frac{1}{3}$ and $CP = \frac{2}{3}$. Determine all possible values that the absolute value of the di... | [
"$$\n\\boxed{\\left\\{\\frac{4}{3},\\ \\frac{5}{3}\\right\\}}\n$$\nWe can draw, as in the figures below, a regular hexagon $ACEFGH$ which is inscribed in the circle given in the statement of the problem. As we are concerned with the absolute value of the difference between $BP$ and $DP$, the point $B$ is chosen to ... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | {4/3, 5/3} | |
0fec | Problem:
En la orilla de un río de $100$ metros de ancho está situada una planta eléctrica y en la orilla opuesta, y a $500$ metros río arriba, se está construyendo una fábrica. Sabiendo que el río es rectilíneo entre la planta y la fábrica, que el tendido de cables a lo largo de la orilla cuesta a $9\,€$ cada metro y ... | [
"Solution:\nCada trayecto tendrá un recorrido formado por un tramo sobre el río, en el que se avanzará una distancia de $b$ metros y uno o dos tramos a lo largo de la orilla que recorrerán los restantes $500 - b$ metros. El recorrido de tal trayecto será $L(b)$ y el gasto $g(b)$.\n\n\n\nLa ... | Spain | null | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 550 m | |
01p8 | Find all triples $(x, y, z)$ of nonnegative integers $x$, $y$, $z$ such that $7^x = 3^z - 2^y$. | [
"Answer: $\\{(x, y, z)\\} = \\{(0, 1, 1), (1, 1, 2), (0, 3, 2), (2, 5, 4)\\}$.\n\nWe rewrite the equation in the form $7^x + 2^y = 3^z$. Note that for $y = 0$ the left-hand side of the equation is an even integer while the right-hand side is an odd integer. So, $y > 0$.\n\nConsider three cases: $y = 1$, $y = 2$ and... | Belarus | BelarusMO 2013_s | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Multiplicative order"
] | null | proof and answer | [[0, 1, 1], [1, 1, 2], [0, 3, 2], [2, 5, 4]] | |
0e0c | Prove that $1005^{\ln 121} = 11^{\ln(1+3+5+\dots+2009)}$. | [
"Since\n$$\n\\begin{aligned}\n1 + 3 + 5 + \\dots + 2009 &= \\\\\n&= (1 + 2009) + (3 + 2007) + \\dots + (1003 + 1007) + 1005 \\\\\n&= 502 \\cdot 2010 + 1005 = 1005^2.\n\\end{aligned}\n$$\nit suffices to show that $1005^{\\ln(121)} = 11^{\\ln(1005^2)}$. From this we get\n$$\n\\ln(121) \\ln(1005) = \\ln(1005^2) \\ln(1... | Slovenia | National Math Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
0414 | Two circles $K_1$ and $K_2$ of different radii intersect at two points $A$ and $B$, let $C$ and $D$ be two points on $K_1$ and $K_2$, respectively, such that $A$ is the midpoint of the segment $CD$. The extension of $DB$ meets $K_1$ at another point $E$, the extension of $CB$ meets $K_2$ at another point $F$. Let $l_1$... | [
"(1) Since $C$, $A$, $B$, $E$ are concyclic, and $D$, $A$, $B$, $F$ are concyclic, $CA = AD$, and by the theorem of power of a point, we have\n$$\nCB \\cdot CF = CA \\cdot CD = DA \\cdot DC = DB \\cdot DE. \\qquad \\textcircled{1}\n$$\nSuppose on the contrary that $l_1$ and $l_2$ do not intersect, then $CD \\parall... | China | China Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0kkq | Problem:
Roger initially has 20 socks in a drawer, each of which is either white or black. He chooses a sock uniformly at random from the drawer and throws it away. He repeats this action until there are equal numbers of white and black socks remaining.
