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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
07z0 | Problem:
Un antiquario ha comprato una coppia di comodini per 1 milione e mezzo. Ne vende uno ad Aldo e uno a Berto per 1 milione l'uno.
Viene Carlo che li aveva visti in vetrina, ed è disposto ad acquistare la coppia per 3 milioni e mezzo.
L'antiquario riesce a riacquistare i comodini da Aldo e Berto, pagandoli 1 m... | [] | Italy | Progetto Olimpiadi di Matematica | [
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | MCQ | D | |
0g1k | Problem:
Der brasilianische IMO-Leader wählt zwei natürliche Zahlen $n$ und $k$ mit $n>k$, und sagt diese dann seinem Deputy und einem Teilnehmer. Dann flüstert der Leader dem Deputy eine binäre Folge der Länge $n$ ins Ohr. Der Deputy schreibt alle binären Folgen der Länge $n$ auf, die sich genau an $k$ Stellen von de... | [
"Solution:\n\nWenn wir eine fixe Stelle der Lösungsfolge $L$ betrachten, dann stimmen $\\binom{n-1}{k}$ der aufgelisteten Folgen an dieser Stelle mit $L$ überein und $\\binom{n-1}{k-1}$ der aufgelisteten Folgen unterscheiden sich an dieser Stelle von $L$. Da der Teilnehmer $n$ und $k$ kennt, kann er diese Werte auc... | Switzerland | IMO-Selektion | [
"Discrete Mathematics > Combinatorics > Algebraic properties of binomial coefficients",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | Minimum guesses = 1 if k ≠ n/2; minimum guesses = 2 if k = n/2. | |
08sr | Currently circulating coins in Japanese currency come in 6 different denominations: 500 yen, 100 yen, 50 yen, 10 yen, 5 yen and 1 yen. Taro had 1 each of 1000 yen note, 100 yen coin, 10 yen coin and 1 yen coin. He made a certain purchase from a merchant and handed him all he had and received some change. Assume that Ta... | [
"To begin with, we note that two or more of 500 yen coins, or of 50 yen coins, or of 5 yen coins could not be included in the change Taro received. This is because if two or more of any of these coins were in the change, two of them could be exchanged for one 1000 yen note, or one 100 yen coin or one 10 yen coin to... | Japan | Japan Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | 8 | |
0kt5 | Equilateral triangle $\triangle ABC$ is inscribed in circle $\omega$ with radius $18$. Circle $\omega_A$ is tangent to sides $\overline{AB}$ and $\overline{AC}$ and is internally tangent to $\omega$. Circles $\omega_B$ and $\omega_C$ are defined analogously. Circles $\omega_A$, $\omega_B$, and $\omega_C$ meet in six po... | [] | United States | 2022 AIME I | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 378 | |
02xc | Problem:
Se $x$, $y$ e $z$ são números reais positivos e
$$
\frac{x y}{x+y}=a, \quad \frac{x z}{x+z}=b, \quad \text{e} \quad \frac{y z}{y+z}=c
$$
a) Verifique que
$$
x=\frac{a y}{y-a}
$$
b) Verifique que
$$
x=\frac{2 a b c}{a c+b c-a b}
$$ | [
"Solution:\na) Da primeira equação do enunciado, temos\n$$\n\\begin{aligned}\n\\frac{x y}{x+y} & =a \\\\\nx y & =a x+a y \\\\\nx(y-a) & =a y \\\\\nx & =\\frac{a y}{y-a} .\n\\end{aligned}\n$$\nNote que a última divisão por $y-a$ é possível, pois se $y-a=0$ segue que $a y=0$ e isso é ímpossível, dado que $a$ e $x$ sã... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof only | null | |
01es | Let $p$ be an odd prime. Find all positive integers $n$ for which $\sqrt{n^2 - np}$ is a positive integer? | [
"Answer: $n = \\left(\\frac{p+1}{2}\\right)^2$.\nAssume that $\\sqrt{n^2 - pn} = m$ is a positive integer. Then $n^2 - pn - m^2 = 0$, and hence\n$$\nn = \\frac{p \\pm \\sqrt{p^2 + 4m^2}}{2}.\n$$\nNow $p^2+4m^2 = k^2$ for some positive integer $k$, and $n = \\frac{p+k}{2}$ since $k > p$. Thus $p^2 = (k+2m)(k-2m)$, a... | Baltic Way | Baltic Way shortlist | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | n = ((p+1)/2)^2 | |
0ed1 | Let $a$ and $b$ be real numbers which satisfy the equality $\frac{a^2}{1+a^2} + \frac{b^2}{1+b^2} = 1$. Determine all possible values of the expression
$$
(a + b) \left( \frac{a}{1 + a^2} + \frac{b}{1 + b^2} \right).
$$ | [
"We multiply the given equation by $(1+a^2)(1+b^2) \\neq 0$ and simplify it to get the equivalent equation $a^2b^2 = 1$. Therefore $ab = \\pm 1$. We also simplify the given expression\n$$\n\\begin{aligned}\n(a+b) \\left( \\frac{a}{1+a^2} + \\frac{b}{1+b^2} \\right) &= (a+b) \\frac{a+ab^2+b+a^2b}{(1+a^2)(1+b^2)} = (... | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | 0, 2 | |
06br | Let $f(x) = x^n + a_1x^{n-1} + \cdots + a_n$ be a polynomial of degree $n \ge 2$ with real roots $b_1, b_2, \ldots, b_n$. If $x$ is greater than $b_1, b_2, \ldots, b_n$, prove that
$$
f(x+1) \left( \frac{1}{x-b_1} + \frac{1}{x-b_2} + \cdots + \frac{1}{x-b_n} \right) \ge 2n^2.
$$ | [
"Let $c_j = x - b_j > 0$ for each $j$. Note that\n$$\nf(y) = (y - b_1)(y - b_2)\\cdots(y - b_n).\n$$\nTherefore, we have\n$$\n\\begin{aligned}\nf(x+1) &= (x+1-b_1)(x+1-b_2)\\cdots(x+1-b_n) \\\\\n&= (1+c_1)(1+c_2)\\cdots(1+c_n).\n\\end{aligned}\n$$\nNow, by the AM-GM inequality, we have\n$$\n\\frac{1}{c_1} + \\frac{... | Hong Kong | IMO HK TST | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials"
] | null | proof only | null | |
0ih9 | Problem:
Let $A_{1} A_{2} \ldots A_{k}$ be a regular $k$-gon inscribed in a circle of radius $1$, and let $P$ be a point lying on or inside the circumcircle. Find the maximum possible value of $\left(PA_{1}\right)\left(PA_{2}\right) \cdots\left(PA_{k}\right)$. | [
"Solution:\nPlace the vertices at the $k$th roots of unity, $1, \\omega, \\ldots, \\omega^{k-1}$, and place $P$ at some complex number $p$. Then\n$$\n\\begin{aligned}\n\\left(\\left(PA_{1}\\right)\\left(PA_{2}\\right) \\cdots\\left(PA_{k}\\right)\\right)^{2} & = \\prod_{i=0}^{k-1}\\left|p-\\omega^{i}\\right|^{2} \\... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 2 | |
05s1 | Problem:
Trouver toutes les fonctions $f: \mathbb{N} \rightarrow \mathbb{N}$ telles que
$$
x f(y)+y f(x)=(x+y) f\left(x^{2}+y^{2}\right)
$$
pour tous les entiers naturels $x$ et $y$.