Suppose that the probability he stops before all socks are gone ... | [
"Solution:\n\nLet $b_{i}$ and $w_{i}$ be the number of black and white socks left after $i$ socks have been thrown out. In particular, $b_{0}+w_{0}=20$.\n\nThe key observation is that the ratio $r_{i}=\\frac{b_{i}}{b_{i}+w_{i}}$ is a martingale (the expected value of $r_{i+1}$ given $r_{i}$ is just $r_{i}$).\n\nSup... | United States | HMMT Spring 2021 Guts Round | [
"Discrete Mathematics > Combinatorics > Expected values"
] | null | proof and answer | 20738 | |
01g7 | Prove that the equation $7^x = 1 + y^2 + z^2$ has no positive integral solution. | [
"Assume that there exist positive integers $x$, $y$, $z$ satisfying the given equation. Let's consider two cases:\n\nIf $x$ is odd then $7^x \\equiv 7 \\pmod{8}$. Since each of $y^2$ and $z^2$ can have $0$, $1$, $4$ as the remainders when divided by $8$. We have contradiction because $1 + y^2 + z^2 \\equiv 1, 2, 3,... | Baltic Way | Baltic Way 2019 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Number Theory > Algebraic Number Theory > Unique factorization"
] | English | proof only | null | |
07ij | In an isosceles triangle $ABC$ ($BC = BA$), point $P$ is arbitrarily chosen on the altitude from vertex $B$. Suppose that the circumcircle of $BAP$ intersects $AC$ for the second time at point $M$. Let $N$ be the reflection of $M$ with respect to the midpoint of $AC$. $NP$ intersects the circumcircle of $BAP$ at $X$ ($... | [
"We denote by $Q$ the intersection of $CZ$ and $AB$. First, we show that quadrilateral $QXYP$ is cyclic. Note that $\\angle PNA = \\angle PYQ$. Therefore, if we prove that $\\angle QXP = \\angle PNA$, it follows that $QXYP$ is cyclic. This is also equivalent to proving that $XQ$ is parallel to $AN$.\n\n![](attached... | Iran | 40th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
05hu | Problem:
Soient $x_{1}, \ldots, x_{n}$ des réels quelconques. Montrer que
$$
\sum_{i=1}^{n} \sum_{j=1}^{n}\left|x_{i}+x_{j}\right| \geqslant n \sum_{i=1}^{n}\left|x_{i}\right| .
$$ | [
"Solution:\nParmi les $n$ nombres, on suppose que $r$ sont positifs (ou nuls), et $s=n-r$ sont strictement négatifs. On note $\\left(a_{i}\\right)_{1 \\leqslant i \\leqslant r}$ la liste des nombres positifs (ou nuls), et $\\left(-b_{j}\\right)_{1 \\leqslant j \\leqslant s}$ la liste des nombres strictement négatif... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof only | null | |
078q | There are $n \ge 3$ particles on a circle situated at the vertices of a regular $n$-gon. All these particles move on the circle with the same constant speed. One of the particles moves in the clockwise direction while all others move in the anti-clockwise direction. When particles collide, that is, they are all at the ... | [
"The answer is $(n-1)m'$ where $m'$ is the smallest number such that $\\frac{m'(n-2)}{2n}$ is an integer.\nMore precisely, the answer is\n* $2n(n-1)$ when $n$ is odd\n* $n(n-1)$ when $n$ is divisible by 4\n* $\\frac{n(n-1)}{2}$ when $n-2$ is divisible by 4. (but the $n/2$-th point will be the one moving in reverse,... | India | IMO TST | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | null | proof and answer | s = 2n(n−1) if n is odd; s = n(n−1) if n is divisible by 4; s = n(n−1)/2 if n−2 is divisible by 4 | |
06v2 | Let $\mathbb{Z}$ be the set of integers. Determine all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that, for all integers $a$ and $b$,
$$
f(2a) + 2f(b) = f(f(a+b)).
$$ | [
"Substituting $a=0$, $b=n+1$ gives $f(f(n+1)) = f(0) + 2f(n+1)$. Substituting $a=1$, $b=n$ gives $f(f(n+1)) = f(2) + 2f(n)$.\nIn particular, $f(0) + 2f(n+1) = f(2) + 2f(n)$, and so $f(n+1) - f(n) = \\frac{1}{2}(f(2) - f(0))$. Thus $f(n+1) - f(n)$ must be constant. Since $f$ is defined only on $\\mathbb{Z}$, this te... | IMO | IMO 2019 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Functional equations"
] | English | proof and answer | All functions are either f(n) = 0 for all integers n, or f(n) = 2n + K for any integer constant K. | |
0chl | a) Prove that there are infinitely many natural numbers $n$ such that $2 \cdot n$ is a perfect square and $3 \cdot n$ is a perfect cube.