Note : on rappelle que $\mathbb{N}$ désigne l'ensemble des entiers naturels, c'est-à-dire $\mathbb{N}=\{0,1,2, \ldots\}$. | [
"Solution:\n\nDans la suite, on notera $\\mathbf{E}_{x, y}$ l'équation de l'énoncé.\n\nTout d'abord, l'équation $E_{x, 0}$ indique que $x f(0)=x f\\left(x^{2}\\right)$, ce qui montre que $f\\left(x^{2}\\right)=f(0)$ pour tout entier $x \\geqslant 1$.\n\nPar conséquent, pour tout entier $y \\geqslant 1$, l'équation ... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Other"
] | null | proof and answer | All constant functions from the natural numbers to the natural numbers; that is, f(n) = c for some fixed nonnegative integer c. | |
0fj8 | Problem:
Consideramos los siguientes 27 puntos de un cubo: el centro (1), los centros de las caras (6), los vértices (8) y los centros de las aristas (12). Coloreamos cada uno de esos puntos de azul o de rojo. ¿Puede hacerse de modo que no haya tres puntos del mismo color alineados? Demuéstralo. | [
"Solution:\nLa respuesta es no. Para demostrarlo, llamaremos a los puntos centrales de las caras de la siguiente forma: F-frontal, P-posterior, I-izquierda, D-derecha, T-tapa y S-suelo; en donde el nombre representa la situación respecto a un observador que mira el cubo desde un punto exterior situado frente a la c... | Spain | Viernes 19 de enero de 2001 | [
"Geometry > Solid Geometry > Other 3D problems",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0hup | Problem:
Suppose that $S_{1}, S_{2}, S_{3}, \ldots$ are sets of integers such that no integer is contained in more than one $S_{n}$; every $S_{n}$ has exactly two elements; and the sum of the elements of $S_{n}$ is $n$. Prove that there exist infinitely many values of $n$ with the following property: one of the elemen... | [
"Solution:\n\nConsider any set $S$ representable as the union of $n$ disjoint pairs of integers such that, if any negative number $-m$ is in $S$, then the other element of its pair (henceforth called its \"partner\") is $\\geq 13 m / 6$. Any such $S$ will be called \"plausible\" of order $n$. We will work with plau... | United States | Berkeley Math Circle | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations ... | null | proof only | null | |
023h | Problem:
$P$ é um ponto no interior do quadrado $ABCD$ tal que $PA = 1$, $PB = 2$ e $PC = 3$. Qual é a medida do ângulo $A \hat{P} B$?
Figura 119.1 | [
"Solution:\n\nSeja $Q$ um ponto tal que os triângulos $CQB$ e $APB$ são congruentes, como mostrado na figura. Isto é equivalente a fazer uma rotação do triângulo $APB$ com centro em $B$ e ângulo $90^{\\circ}$ no sentido horário. Em particular, temos que $\\angle PBQ = 90^{\\circ}$. Assim, $PQ^{2} = PB^{2} + BQ^{2} ... | Brazil | Desafios | [
"Geometry > Plane Geometry > Transformations > Rotation",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 135° | |
0ahl | Let $\mathbb{Z}^+$ be the set of all positive integers. Find all functions $f: \mathbb{Z}^+ \to \mathbb{Z}^+$ such that $f(n!) = f(n)!$, and $m-n$ divides $f(m)-f(n)$ for all distinct positive integers $m$ and $n$. | [
"There are three functions: the constant functions $1$, $2$ and the identity function $\\text{id}_{\\mathbb{Z}^+}$. These functions clearly satisfy the conditions in the hypothesis. Let us prove that there are only ones.\n\nConsider such a function $f$ and suppose that it has a fixed point $a \\ge 3$, that is $f(a)... | North Macedonia | 29-th Balkan Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | The identity function, the constant function equal to one, and the constant function equal to two. | |
06b1 | Let $ABC$ be a triangle with $AB < AC < BC$ and circumcircle $\Gamma_1$ of center $O$. We consider the circle $\Gamma_2$ with center $D$ lying on the circle $\Gamma_1$, and tangent to the line $BC$ at the point $E$ and tangent to the extension of the side $AB$ at point $F$. The circles $\Gamma_1$ and $\Gamma_2$ interse... | [
"We have $DE \\perp BC$ and $DF \\perp AB$ (because $\\Gamma_2$ is tangent to the side $BC$ at $E$ and to the line $AB$ at $F$).\n\n\nFigure 4\n\n\nFigure 5\n\nHence the quadrilateral $BEDF$ is cyclic and let $\\Gamma_3$ be its circumcircle.\nMoreover we have the eq... | Greece | Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
002i | Sea $ABC$ un triángulo cuyos ángulos $\angle CAB$ y $\angle ABC$ son mayores que $45^\circ$. $PQRS$ es un cuadrado tal que $P$ y $Q$ están en el interior del lado $AB$ en el orden $APQB$, $R$ está en el interior del lado $BC$, $S$ está en el interior del lado $CA$. Sean $M$ y $N$ los pies de las perpendiculares trazada... | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals"
] | Español | proof only | null | |
0evx | Problem:
120 unit squares are arbitrarily arranged in a $20 \times 25$ rectangle (both position and orientation is arbitrary). Prove that it is always possible to place a circle of unit diameter inside the rectangle without intersecting any of the squares. | [
"Solution:\n\nIf a circle with unit diameter intersects a unit square, then its center must lie inside an area $3 + \\pi /4$, namely an oval centered on the square and comprising: the original square, area $1$; four $1 \\times 1/2$ rectangles on the sides, total area $2$; and four quarter circles at the corners, to... | Soviet Union | 1st ASU | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
089j | Problem:
In un trapezio isoscele il punto d'incontro delle diagonali vede la base minore sotto un angolo di $150^{\circ}$ e ciascuna diagonale è lunga $6~\mathrm{cm}$. Qual è l'area del trapezio?
(A) $9~\mathrm{cm}^{2}$
(B) $16~\mathrm{cm}^{2}$
(C) $24~\mathrm{cm}^{2}$
(D) i dati sono insufficienti
(E) nessuna delle ... | [
"Solution:\n\nLa risposta è (A). Indichiamo con $A, B, C, D$ i vertici del trapezio, con $O$ il punto d'incontro delle diagonali, con $H$ e $K$ i piedi delle perpendicolari ad $AC$ condotte, rispettivamente, da $B$ e $D$. L'area $S$ del trapezio $ABCD$ è pari alla somma delle aree dei triangoli $ACB$ e $ACD$, che h... | Italy | Olimpiadi di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | A | |
068u | Let $\triangle AB\Gamma$ with $AB < A\Gamma < B\Gamma$ be a triangle inscribed in a circle $c$ with center $O$. Let $\Delta$ be the diametrically opposite to $A$ and $c_1$ is a circle which has center $K$ on $B\Delta$ and passes through $B$ and $\Gamma$. If $c_1$ meets $A\Gamma$ at the point $E$, prove that the circumc... | [
"Let $Z$ be the intersection of $c_2$ with $AB$ and $M$ the intersection of $KZ$ with $BE$. The angle $A\\hat{B}\\Delta$ (and so the angle $Z\\hat{B}K$) is right since it sees the diameter $A\\Delta$ of the circle $c(O, R)$.\nSince $Z\\hat{B}K = 90^\\circ$, $ZK$ is a diameter of $c_1$ and thus $ZK$ is the midperpen... | Greece | Hellenic Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
02xb | Problem:
No quadrado mágico a seguir, todo inteiro de $1$ a $25$ pode ser colocado nos quadradinhos de modo que as somas em toda linha e coluna, bem como nas diagonais, é a mesma. Dado que o número no centro do quadrado é $18$, qual é o valor da soma dos números escritos nos quadradinhos sombreados?