b) Prove that there is no natural number $m$ such that $2 + m$ is a perfect square and $3 \cdot m$ is a perfect cube. | [
"a) Consider the numbers $n = 72 \\cdot a^6$, with natural $a$. Then $2 \\cdot n = (12 \\cdot a^3)^2$ and $3 \\cdot n = (6 \\cdot a^2)^3$, which shows that every such $n$ is 'good'.\n\nb) If $3 \\cdot m$ is a perfect cube, then $3 \\cdot m$ is a multiple of $27$. In this case $m$ is a multiple of $9$, so $m + 2 = M... | Romania | 74th Romanian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof only | null | |
089x | Problem:
Sui vertici di un poligono con $n \geq 3$ lati sono scritti dei numeri interi, in modo tale che il numero scritto su ciascun vertice abbia la stessa parità della somma dei numeri scritti sui due vertici adiacenti (cioè se il numero sul vertice è pari, anche la somma dei numeri che compaiono sui vertici adiace... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Osserviamo infatti che ogni vertice su cui è scritto un numero dispari ha esattamente un vicino contrassegnato con un numero dispari, dunque i dispari si presentano in coppie (ed il loro numero totale è quindi pari, eventualmente zero)."
] | Italy | Progetto Olimpiadi della Matematica - Gara di Febbraio | [
"Number Theory > Other",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Other"
] | null | MCQ | C | |
0lfk | Let $ABC$ be an acute triangle with circumcenter $O$. Let $A'$ be the center of the circle passing through $C$ and tangent to $AB$ at $A$, let $B'$ be the center of the circle passing through $A$ and tangent to $BC$ at $B$, let $C'$ be the center of the circle passing through $B$ and tangent to $CA$ at $C$.
a) Prove t... | [
"a) Let $(A'), (B'), (C')$ respectively represent the circle passing through point $C$ and touching the line $AB$ at point $A$, the circle passing through point $B$ and touching the line $BC$ at point $B$, and the circle passing through point $C$ and touching the line $CA$ at point $C$.\n\nLet $K$ be the second int... | Vietnam | Vietnamese MO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Transformations > Rotation",
"Ge... | English | proof only | null | |
088j | Problem:
Due numeri $a$ e $b$ sono tali che $\frac{3a+b}{a-b}=2$. Quanto vale $\frac{a^{3}}{b^{3}}$?
(A) $-27$
(B) $-8$
(C) $1$
(D) $8$
(E) $27$. | [
"Solution:\n\nLa risposta è $(\\mathbf{A})$. Vale $3a+b=2a-2b$, quindi $a=-3b$, e perciò $\\frac{a^{3}}{b^{3}}=(-3)^{3}=-27$."
] | Italy | Olimpiadi della Matematica - Gara di Febbraio | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | MCQ | A | |
062a | Problem:
Der Punkt $P$ liege im Inneren des Dreiecks $A B C$ und erfülle
$$
\varangle B P C-\varangle B A C=\varangle C P A-\varangle C B A=\varangle A P B-\varangle A C B .
$$
Man beweise, dass dann gilt:
$$
\overline{P A} \cdot \overline{B C}=\overline{P B} \cdot \overline{A C}=\overline{P C} \cdot \overline{A B}
$$... | [
"Solution:\n\n1.\nDie Verlängerungen von $A P, B P, C P$ mögen den Umkreis des Dreiecks $A B C$ erneut in $A'$, $B'$, $C'$ treffen. Mit Hilfe des Peripheriewinkelsatzes und einem einfachen Winkelsummenargument finden wir $\\varangle B' A' C' = \\varangle B' A' A + \\varangle A A' C' = \\varangle B' B A + \\varangle... | Germany | 1. IMO-Auswahlklausur | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof only | null | |
0lae | Let $AD$ be the median from $A$ of a triangle $ABC$. Let $M$ be a variable point on the given line $d$ perpendicular to $AD$ and denote by $E$ and $F$ respectively the midpoints of $MB$ and $MC$. The line through $E$ and perpendicular to $d$ meets $AB$ at $P$, the line through $F$ and perpendicular to $d$ meets $AC$ at... | [] | Vietnam | Vijetnam 2008 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
08yr | In an acute triangle $ABC$, points $D$ and $E$ lie on sides $AB$ and $AC$ respectively which satisfy $BD = CE$. Point $P$ is on line segment $DE$ and point $Q$ lie on arc $BC$, not containing $A$, of the circumcircle of triangle $ABC$. These points satisfy $BP : PC = EQ : QD$ and points $A, B, C, D, E, P, Q$ are all di... | [
"Let $F, G$ be the intersection points of lines $QD, QE$ and triangle $ABC$ other than $Q$ respectively. Let line $BG$ and $CF$ meet at $R$. Then, application of Pascal's theorem for six points $A, B, G, Q, F, C$ which are concyclic shows that $D, E, R$ are colinear. These six points are on the same circle in the o... | Japan | Japan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
00av | Let $AB$ a line segment of length $1$. Several elementary particles start moving simultaneously at constant speeds from $A$ to $B$. As soon as a particle reaches $B$, it turns around and heads to $A$; when reaching $A$, it starts moving to $B$ again, and so on indefinitely.