![](attached_imag... | [
"Solution:\n\nComo a soma de todos os números do quadrado mágico é $1+2+\\ldots+25=325$, a soma dos números em cada linha, coluna ou diagonal é $325 / 5=65$. Ao somar todos os números das duas diagonais, da linha do meio e da coluna do meio, estamos somando todos os números que não estão sombreados e assim obtemos ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 119 | |
009j | In the convex quadrilateral $ABCD$ the angles at $A$ and $C$ are equal and the bisector of $B$ passes through the midpoint of side $CD$. Given that $CD = 3AD$, find the ratio $AB/BC$. | [
"Let $M$ be the midpoint of $CD$. Since $BM$ is the bisector of $B$, the reflection $E$ of $C$ in $BM$ lies on the ray $BA$. Because $\\angle MEB = \\angle MCB$, by reflection, and $\\angle MECB = \\angle DAB$ by hypothesis, we have $\\angle MEB = \\angle DAB$.\n\nHence $ME \\parallel DA$; in particular $E$ is on t... | Argentina | NATIONAL XXX OMA | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 5/3 | |
0ikv | Problem:
Triangle $A B C$ has a right angle at $B$. Point $D$ lies on side $B C$ such that $3 \angle B A D = \angle B A C$. Given $A C = 2$ and $C D = 1$, compute $B D$. | [
"Solution:\n\nLet $B D = x$. We reflect $D$ over $A B$ to $D'$. Then $D D' = 2x$, but $A D$ bisects $C A D'$, so $4x = A D' = A D$. Also, $A D = \\sqrt{x^2 + A B^2} = \\sqrt{x^2 + A C^2 - B C^2} = \\sqrt{x^2 + 4 - (x + 1)^2} = \\sqrt{3 - 2x}$. We have the quadratic $16x^2 = 3 - 2x$ which gives $x = 3/8$."
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | final answer only | 3/8 | |
0din | Let $ABC$ be a right-angled triangle with $\angle BAC = 90^\circ$, and let $E$ be the foot of the perpendicular from $A$ to $BC$. Let $Z \neq A$ be a point on the line $AB$ with $AB = BZ$. Let $(c_1)$ be the circumcircle of the triangle $BEZ$ and $(c_2)$ be an arbitrary circle passing through the points $A$ and $E$. Su... | [
"Since triangles $AEB$ and $CAB$ are similar, we have $\\frac{AB}{EB} = \\frac{CB}{AB}$. Note that $AB = BZ$, thus\n$$\n\\frac{BZ}{EB} = \\frac{CB}{AB},\n$$\nfrom which it follows that triangles $ZBE$ and $CBZ$ are also similar. Since $FEBZ$ is cyclic, then $\\angle BEZ = \\angle BFZ$. So by the similarity of trian... | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ewg | Problem:
Given a number with 1998 digits which is divisible by 9. Let $x$ be the sum of its digits, let $y$ be the sum of the digits of $x$, and $z$ the sum of the digits of $y$. Find $z$. | [
"Solution:\n$x \\leq 9 \\times 1998 = 17982$. Hence $y \\leq$ the greater of $1 + 7 + 9 + 9 + 9 = 35$ and $9 + 9 + 9 + 9 = 36$. But $9$ divides the original number and hence also $x$, $y$ and $z$. Hence $z = 9$."
] | Soviet Union | 2nd ASU | [
"Number Theory > Divisibility / Factorization",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 9 | |
0ggf | Let $\mathbb{R}$ denote the set of all real numbers. Determine all functions $f: \mathbb{R} \setminus \{-1\} \to \mathbb{R}$ such that for all real numbers $x, y$ satisfying $(x+1)(y+1)(xy+1) \neq 0$, the equation
$$
f(x) f\left(f\left(\frac{1-y}{1+y}\right)\right) = f\left(\frac{x+y}{xy+1}\right)
$$
holds. | [] | Taiwan | 2022 數學奧林匹亞競賽第五階段培訓營, 國際競賽實作(二) | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | Chinese; English | proof and answer | f(x) = 0 for all x ∈ ℝ \ {−1} and f(x) = 1 for all x ∈ ℝ \ {−1} | |
0bwr | A function $f: \mathbb{R} \to \mathbb{R}$ is said to have property ($\mathcal{P}$) if for any sequence of reals $(x_n)_{n \ge 1}$, such that $(f(x_n))_{n \ge 1}$ is convergent, the sequence $(x_n)_{n \ge 1}$ is also convergent. Prove that a surjective function that posses property ($\mathcal{P}$), is continuous.
Mihai ... | [
"Consider $a, b \\in \\mathbb{R}$ such that $f(a) = f(b)$. The sequence $(x_n)_{n \\ge 1}$ defined by $x_n = \\begin{cases} a, & n \\text{ even} \\\\ b, & n \\text{ odd} \\end{cases}$ converges as $(f(x_n))_{n \\ge 1}$ is convergent, concluding that $f$ is injective. $f$ is surjective by definition, $f$ is invertib... | Romania | THE 68th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | English | proof only | null | |
03y3 | Let $k$ be an integer and $k > 1$. Define a sequence $\{a_n\}$ as follows: $a_0 = 0$, $a_1 = 1$ and $a_{n+1} = k a_n + a_{n-1}$ for $n = 1, 2, \dots$
Determine, with proof, all possible $k$ for which there exist non-negative integers $\ell, m$ ($\ell \ne m$) and positive integers $p, q$ such that $a_\ell + k a_p = a_m ... | [
"The answer is $k = 2$.\nIf $k = 2$, then $a_0 = 0$, $a_1 = 1$, $a_2 = 2$, so $a_0 + 2 a_2 = a_2 + 2 a_1 = 4$. Hence, $(\\ell, m) = (0, 2)$ and $(p, q) = (2, 1)$.\n\nFor $k \\ge 3$, it follows from the recurrent relation that the sequence $\\{a_n\\}$ is strictly increasing, and $k \\mid a_{n+1} - a_{n-1}$ for all $... | China | China Western Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | k = 2 | |
0cay | Let $a_n = \{\sqrt{n}\} - \{\sqrt{n+1}\} + \{\sqrt{n+2}\} - \{\sqrt{n+3}\}$, where $\{x\}$ denotes the fractional part of the real number $x$.
a) Prove that $a_1 > 0.2$.
b) Prove that $a_n < 0$, for infinitely many $n$, and $a_n > 0$, for infinitely many $n$. | [
"a) $a_1 = 0 - \\{\\sqrt{2}\\} + \\{\\sqrt{3}\\} + 0 = \\sqrt{3} - 1 - (\\sqrt{2} - 1) = \\sqrt{3} - \\sqrt{2}$. Since $\\sqrt{3} > 1.7$ and $\\sqrt{2} < 1.5$, the conclusion follows.\n\nb) Notice that, if $m$ is a positive integer and $m^2 \\leq a < b < (m+1)^2$, then $\\lfloor\\sqrt{a}\\rfloor = \\lfloor\\sqrt{b}... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
06u2 | Denote by $\mathbb{N}$ the set of all positive integers. Find all functions $f: \mathbb{N} \rightarrow \mathbb{N}$ such that for all positive integers $m$ and $n$, the integer $f(m)+f(n)-m n$ is nonzero and divides $m f(m)+n f(n)$. | [
"It is given that\n$$\nf(m)+f(n)-m n \\mid m f(m)+n f(n) . \\tag{1}\n$$\nTaking $m=n=1$ in (1), we have $2 f(1)-1 \\mid 2 f(1)$. Then $2 f(1)-1 \\mid 2 f(1)-(2 f(1)-1)=1$ and hence $f(1)=1$.\nLet $p \\geqslant 7$ be a prime. Taking $m=p$ and $n=1$ in (1), we have $f(p)-p+1 \\mid p f(p)+1$ and hence\n$$\nf(p)-p+1 \\... | IMO | IMO 2016 Shortlisted Problems | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | f(n) = n^2 for all positive integers n | |
0frf | Tenemos 2021 colores y 2021 fichas de cada color. Colocamos las $2021^2$ fichas en fila. Se dice que una ficha, $F$, es "mala" si a cada lado de $F$ quedan un número impar de las $2020 \times 2021$ fichas que no comparten color con $F$.
a) Determina cuál es el mínimo número posible de fichas malas.
b) Si se impone la c... | [
"(a) Como $2020 \\times 2021$ es par, para decidir si una ficha es mala es suficiente comprobar que el número de fichas a su izquierda con las que no comparte color es impar. Sea $A$ el conjunto de fichas que ocupan una posición impar en la fila, es decir, las fichas que tienen un número par de fichas a su izquierd... | Spain | LVII Olimpiada Matemática Española Concurso Final Nacional | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | a) 1010, b) 3030 | |
0j1b | Problem:
How many positive integers less than or equal to $240$ can be expressed as a sum of distinct factorials? Consider $0!$ and $1!$ to be distinct. | [
"Solution:\n\nAnswer: $39$\n\nNote that $1 = 0!$, $2 = 0! + 1!$, $3 = 0! + 2!$, and $4 = 0! + 1! + 2!$. These are the only numbers less than $6$ that can be written as the sum of factorials. The only other factorials less than $240$ are $3! = 6$, $4! = 24$, and $5! = 120$. So a positive integer less than or equal t... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Other",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 39 | |
0f1h | Problem:
What is the smallest perimeter for a convex $32$-gon whose vertices are all lattice points? | [] | Soviet Union | ASU | [
"Geometry > Plane Geometry > Combinatorial Geometry > Convex hulls",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 24 + 4(√2 + √5 + √10 + √13 + √17 + √26) | |
05b0 | Non-negative integers $a$, $b$, $q$, $r$ are all less than $5$ and satisfy conditions $q < a$ and $r < b \le a$. Dividing $qb + r$ by $5$ gives a remainder of $a$. Can we be certain that dividing $a$ by $b$ gives a remainder of $r$? | [
"*Solution 1:* The numbers $a = b = 3$ and $q = r = 2$ satisfy the conditions of the problem, but dividing $3$ by $3$ gives a remainder of $0$, not $2$.\n\n\n*Solution 2:* The numbers $a = b = 4$, $q = 2$ and $r = 1$ satisfy the conditions of the problem, but dividing $4$ by $4$ gives a remainder of $0$, not $1$.\n... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | No | |
012h | Problem:
Find all integers $n > 1$ such that any prime divisor of $n^{6} - 1$ is a divisor of $\left(n^{3} - 1\right)\left(n^{2} - 1\right)$. | [
"Solution:\nConsider the equality\n$$\nn^{6}-1=\\left(n^{2}-n+1\\right)(n+1)\\left(n^{3}-1\\right) \\text{.}\n$$\nThe integer $n^{2}-n+1=n(n-1)+1$ clearly has an odd divisor $p$. Then $p \\mid n^{3}+1$. Therefore, $p$ does not divide $n^{3}-1$ and consequently $p \\mid n^{2}-1$. This implies that $p$ divides $\\lef... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 2 | |
0gx9 | $BB_1$ is an altitude of the acute-angled scalene triangle $ABC$. Point $D$ is placed on the side $BC$ so that $\angle BAD = \angle CBB_1$. Segments $AD$ and $BB_1$ intersect at point $F$. Line $l$ is drawn through point $B$ at right angle to the side $AB$. This line intersects line $CF$ at point $K$. Prove that line $... | [
"Let line $AD$ intersect a circle circumscribed about triangle $\\triangle ABC$ at point $N$. Then $\\angle BCN = \\angle BAN = \\angle CBB_1$. This implies that $BB_1 \\parallel CN$ (fig.2), and therefore $\\angle ACN = 90^\\circ \\Rightarrow AN$ is a diameter. Thus $\\angle NBA = 90^\\circ$, which implies that $N... | Ukraine | Ukrajina 2008 | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0ggx | 設 $ABCD$ 為平行四邊形, 其中 $AC = BC$。在直線 $AB$ 上取一點 $P$, 使得 $B$ 介於 $A, P$ 之間。設三角形 $ACD$ 的外接圓與線段 $PD$ 再交於點 $Q$, 而三角形 $APQ$ 的外接圓與線段 $PC$ 再交於點 $R$。
證明：三直線 $CD, AQ, BR$ 共點。 | [
"在以下的各種解法中,都用到以下的事實。\n因為 $AC = BC = AD$, 所以 $\\angle ABC = \\angle BAC = \\angle ACD = \\angle ADC$。由於四邊形 $APRQ$, $AQCD$ 都有外接圓, 可得\n$$ \\angle CRA = 180^\\circ - \\angle ARP = 180^\\circ - \\angle AQP = \\angle DQA = \\angle DCA = \\angle CBA, $$\n故 $A, B, C, R$ 四點共一圓, 記該圓為 $\\gamma$。\n\n\n... | Taiwan | 2022 數學奧林匹亞競賽第一階段選訓營, 獨立研究 (二) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals"
] | Chinese; English | proof only | null | |
0d75 | Let $ABC$ be a triangle with $AB \neq AC$. The incircle of triangle $ABC$ is tangent to $BC$, $CA$, $AB$ at $D$, $E$, $F$, respectively. The perpendicular line from $D$ to $EF$ intersects $AB$ at $X$. The second intersection point of circumcircles of triangles $AEF$ and $ABC$ is $T$. Prove that $TX \perp TF$. | [
"Let $G$ be the intersection point of $DX$ and $EF$.\n\n\n\nConsider the inversion $I_{I}^{ID^{2}}$, then the circumcircle of $\\triangle ABC$ is sent to the nine-point circle of $\\triangle DEF$ and the line $EF$ is sent to the circumcircle of $\\triangle AEF$. Hence $I$, $T$, and $G$ are ... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane G... | English | proof only | null | |
09jd | Find the smallest positive integer $N$ with the following property:
For any degree five polynomial $P(x)$ with integer coefficients, there is $0 \le x \le 1$ such that $|P(x)| > \frac{1}{N}$. | [
"Answer: $N = 56$.\nFirst we prove $N = 56$ has the property. Suppose, on the contrary, that for a degree five polynomial $P(x)$ with integer coefficients, we have $|P(x)| \\le \\frac{1}{56}$ for all $0 \\le x \\le 1$.\nThen the nonnegative integers $|P(0)|, |P(1)|, 32|P(1/2)|$ are strictly smaller than 1, thus are... | Mongolia | Mongolian Mathematical Olympiad Round 3 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Prealgebra / Basic Algebra > Integers",
"N... | null | proof and answer | 56 | |
0f4h | Problem:
$ABCD$ is a convex quadrilateral. $K$ is the midpoint of $AB$ and $M$ is the midpoint of $CD$. $L$ lies on the side $BC$ and $N$ lies on the side $AD$. $KLMN$ is a rectangle. Show that its area is half that of $ABCD$. | [] | Soviet Union | 15th ASU | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0dt6 | Find all functions $f: \mathbb{Z} \to \mathbb{Z}$ such that $f(x+y)[(f(x)-f(y))^2+f(xy)] = f(x^3)+f(y^3)$. | [
"There are 4 solutions, namely\n$$\nf(x) = x, \\quad f(x) = 0, \\quad f(x) = 2, \\quad f(x) = \\begin{cases} 1 & \\text{if } x \\text{ odd} \\\\ 2 & \\text{if } x \\text{ even} \\end{cases}\n$$\nLet (1) be the original equation. The first 3 solutions can be easily verified. The last solution can be verified by cons... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Algebra > Algebraic Expressions > Functional Equations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | All solutions are: (1) f(x) = x for all integers x; (2) f(x) = 0 for all integers x; (3) f(x) = 2 for all integers x; (4) f(x) = 1 for odd integers x and f(x) = 2 for even integers x. | |