Find all rational numbers $r > 1$ with the fo... | [
"The values in question are all integers $r$ greater than $1$.\n\nWe start with a general observation about two particles $P_1$ and $P_2$ moving on $AB$ by the given rules, with different constant speeds $v_1$ and $v_2$, $v_1 > v_2$. Suppose that they are at the same point $Q$ of $AB$ at a certain moment $t$. There... | Argentina | Argentine National Olympiad 2016 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | all integers r > 1 | |
0a2w | We place the digits $1$ through $9$ one by one in a $3 \times 3$ grid. The digit $1$ may be placed in an arbitrarily chosen box; each subsequent digit comes in a box that is horizontally or vertically adjacent to the box that contains the previous digit. See, for example, the picture on the right. We call such a grid a... | [] | Netherlands | Dutch Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 5 | |
0evv | Problem:
A and $B$ play the following game with $N$ counters. $A$ divides the counters into $2$ piles, each with at least $2$ counters. Then $B$ divides each pile into $2$ piles, each with at least one counter. $B$ then takes $2$ piles according to a rule which both of them know, and $A$ takes the remaining $2$ piles. ... | [
"Solution:\nAnswers: $[N/2]$, $[(N+1)/2]$, $[N/2]$.\n\nSuppose $A$ leaves piles $n$, $m$ with $n \\leq m$.\n\nUnder $R1$, $B$ can certainly secure $m$ by dividing the larger pile into $1$ and $m-1$. He cannot do better, because if $b$ is the biggest of the $4$ piles, then the smallest is at most $m-b$. Hence $A$'s ... | Soviet Union | 1st ASU | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | [N/2], [(N+1)/2], [N/2] | |
09qq | Problem:
Zij $n \geq 2$ een geheel getal. Zij $a$ het grootste positieve gehele getal waarvoor geldt $2^{a} \mid 5^{n}-3^{n}$. Zij $b$ het grootste positieve gehele getal waarvoor geldt $2^{b} \leq n$. Bewijs dat $a \leq b+3$. | [
"Solution:\n\nWe bewijzen dit allereerst voor oneven getallen $n$. Hiervoor geldt modulo 4 dat $5^{n} \\equiv 1^{n}=1$ en $3^{n} \\equiv(-1)^{n} \\equiv-1$, dus $5^{n}-3^{n} \\equiv 2 \\bmod 4$. Dus als $n$ oneven is, geldt $a=1$. Omdat $b \\geq 1$, is nu voldaan aan $a \\leq b+3$.\n\nStel nu dat $n \\equiv 2 \\bmo... | Netherlands | Dutch TST | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | null | proof only | null | |
0kyh | Problem:
The integers from $1$ to $9$ are arranged in a $3 \times 3$ grid. The rows and columns of the grid correspond to $6$ three-digit numbers, reading rows from left to right, and columns from top to bottom. Compute the least possible value of the largest of the $6$ numbers. | [
"Solution:\n\nThe $5$ cells that make up the top row and left column are all leading digits of the three-digit numbers. Therefore, the largest number has leading digit at least $5$, achievable only if $6,7,8$, and $9$ are placed in the bottom right $2 \\times 2$ square. Then, the only three-digit numbers with tens ... | United States | HMMT November | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | null | proof and answer | 523 | |
0d0u | Let $P$ be a polynomial with real coefficients and odd degree. Suppose that the number of real solutions to
$$
P(P(x)) = P(x), \quad P(x) \neq x
$$
is finite and odd. Show that there exists a real $c$ such that $P(c) = c$ and the polynomial $P(x) - c$ has a real root of multiplicity at least two. (That is, it is divisi... | [
"Let $S$ be the set of all $a$ with $P(a) = a$. For each $a \\in S$, let $T_a$ be the set of solutions to $P(P(x)) = P(x) = a$ with $x \\neq a$. Since $P(P(x)) = P(x)$ and $P(x) \\neq x$ has an odd number of solutions, the union of all $T_a$ has an odd number of elements. Therefore one of the $T_a$ has an odd numbe... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof only | null | |
04sq | In any triangle $ABC$, in which the median from $C$ is not perpendicular to any of the sides $CA$ nor $CB$, let us denote $X$ and $Y$ intersections of this median's axis with lines $CA$ and $CB$. Find all such triangles $ABC$ for which points $A, B, X, Y$ lie on the same circle. (Ján Mazák) | [] | Czech Republic | Czech and Slovak Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof and answer | All triangles with CA = CB (i.e., triangles isosceles at C). | |
05ms | Problem:
Les sept nains ont des tailles deux à deux distinctes. Ils se rendent à la mine en colonne dans un certain ordre, de telle manière que le nain en tête est plus grand que le deuxième, qui est plus petit que le troisième, qui est plus grand que le quatrième et ainsi de suite...
Combien y a-t-il de telles maniè... | [
"Solution:\n\nOn note $u_{3}$ le nombre de manières d'arranger 3 nains de manière à ce que le premier est plus grand que le second, qui est plus grand que le troisième, $u_{5}$ le nombre de manières similaires d'arranger 5 nains et $u_{7}$ la solution de l'exercice.\n\nAvec 3 nains, on est obligé de mettre le nain ... | France | OCympiades Françaises de Mathématiques - Envoi Numéro 4 - Combinatoire | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 272 | |
08hd | Problem:
Let $a > 1$ be a non-integer number and $a \neq \sqrt[p]{q}$ for every positive integers $p \geq 2$ and $q \geq 1$, $k = [\log_{a} n] \geq 1$, where $[x]$ is the integral part of the real number $x$. Prove that for every positive integer $n \geq 1$ the equality
$$
[\log_{a} 2] + [\log_{a} 3] + \ldots + [\log_{... | [] | JBMO | THE 47-th MATHEMATIAL OLYMPIAD OF REPUBLIC OF MOLDOVA | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
0jgf | Problem:
Let $AB$ and $CD$ be two nonperpendicular diameters of a circle centered at $O$, and let $Q$ be the reflection of $D$ about $AB$. The tangent at $B$ meets $AC$ at $P$, and $DP$ meets the circle again at $E$. Prove that lines $AE$, $BP$, and $CQ$ are concurrent. | [
"\nLet $X$ be the intersection of $CQ$ and $BP$. We first note that $CQ \\parallel AB$ since\n$$\n\\angle CQA = \\angle CBA = \\angle BAD = \\angle BAQ\n$$\nNote that $\\triangle CPX \\sim \\triangle CDA$ since the angles at $X$ and $A$ are right and\n$$\n\\angle CPX = 90 - \\angle CAB = \\... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Concurrency and Collinearity",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
00sh | Suppose that the numbers $\{1, 2, \dots, 25\}$ are written in some order in an $5 \times 5$ array. Find the maximal positive integer $k$, such that the following holds. There is always an $2 \times 2$ subarray whose numbers have a sum not less than $k$.
An $5 \times 5$ array must be completed with all numbers $\{1, 2,... | [
"We will prove that $k_{\\max} = 45$.\n\nWe number the columns and the rows and we select all possible $3^2 = 9$ choices of an odd column with an odd row.\nCollecting all such pairs of an odd column with an odd row, we double count some squares. Indeed, we take some $3^2$ squares 5 times, some 12 squares 3 times an... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 45 | |
07w8 | For any two numbers $x, y$, we denote $N(x, y) = x^2 - xy + y^2$.
a. Prove the product formula $N(a, b)N(x, y) = N(ax - by, ay + bx - by)$.
b. Show that the equation $N(x, y) = 2023 - 490z^2$ has at least 16 distinct solution triples $(x, y, z)$ with integers $x, y, z \ge 0$. | [
"a. The formula in part (a) can be checked by direct calculation\n$$\n(x^2 - xy + y^2)(a^2 - ab + b^2) = (xa - by)^2 - (xa - by)(ay + bx - by) + (ay + bx - by)^2.\n$$\nAlternatively, we may let $\\xi, \\bar{\\xi}$ be the complex roots of the quadratic $x^2 - x + 1$, so that $\\xi^2 = \\xi - 1$, as well as $\\bar{\\... | Ireland | IRL_ABooklet_2023 | [
"Number Theory > Algebraic Number Theory > Quadratic fields",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | Part (a): N(a, b) N(x, y) = N(ax − by, ay + bx − by).