0cu6 | Let $ABCD$ be a convex quadrilateral with $\angle DAB = 90^\circ$. Let $M$ be the midpoint of $BC$. Given that $\angle ADC = \angle BAM$, prove that $\angle ADB = \angle CAM$. (E. Bakaev)
Дан выпуклый четырёхугольник $ABCD$, в котором $\angle DAB = 90^\circ$. Пусть $M$ — середина стороны $BC$. Оказалось, что $\angle A... | [
"Choose a point $K$ such that $A$ is the midpoint of $BK$ (see Fig. 2). Then the quadrilateral $AKDC$ is cyclic, due to\n$$\n\\angle BKC = \\angle BAM = \\angle ADC.\n$$\nHence $\\angle CAM = \\angle ACK = \\angle ADK = \\angle ADB$.\n\n\nПервое решение.\n\nНа продолжении отрезка $AB$ за точку $A$ отметим точку $K$... | Russia | Russian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneou... | English; Russian | proof only | null | |
08qv | Problem:
Let $ABC$ be an acute triangle such that $AH = HD$, where $H$ is the orthocenter of $ABC$ and $D \in BC$ is the foot of the altitude from the vertex $A$. Let $\ell$ denote the line through $H$ which is tangent to the circumcircle of the triangle $BHC$. Let $S$ and $T$ be the intersection points of $\ell$ with... | [
"Solution:\n\nIn order to prove that $SM$ and $TN$ are parallel, it suffices to prove that both of them are perpendicular to $ST$. Due to symmetry, we will provide a detailed proof of $SM \\perp ST$, whereas the proof of $TN \\perp ST$ is analogous. In this solution we will use the following notation: $\\angle BAC ... | JBMO | JBMO | [
"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 | |
043j | In a plane rectangular coordinate system $xOy$, given ellipse
$$\Gamma : \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \quad (a > b > 0),$$
find the range of the eccentricity of $\Gamma$. | [
"By symmetry, it is useful to set $A(a, 0)$, $B(0, b)$. Denote $c = |OF| = \\sqrt{a^2 - b^2}$ and the eccentricity $e = \\frac{c}{a}$.\nLet $|OP| = r$. By the properties of ellipse, we know that the range of $r$ is $[b, a]$.\nNote that $O$ is the midpoint of $PQ$, and thus\n$$\n\\overrightarrow{FP} \\cdot \\overrig... | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | [3/4, (-1 + sqrt(37))/6] | |
076i | For positive real numbers $a$, $b$, $c$, which of the following statements necessarily implies $a = b = c$:
(I) $a(b^3 + c^3) = b(c^3 + a^3) = c(a^3 + b^3)$,
(II) $a(a^3 + b^3) = b(b^3 + c^3) = c(c^3 + a^3)$?
Justify your answer. | [
"We show that (I) need not imply that $a = b = c$ whereas (II) always implies $a = b = c$.\n\nObserve that $a(b^3 + c^3) = b(c^3 + a^3)$ gives $c^3(a - b) = ab(a^2 - b^2)$. This gives either $a = b$ or $ab(a + b) = c^3$. Similarly, $b = c$ or $bc(b + c) = a^3$. If $a \\neq b$ and $b \\neq c$, we obtain\n$$\nab(a + ... | India | IND_National | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Only (II) necessarily implies a = b = c; (I) does not (e.g., a = b = ((1 + √5)/2) c gives (I) without a = b = c). | |
0bo7 | Let $A$ be a finite set of real numbers. We define the sets
$$
S = \{x + y \mid x, y \in A\}, \quad D = \{x - y \mid x, y \in A\}.
$$
Prove that $\mathrm{card}(A) \cdot \mathrm{card}(D) \leq (\mathrm{card}(S))^2$. | [
"We will define a one-to-one function $f : A \\times D \\to S \\times S$ in the following manner: for $a \\in A$ and $d \\in D$, let\n$$\nf(a, d) = (a + x_d, a + y_d),\n$$\nwhere $x_d, y_d \\in A$ are chosen such that $x_d - y_d = d$ and $x_d$ is maximal. Clearly, for a given $d$, $x_d$ and $y_d$ exist and are uniq... | Romania | 66th ROMANIAN MATHEMATICAL OLYMPIAD | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof only | null | |
03fx | Do there exist $2024$ non-zero reals $a_1, a_2, \dots, a_{2024}$, such that
$$
\sum_{i=1}^{2024} \left(a_i^2 + \frac{1}{a_i^2}\right) + 2 \sum_{i=1}^{2024} \frac{a_i}{a_{i+1}} + 2024 = 2 \sum_{i=1}^{2024} \left(a_i + \frac{1}{a_i}\right)?
$$ | [
"Suppose that such $a_i$ exist. The condition rewrites as $\\sum_{i=1}^{2024} \\left(a_i + \\frac{1}{a_{i+1}} - 1\\right)^2 = 0$, which means that $a_i + \\frac{1}{a_{i+1}} = 1$ for all $i = 1, 2, \\dots, 2024$. It's easy to obtain consecutively that for any $i = 1, 2, \\dots, 2024$, $a_{i+1} = \\frac{1}{1-a_i}$, $... | Bulgaria | Bulgarian National Olympiad - Final Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | No | |
0in7 | An *animal* with $n$ cells is a connected figure consisting of $n$ equal-sized square cells. Animals are also called *polyominoes*. They can be defined inductively. Two cells are *adjacent* if they share a complete edge. A single cell is an animal, and given an animal with $n$-cells, one with $n+1$ cells is obtained by... | [
"**First Solution:** Let $s$ denote the minimum number of cells in a dinosaur; the number this year is $s = 2007$.\n\n**Claim:** The maximum number of cells in a primitive dinosaur is $4(s - 1) + 1 = 8025$.\n\nFirst, a primitive dinosaur can contain up to $4(s - 1) + 1$ cells. To see this, consider a dinosaur in th... | United States | USAMO | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof and answer | 8025 | |
0660 | In the Cartesian plane $Oxy$ we consider the points $A_1(40,1)$, $A_2(40,2)$, ..., $A_{40}(40,40)$, as well as the line segments $OA_1, OA_2, ..., OA_{40}$. A point of the Cartesian plane will be called "good" if its coordinates are integers and lies in the interior of a line segment $OA_i$, $i = 1, 2, 3, ..., 40$. Als... | [
"\nFigure 2\n\nA point $M(k,l)$ will belong in the interior of the line segment $OA_i$, if and only if $\\overrightarrow{OM}$ and $\\overrightarrow{OA_i}$ have the same slope (with $k,l$ integers and $0 < k \\le 40$), that is\n$$\n\\frac{i}{40} = \\frac{l}{k} \\quad (\\text{with } k,l \\tex... | Greece | 28th Hellenic Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > φ (Euler's totient)",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | English | proof and answer | 140 good points and 24 good segments | |
0djc | Let $ABC$ be a non-isosceles triangle with altitudes $AD$, $BE$, $CF$ concurrent at $H$. Denote $K$, $M$ as the midpoints of $AH$, $BC$ and $BK$ meets $AM$ at $L$. On $AC$, take $I$ such that $OI \parallel BC$ and $DE$ meets $IK$ at $S$. Prove that $L$, $K$, $S$, $E$ are concyclic. | [] | Saudi Arabia | SAUDI ARABIAN IMO Booklet 2023 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof only | null | |
0cbu | Find all the triples $(a, b, c)$ of integer numbers, fulfilling the relations
$$
a^2 + a = b + c, \quad b^2 + b = a + c, \quad c^2 + c = a + b.