Part (b): Sixteen distinct nonnegative integer solutions (x, y, z) include:
- z = 2: (3, 9, 2), (6, 9, 2), (9, 3, 2), (9, 6, 2)
- z = 1: (4, 41, 1), (41, 4, 1), (31, 44, 1), (44, 31, 1), (44, 13, 1), (13, 44, 1), (37, 41, 1), (41, 37, 1)
- z = 0: (17, 51, 0), (51, 1... | |
0bx5 | Let $n$ and $k$ be two positive integers such that $1 \le n \le k$. Prove that, if $d^k + k$ is a prime number for each positive divisor $d$ of $n$, then $n + k$ is a prime number. | [
"For $d=1$ it follows that $1+k$ is a prime.\n\nFor $d=n$ it follows that $n^k + k$ is a prime. As $k+1$ does not divide $n$ (being larger than $n$), from Fermat's Theorem we get $n^k \\equiv 1 \\pmod{k+1}$, hence $n^k + k \\equiv 0 \\pmod{k+1}$. It follows that $n^k + k = k+1$, hence $n=1$. We have seen at the beg... | Romania | THE 68th NMO SELECTION TESTS FOR THE JUNIOR BALKAN MATHEMATICAL OLYMPIAD | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof only | null | |
0c59 | Let $f : [0, 1] \to \mathbb{R}$ be an integrable function, and let $(a_n)_{n \ge 1}$ be a bounded below sequence of real numbers, so that $\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k = a < \infty$. Prove that
$$
\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} a_k f\left(\frac{k}{n}\right) = a \int_{0}^{1} f(x) dx... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Algebra > Algebraic Expressions > Sequences and Series > Abel summation",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0fko | Problem:
En el interior de un paralelogramo $ABCD$ se dibujan dos circunferencias. Una es tangente a los lados $AB$ y $AD$, y la otra es tangente a los lados $CD$ y $CB$. Probar que si estas circunferencias son tangentes entre sí, el punto de tangencia está en la diagonal $AC$. | [
"Solution:\n\nVeremos que los puntos $A$, $K$ y $C$ están alineados.\n\n\n\nSean $O_{1}$ y $O_{2}$ los centros de la primera y segunda circunferencia, respectivamente. Notar que $A O_{1}$ biseca el ángulo $DAB$, y análogamente $C O_{2}$ biseca el ángulo $DCB$. Como los lados son paralelos d... | Spain | XLV Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Homothety"
] | null | proof only | null | |
07q0 | Let $a$, $b$, $c$, $d$ be real numbers, with at least one of $a$ or $c$ non-zero, such that
$$
a^2 + ac + c^2 + 1 = b^2 + bd + d^2, \quad \text{and} \quad 2ab + ad + 2cd + bc = 0.
$$
Show that $ac < 0$ or $(3a^2 - b^2 + 1)(3c^2 - d^2 + 1) \le 0$. | [
"Notice that $b \\neq 0$ or $d \\neq 0$ as $a^2 + ac + c^2 + 1 = (a + c/2)^2 + (3/4)c^2 + 1 > 0$. Suppose that $c = 0$. By assumption, we then have $a \\neq 0$ and the second equation becomes $2ab + ad = 0$, hence $b = -d/2$. Substituting this into the first equation gives $d^2 = (4/3)(a^2 + 1)$. So $b^2 = (a^2 + 1... | Ireland | Ireland | [
"Algebra > Intermediate Algebra > Complex numbers",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
0dxg | Problem:
Dokaži, da za vsa realna števila $x$ in $y$ velja neenakost
$$
\cos \left(x^{2}\right)+\cos \left(y^{2}\right)-\cos (x y)<3
$$ | [
"Solution:\n\nKer za vsako realno število $\\alpha$ velja $-1 \\leq \\cos \\alpha \\leq 1$, je $\\cos x^{2} \\leq 1$, $\\cos y^{2} \\leq 1$ in $-\\cos x y \\leq 1$. Sledi\n$$\n\\cos x^{2}+\\cos y^{2}-\\cos x y \\leq 3\n$$\n\nPokazati moramo, da ne more veljati enačaj. Denimo, da velja. Potem je $\\cos x^{2}=1$, $\\... | Slovenia | 51. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
0kai | Problem:
Determine the second smallest positive integer $n$ such that $n^{3}+n^{2}+n+1$ is a perfect square. | [