$$ | [
"Adding the relations yields $a^2 + b^2 + c^2 = a + b + c$. (*)\n\nNotice that, if $x$ is an integer, then $x^2 \\ge x$, with equality only if $x = 0$ or $x = 1$.\nIndeed, if $x \\le 0$, then $x^2 \\ge 0 \\ge x$, and if $x \\ge 1$, then $x^2 = x \\cdot x \\ge x \\cdot 1 = x$.\nTherefore, the equality (*) is possibl... | Romania | THE 73rd ROMANIAN MATHEMATICAL OLYMPIAD - FINAL ROUND | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (0, 0, 0) and (1, 1, 1) | |
0knf | Problem:
Let $f$ be a monic cubic polynomial satisfying $f(x)+f(-x)=0$ for all real numbers $x$. For all real numbers $y$, define $g(y)$ to be the number of distinct real solutions $x$ to the equation $f(f(x))=y$. Suppose that the set of possible values of $g(y)$ over all real numbers $y$ is exactly $\{1,5,9\}$. Comput... | [
"Solution:\nWe claim that we must have $f(x)=x^{3}-3 x$.\n\nFirst, note that the condition $f(x)+f(-x)=0$ implies that $f$ is odd. Combined with $f$ being monic, we know that $f(x)=x^{3}+a x$ for some real number $a$. Note that $a$ must be negative; otherwise $f(x)$ and $f(f(x))$ would both be increasing and $1$ wo... | United States | HMMT Spring 2021 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | 970 | |
089a | Problem:
Si sa che $p(x)$ è un polinomio monico di grado $5$. Inoltre, si sa che le soluzioni dell'equazione $p(x)=0$ sono esattamente $x=0, 1, 2, 4$. Determinare il massimo valore che può assumere il coefficiente del termine di primo grado.
Nota: un polinomio è monico se il coefficiente del suo termine di grado più ... | [
"Solution:\n\nLa risposta è $(\\mathbf{C})$. Il polinomio si può scrivere come prodotto di cinque fattori di primo grado: $x(x-1)(x-2)(x-4)(x-k)$, dove $k \\in \\{0,1,2,4\\}$. Il termine di primo grado ha coefficiente $(-1)(-2)(-4)(-k)=8k$, che è massimo per $k=4$ e vale in tal caso $32$."
] | Italy | Italian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | MCQ | C | |
0e2y | Problem:
Poenostavi izraz
$$
(x-1)\left(\left(x^{-3}+x^{-2}+x^{-1}\right)\left(x^{-1}-1\right)-\left(x^{-2}-1\right)\left(x^{-2}+1\right)\right)^{-1}
$$ | [
"Solution:\n\nPotence z negativnim eksponentom v oklepajih zapišemo z ulomki\n$$(x-1)\\left(\\left(\\frac{1}{x^{3}}+\\frac{1}{x^{2}}+\\frac{1}{x}\\right)\\left(\\frac{1}{x}-1\\right)-\\left(\\frac{1}{x^{2}}-1\\right)\\left(\\frac{1}{x^{2}}+1\\right)\\right)^{-1}.$$\n\nVsote oziroma razlike razširimo na skupni imeno... | Slovenia | 10. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol, Državno tekmovanje | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | x | |
04ac | Matija and Tomislav play the following game:
Each of them throws a pair of dice. If at least one of them gets numbers whose sum is divisible by $3$, then Matija wins; otherwise, Tomislav wins.
What is the probability that Matija wins? | [] | Croatia | Hrvatska 2011 | [
"Statistics > Probability > Counting Methods > Other"
] | English | proof and answer | 5/9 | |
0l73 | Problem:
The number
$$
\frac{9^{9} - 8^{8}}{1001}
$$
is an integer. Compute the sum of its prime factors. | [
"Solution:\n\nObserve\n$$\n9^{9} - 8^{8} = 27^{6} - 16^{6}\n$$\n$$\n= (27^{2} - 16^{2})(27^{4} + 27^{2} \\cdot 16^{2} + 16^{4})\n$$\n$$\n= (27 - 16)(27 + 16)(27^{2} - 27 \\cdot 16 + 16^{2})(27^{2} + 27 \\cdot 16 + 16^{2})\n$$\n$$\n= 11 \\cdot 43 \\cdot 553 \\cdot 1417.\n$$\nThe remaining factorizations are motivate... | United States | HMMT February | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | final answer only | 231 | |
0gcx | A 和 B 在一個直角座標平面下棋。一開始 A 先選一個位置 $(x_0, y_0)$ 下第一顆棋。接著從 B 開始, A 和 B 輪流放棋子:
i) 如果 A 上一回合下的棋子位於座標 $(x, y)$, 那接下來 B 只能把棋下在座標 $(x + 2, y + 1)$, $(x + 2, y - 1)$, $(x - 2, y + 1)$, $(x - 2, y - 1)$ 中的其中一個。
ii) 如果 B 上一回合下的棋子位於座標 $(x, y)$, 那接下來 A 只能把棋下在座標 $(x + 1, y + 2)$, $(x + 1, y - 2)$, $(x - 1, y + 2)$, $(x - 1, y - 2)$ ... | [
"$n = 2018$ 時, B 有必勝策略; $n = 2019$ 時, A 有必勝策略。\n\n首先, 考慮一個 $n \\times n$ 的棋盤, 棋盤的上方和下方連通, 左方和右方連通。由於對所有的 $1 \\le a, b \\le n$, $\\{(nx + a, ny + b) \\mid x, y \\in \\mathbb{Z}\\}$ 中只能下一顆棋子, 因此我們可以用這樣的棋盤取代座標平面, 棋盤中的方格 $(a, b)$ 代表座標平面中的點集 $\\{(nx + a, ny + b) \\mid x, y \\in \\mathbb{Z}\\}$。\n\n(約定棋盤的列由下至上編號為 $1$ 到 $... | Taiwan | 二〇一九數學奧林匹亞競賽第一階段選訓營 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | For n equal to 2018, player B has a winning strategy; for n equal to 2019, player A has a winning strategy. | |
00vt | Let $I$ be the incenter of triangle $\triangle ABC$, and $D$ be an arbitrary point on arc $\widehat{AB}$, not containing point $C$, of the circumcircle of $\triangle ABC$. Let $I_A$ and $I_B$ be the incenters of $\triangle BCD$ and $\triangle ACD$ respectively. Lines $DI_A$ and $II_B$ intersect at point $X_A$, and line... | [
"Let $M_A$ and $M_B$ be the midpoints of arcs $\\widehat{BC}$ and $\\widehat{CA}$ (not containing points $A$ and $B$ respectively), and $S_C$ be the midpoint of arc $\\widehat{ACB}$ of the circumcircle of $\\triangle ABC$. We will prove that lines $X_A X_B$ and $I_A I_B$ intersect on $CS_C$.\nBy the Incenter/Excent... | Balkan Mathematical Olympiad | 42nd Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Concurrency and Collinearity > Desargues theorem",
"Geometry > Plane Geom... | English | proof only | null | |
00d6 | Alrededor de una circunferencia están escritos 20 números enteros positivos distintos. Alex divide cada número por el número vecino, recorriendo la circunferencia en el sentido de las agujas del reloj, y anota los restos que obtiene en cada caso. Teo divide cada número por el número vecino, recorriendo la circunferenci... | [
"Numeramos los 20 números en sentido horario como $a_1, a_2, \\dots, a_{20}$ de modo que $a_{20}$ sea el menor de los 20 números escritos en la circunferencia. Sea $r_{20}$ el resto de dividir $a_{20}$ por $a_1$, es decir, $a_{20} = a_1 \\cdot q + r_{20}$. Como $a_{20}$ es el menor de todos los números escritos, va... | Argentina | Nacional OMA | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization"
] | Spanish | proof and answer | 20 | |
0k31 | Problem:
Four cars $A$, $B$, $C$, and $D$ travel at constant speeds on the same road (not necessarily in the same direction). Car $A$ passed $B$ and $C$ at 8am and 9am, respectively, and met $D$ at 10am. Car $D$ met $B$ and $C$ at $12 \mathrm{pm}$ and $2 \mathrm{pm}$, respectively. Determine at what time $B$ passed $C... | [
"Solution:\n\nDraw lines $A$, $B$, $C$, $D$ to graph the movement of the four cars, with time on the $x$-axis and distance on the $y$-axis, and let $(X Y)$ be the point where lines $X$ and $Y$ meet. Then $(A C)$ is the midpoint of the line from $(A B)$ to $(A D)$, and $(D B)$ is the midpoint of the line from $(D A)... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates"
] | null | proof and answer | 10:40 am | |
0dzf | Problem:
Dane so dolžine stranic trikotnika $ABC$: $|AB| = 15~\mathrm{cm}$, $|BC| = 14~\mathrm{cm}$ in $|CA| = 13~\mathrm{cm}$. Naj bo $D$ nožišče višine na stranico $BC$, $E$ pa taka točka na tej višini, da je $\angle BAD = \angle DEC$. Presečišče premic $AB$ in $CE$ označimo s $F$. Izračunaj $|EF|$. | [
"Solution:\n\nNajprej izračunajmo dolžini $|AD|$ in $|CD|$. Označimo $|AD| = v$ in $|CD| = x$. Po Pitagorovem izreku je $v^{2} = |AC|^{2} - x^{2} = |AB|^{2} - (|BC| - x)^{2}$, od koder dobimo $13^{2} - x^{2} = 15^{2} - (14 - x)^{2}$ oziroma $13^{2} = 15^{2} - 14^{2} + 2 \\cdot 14 \\cdot x$. Torej je $x = 5$ in pote... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 10/3 | |
0dza | Problem:
Za katera naravna števila $n > 1$ doseže izraz
$$
\frac{\log_{10} 2 \cdot \log_{10} 3 \cdots \log_{10} n}{10^{n-1}}
$$
najmanjšo vrednost? Kolikšna je ta vrednost? | [
"Solution:\nPrimerjajmo izraza $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10}(n-1)}{10^{n-2}}$ in $\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{10^{n-1}}$. Neenakost\n$$\n\\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10}(n-1)}{10^{n-2}} \\geq \\frac{\\log_{10} 2 \\log_{10} 3 \\cdots \\log_{10} n}{... | Slovenia | 52. matematično tekmovanje srednješolcev Slovenije | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | The minimum occurs at n = 10^10 − 1 and n = 10^10. The minimal value is (log_{10} 2 · log_{10} 3 · … · log_{10}(10^{10})) / 10^{10^{10} − 1}. | |
0j6i | Problem:
Let $n$ be an integer greater than $3$. Let $R$ be the set of lattice points $(x, y)$ such that $0 \leq x, y \leq n$ and $|x-y| \leq 3$. Let $A_{n}$ be the number of paths from $(0,0)$ to $(n, n)$ that consist only of steps of the form $(x, y) \rightarrow (x, y+1)$ and $(x, y) \rightarrow (x+1, y)$ and are co... | [
"Solution:\n\nAnswer: $2+\\sqrt{2}$\n\nWe first find $A_{n}$ in terms of $n$. Let $a_{n}$ be the number of ways to get to the point $(n, n+3)$, and let $b_{n}$ be the number of ways to get to the point $(n+1, n+2)$. By symmetry, $a_{n}$ is also the number of ways to get to $(n+3, n)$ and $b_{n}$ is also the number ... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | null | proof and answer | 2+√2 | |
0j3f | Problem:
How many polynomials of degree exactly $5$ with real coefficients send the set $\{1,2,3,4,5,6\}$ to a permutation of itself? | [
"Solution:\n\nAnswer: $714$\n\nFor every permutation $\\sigma$ of $\\{1,2,3,4,5,6\\}$, Lagrange Interpolation gives a polynomial of degree at most $5$ with $p(x) = \\sigma(x)$ for every $x = 1,2,3,4,5,6$. Additionally, this polynomial is unique: assume that there exist two polynomials $p, q$ of degree $\\leq 5$ suc... | United States | 13th Annual Harvard-MIT Mathematics Tournament | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 714 | |
0hp3 | Problem:
Determine, with proof, whether the following statement is true or false: Out of any six natural numbers, one can find either three which are pairwise relatively prime or three whose greatest common divisor is greater than $1$. | [
"Solution:\nThe statement is false. To avoid having three pairwise relatively prime numbers, we should give each pair of numbers a prime factor in common; to avoid having three numbers with a nontrivial gcd, we should not use any factor to cover more than two numbers. Utilizing the first fifteen primes for the $\\b... | United States | Berkeley Math Circle Monthly Contest 4 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | false | |
0dai | Find all functions $f: \mathbb{Z} \rightarrow \mathbb{Z}$ such that
$$
x f(2 f(y)-x)+y^{2} f(2 x-f(y))=\frac{(f(x))^{2}}{x}+f(y f(y)),
$$
for all $x, y \in \mathbb{Z}, x \neq 0$. | [
"Let $a=f(0)$ and denote ($*$) as the given condition. Assume that $a \\neq 0$, substituting $x=2a$ and $y=0$ in ($*$), we get\n$$\n2a^{2}=\\frac{(f(2a))^{2}}{2a}+a\n$$\nor\n$$\n4a-2=\\left(\\frac{f(2a)}{2a}\\right)^{2}.\n$$\nThis follows that $4a-2$ is a perfect square which is a contradiction since there is no pe... | Saudi Arabia | Team selection tests for BMO 2018 | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Modular Arithmetic",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English | proof and answer | f(x) = 0 for all x ∈ ℤ; f(x) = x^2 for all x ∈ ℤ | |
0aeb | Определи ги аглите на триаголникот чии страни $a$, $b$ и $c$ го задоволуваат равенството
$$
a - 4\sqrt{bc} + 2b = 2\sqrt{ac} - 3c.
$$ | [
"Равенството ќе го запишеме во облик\n$$\na - 2\\sqrt{ac} + c + 2c - 4\\sqrt{bc} + 2b = 0, \\\\\n(\\sqrt{a})^2 - 2\\sqrt{a}\\sqrt{c} + (\\sqrt{c})^2 + 2[(\\sqrt{b})^2 - 2\\sqrt{b}\\sqrt{c} + (\\sqrt{c})^2] = 0, \\\\\n(\\sqrt{a} - \\sqrt{c})^2 + 2(\\sqrt{b} - \\sqrt{c})^2 = 0.\n$$\nЗбир на два ненегативни броја е ну... | North Macedonia | Регионален натпревар по математика за основно образование | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | Macedonian, English | proof and answer | Each angle is 60 degrees. | |
0cse | The sides $AD$ and $BC$ of a quadrilateral $ABCD$ are parallel. Assume that the angle bisectors of the angles $DAC$, $DBC$, $ACB$, and $ADB$ form a rhombus. Prove that $AB = CD$. (L. Emelyanov) | [
"Let $O$ be the intersection point of the diagonals $AC$ and $BD$ (see the figure). The bisectors of the angles $ADB$ and $DAC$ intersect at the center $O_1$ of the incircle of triangle $AOD$, and the bisectors of the angles $ACB$ and $DBC$ intersect at the center $O_2$ of the incircle of triangle $BOC$. Thus, the ... | Russia | XL Russian mathematical olympiad | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle... | null | proof only | null | |
0dta | Colour a $20000 \times 20000$ square grid using $2000$ different colours with $1$ colour in each square. Two squares are neighbours if they share a vertex. A path is a sequence of squares so that $2$ successive squares are neighbours. Mark $k$ of the squares. For each unmarked square $x$, there is exactly $1$ marked sq... | [
"The maximum value is $k = 20000$. Define a region as a maximal set of same-colour squares such that there is a path of the same colour between any $2$ squares in the region. From the first condition, there is exactly one marked square in a region and so there are $k$ regions. Two regions are neighbours if they hav... | Singapore | Singapore Mathematical Olympiad (SMO) | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 20000 | |
04wf | A line segment $BC$ is given in the plane. Consider all acute-angled triangles $ABC$ with $|\angle BAC| = 45^\circ$. In each such triangle, denote by $D$ and $E$ those points on the sides $AB$ and $AC$, respectively, such that $BC$ is a common tangent of the circumcircles of triangles $ACD$ and $ABE$. Finally, denote b... | [] | Czech Republic | District Round | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates, barycentric coordinates",
"Geometry > Plane Geometry > Transformations > Spiral similarity",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geo... | English | proof only | null | |
03e8 | Solve in positive integers the equation
$$
m^{\frac{1}{n}} + n^{\frac{1}{m}} = 2 + \frac{2}{mn(m+n)^{\frac{1}{m}+\frac{1}{n}}}.