"Solution:\n$n^{3}+n^{2}+n+1 = (n+1)(n^{2}+1)$. Note that $\\gcd(n^{2}+1, n+1) = \\gcd(2, n+1) = 1$ or $2$, and since $n^{2}+1$ is not a perfect square for $n \\geq 1$, we must have $n^{2}+1 = 2p^{2}$ and $n+1 = 2q^{2}$ for some integers $p$ and $q$. The first equation is a variant of Pell's equation, which (either... | United States | HMMT February 2019 | [
"Number Theory > Diophantine Equations > Pell's equations",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | 7 | |
09bt | $x, y, z, t > 0$ ба $x + y + z + t = 1$ бол
$$
\frac{xy(z+t)}{1-4zt} + \frac{yz(t+x)}{1-4tx} + \frac{zt(x+y)}{1-4xy} + \frac{tx(y+z)}{1-4yz} \ge \frac{128}{3}xyzt
$$ | [
"$$\nA = \\frac{z+t}{zt(1-4zt)} + \\frac{t+x}{tx(1-4tx)} + \\frac{x+y}{xy(1-4xy)} + \\frac{y+z}{yz(1-4yz)} \\ge \\frac{128}{3}\n$$\nболно.\n---\n$$\n\\begin{aligned}\n\\frac{z+t}{zt(1-4zt)} &= \\frac{z+t}{zt(1-2\\sqrt{zt})(1+2\\sqrt{zt})} \\ge \\\\\n&\\ge \\frac{2\\sqrt{zt}}{zt(1-2\\sqrt{zt})(1+2\\sqrt{zt})} = \\fr... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | Mongolian | proof only | null | |
0i1s | Problem:
All subscripts in this problem are to be considered modulo $6$, that means for example that $\omega_{7}$ is the same as $\omega_{1}$. Let $\omega_{1}, \ldots, \omega_{6}$ be circles of radius $r$, whose centers lie on a regular hexagon of side length $1$. Let $P_{i}$ be the intersection of $\omega_{i}$ and $\... | [
"Solution:\n\nConsider two consecutive circles $\\omega_{i}$ and $\\omega_{i+1}$. Let $Q_{i}, Q_{i}'$ be two points on $\\omega_{i}$ and $Q_{i+1}, Q_{i+1}'$ on $\\omega_{i+1}$ such that $Q_{i}, P_{i}$ and $Q_{i+1}$ are colinear and also $Q_{i}', P_{i}$ and $Q_{i+1}'$. Then $Q_{i} Q_{i}' = 2 \\angle Q_{i} P_{i} Q_{i... | United States | Harvard-MIT Math Tournament | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Transformations > Rotation"
] | null | proof and answer | 5 | |
0dxy | Let $a$ and $b$ be real numbers such that $\frac{4a}{a+2b} - \frac{5b}{2a+b} = 1$. Find all possible values of the expression $\frac{a-2b}{4a+5b}$. | [
"First note that $a + 2b$ and $2a + b$ cannot both be zero, so $a$ and $b$ cannot both be zero. Eliminating the fractions in $\\frac{4a}{a+2b} - \\frac{5b}{2a+b} = 1$ we get $8a^2 + 4ab - 5ab - 10b^2 = 2a^2 + 5ab + 2b^2$, which we then rewrite as $6(a + b)(a - 2b) = 0$.\n\nIf $a = -b$, then $b$ must be non-zero and... | Slovenia | Slovenija 2008 | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | proof and answer | 0 and -3 | |
0545 | On the plane three different points *P*, *Q*, and *R* are chosen. It is known that however one chooses another point *X* on the plane, the point *P* is always either closer to *X* than the point *Q* or closer to *X* than the point *R*. Prove that the point *P* lies on the line segment *QR*. | [
"We show that if the point $P$ lies outside the segment $QR$, then the conditions of the problem are not satisfied.\nIf $P$ lies on the line $QR$ but outside the segment $QR$ (Fig. 3), then we can\n\n\nFig. 3\n\n\nFig. 4\n\ntake the point $X$ on the line $QR$ on the... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09tx | Problem:
In een niet-gelijkbenige driehoek $ABC$ is $I$ het middelpunt van de ingeschreven cirkel. De bissectrice van $\angle BAC$ snijdt de omgeschreven cirkel van $\triangle ABC$ nogmaals in $D$. De lijn door $I$ loodrecht op $AD$ snijdt $BC$ in $F$. Het midden van boog $BC$ waar $A$ op ligt, noemen we $M$. De lijn ... | [
"Solution:\n\nOmdat $\\angle DAC = \\angle BAD$ is $D$ het midden van de boog $BC$ waar $A$ niet op ligt. De lijn $DM$ is dus een middellijn van de cirkel. Met Thales zien we dan dat $\\angle DBM = 90^{\\circ}$. Zij nu $K$ het snijpunt van $BC$ en $DM$. Omdat $DM$ de middelloodlijn van $BC$ is, geldt nu $\\angle BK... | Netherlands | IMO-selectietoets III | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle c... | null | proof only | null | |
0giy | To each vertex of a regular pentagon an integer is assigned, so that the sum of all five numbers is positive. If three consecutive vertices are assigned the numbers $x, y, z$ respectively, and $y < 0$, then the following operation is allowed: $x, y, z$ are replaced by $x+y, -y, z+y$ respectively. Such an operation is p... | [
"The algorithm always stops. Let $S = \\sum x_i > 0$ and consider the function\n$$\nf(x_1, x_2, x_3, x_4, x_5) = \\sum_{i=1}^{5} (x_i - x_{i+2})^2, \\quad x_6 = x_1, \\ x_7 = x_2.\n$$\nClearly $f > 0$ always and $f$ is integer valued. Suppose, WLOG, that $y = x_4 < 0$. Then $f_{new} - f_{old} = 2Sx_4 < 0$ since $S ... | Taiwan | APMO Taiwan Preliminary Round 2 | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | Chinese; English | proof and answer | Yes, the process necessarily terminates after a finite number of steps. | |
05mt | Problem:
2015 droites deux à deux distinctes sont tracées dans le plan. On suppose qu'elles délimitent moins de 8000 régions (les régions peuvent être non bornées). Montrer que le nombre de régions est égal à 2016, 4030, 6042 ou 6043, et donner un exemple de configuration dans chaque cas. | [
"Solution:\n\nNotons $n = 2015$ le nombre de droites, $f$ le nombre de régions, $p$ le nombre maximal de droites parallèles et $q$ le nombre maximal de droites concourantes.\n\nMontrons d'abord que $f \\geqslant (p+1)(n-p+1)$. En effet, plaçons d'abord les $p$ droites parallèles. Elles forment $p+1$ régions. Il res... | France | Olympiades Françaises de Mathématiques | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 2016, 4030, 6042, 6043 | |
0eho | Problem:
Podani sta premica $p$ z enačbo $-x+2y=1$ in premica $q$ z enačbo $-4x+3y=16$. Naj bo točka $A$ presečišče premice $p$ z osjo $x$, naj bo točka $B$ presečišče premice $q$ z osjo $x$, naj bo točka $C$ presečišče premic $p$ in $q$ ter naj bo točka $D$ pravokotna projekcija točke $C$ na os $x$.
a) Izračunaj in ... | [
"Solution:\n\na. Presečišče premice $p$ z osjo $x$ ima $y$ koordinato $0$. Tako dobimo $A(-1, 0)$ in $B(4, 0)$. Rešimo sistem enačb za $p$ in $q$, da dobimo koordinate točke $C\\left(-\\frac{29}{5}, -\\frac{12}{5}\\right)$. Točka $D$ ima koordinati $D\\left(-\\frac{29}{5}, 0\\right)$.\n\nb. Kot $\\beta$ iz trikotni... | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Odbirno tekmovanje | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | final answer only | a) A = (−1, 0), B = (−4, 0), C = (−29/5, −12/5), D = (−29/5, 0). b) β (at vertex B) ≈ 126.87° (since tan of the acute angle is 4/3, the interior angle is 180° − arctan(4/3)). c) Area S = 18/5 = 3.6. | |
0jg6 | Problem:
Five consecutive vertices of a regular 2013-gon are given. Prove that one can reconstruct the entire 2013-gon using straightedge alone. | [
"Solution:\n\nLet $A, B, C, D, E$, and $F$ be six consecutive vertices of the polygon. We prove that, given $A, B, C, D$, and $E$, it is possible to construct $F$ with straightedge alone. Then, continuing around the polygon, we can construct all the vertices and then fill in the sides.\n\nOur proof is based on the ... | United States | Berkeley Math Circle Monthly Contest 6 | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null |
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