$$ | [
"There are no solutions in positive integers. Without loss of generality, assume that $m \\ge n$. If $n \\ge 3$, then\n$$\n\\left(1 + \\frac{1}{m}\\right)^m = \\sum_{k=0}^{m} \\binom{m}{k} \\frac{1}{m^k} < \\sum_{k=0}^{m} \\frac{1}{k!} < \\sum_{k=0}^{\\infty} \\frac{1}{k!} = e < n \\implies n^{\\frac{1}{m}} > 1 + \... | Bulgaria | Autumn tournament | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Exponential functions"
] | English | proof and answer | No positive integer solutions | |
0fhz | Problem:
Recortamos varios círculos de papel (no necesariamente iguales) y los extendemos sobre una mesa de modo que haya algunos solapados (con parte interior común), pero de tal forma que no haya ningún círculo dentro de otro.
Probar que es imposible ensamblar las piezas que resultan de recortar las partes no solap... | [
"Solution:\n\nLa frontera de las piezas recortadas (que no sean círculos completos) está formada por arcos cóncavos y convexos (vistos desde fuera) que se cortan en puntos que llamaremos vértices. En un vértice pueden concurrir dos arcos cóncavos o uno cóncavo y otro convexo, pero nunca dos convexos ya que éstos ún... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Combinatorial Geometry"
] | null | proof only | null | |
0ieq | Problem:
Suppose $S$ tiles the natural numbers $\mathbf{N}$. Show that $S$ tiles the set $\{1,2, \ldots, k\}$ for some positive integer $k$. | [
"Solution:\nUsing the notation from above, we can find $l_{1}<l_{2}$ such that $T_{l_{1}} \\sim T_{l_{2}}$. By the same argument as in problem 12, as long as $T_{l_{1}} \\neq \\mathbf{N}$, there is a unique choice for $S_{l_{1}-1}$ that contains the largest integer not in $T_{l_{1}}$. Since the same can be said for... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
058q | In a math period, the teacher asks pupils to solve quadratic equations of the form $x^2 + px + q = 0$ where $p$ and $q$ are some integers. The teacher obtains every new equation by either increasing by 1 or decreasing by 1 the value of either $p$ or $q$ in the equation just solved. In the initial equation, $p = 2020$ a... | [
"In the first equation, one has $p - q = 10$, while in the last equation, one has $p - q = -10$. At each step, either $p$ or $q$ changes exactly by 1, whence also $p - q$ changes exactly by 1. Thus at some step one must have an equation where $p - q = 1$, or equivalently, $p = q + 1$. The equation $x^2 + (q + 1)x +... | Estonia | Estonian Math Competitions | [
"Algebra > Intermediate Algebra > Quadratic functions",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Yes | |
01xg | The sequence $a_1, a_2, a_3, \dots$ of positive integers is defined in the following way: $a_1$ is given, and for each $n \ge 2$ the number $a_n$ is the smallest positive integer divisible by $n$, which is not less than $a_{n-1}$. (For example, if $a_5 = 115$, then $a_6 = 120$, $a_7 = 126$, $a_8 = 128$.)
Prove that fro... | [] | Belarus | 69th Belarusian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English | proof only | null | |
09ae | Let $ABCD$ be a tangential quadrilateral. Let $\omega$ be externally inscribed circle in $ABCD$, tangent to $AD$, $BC$ and $AB$. Denote by $X_{AB}$ the point of tangency of $\omega$ and the circle that passes through $A$ and $B$ and internally tangent to $\omega$. Let us define $X_{BC}$, $X_{CD}$ and $X_{DA}$, analogou... | [
"Let $I$ be center of incircle of $ABCD$, and $AB$ touches incircle at $Q$ and $\\omega$ at $P$. It will be sufficient to show that $X_{AB}$, $P$, $I$ are collinear and $\\angle BX_{AB}P = \\angle AX_{AB}P$ (all passes through $I$). Let $(IP) \\cap \\omega = X$, $PP'$ and $QQ'$ are diameters and $(PQ') \\cap \\omeg... | Mongolia | 46th Mongolian Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Advanced Configurations > Polar triangles, harmonic conjugates",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0efd | Problem:
Krogu očrtamo in včrtamo enakostranični trikotnik. Koliko je razmerje dolžin stranic krogu očrtanega in včrtanega trikotnika?
(A) $1: 3$
(B) $2: 1$
(C) $4: 1$
(D) $5: 2$
(E) $\sqrt{3}: 1$ | [
"Solution:\n\nNaj bo $v$ dolžina višine krogu očrtanega enakostraničnega trikotnika, $v_1$ pa dolžina višine krogu včrtanega enakostraničnega trikotnika in $r$ polmer kroga. Za očrtani trikotnik velja $\\frac{1}{3} v = r$ in za včrtani trikotnik velja $\\frac{2}{3} v_1 = r$. Iz obeh enakosti sledi, da je razmerje v... | Slovenia | 17. tekmovanje dijakov srednjih tehniških in strokovnih šol v znanju matematike | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents"
] | null | MCQ | (B) | |
06e7 | Find, with reasons, all functions $f : \mathbb{N} \to \mathbb{N}$ such that
(i) $f(n) = 1$ if and only if $n = 1$,
(ii) if $d = \text{gcd}(n, m)$, then $f(nm) = \frac{f(n)f(m)}{f(d)}$, and
(iii) $f^{2004}(n) = n$ for every $n \in \mathbb{N}$, where $f^{2004}(n) = \underbrace{f(f(\cdots f(n)))}_{2004}$,
or show that the... | [
"There is no such function.\nPutting $n = m = 2$ in (ii), we obtain\n$$\nf(4) = \\frac{f(2)f(2)}{f(2)} = f(2).\n$$\nThen by condition (iii), we have\n$$\n2 = f^{2004}(2) = f^{2003}(f(2)) = f^{2003}(f(4)) = f^{2004}(4) = 4.\n$$\nThis is a contradiction. So there is no such function."
] | Hong Kong | IMO HK TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof only | null | |
0fnf | Por los puntos medios de dos lados de un triángulo $ABC$ trazamos las medianas y unimos los puntos que trisecan el tercer lado con el vértice opuesto. Así, en el interior, se obtiene una pajarita (dos triángulos unidos por un vértice). Se pide calcular la fracción de superficie total del triángulo que representa la paj... | [
"Supongamos, sin pérdida de generalidad, que el área del triángulo $ABC$ es uno. Es decir, $[ABC] = 1$. Tenemos\n\n\n\n\nLas medianas dividen al triángulo en seis partes de igual área, luego: $[AMF] = [FNC] = \\frac{1}{6}$ y $[AFC] = 2 \\cdot \\frac{1}{6} = \\frac{1... | Spain | Olimpiada Matemática Española | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Miscellaneous > Distan... | Spanish | proof and answer | 1/30 | |
0jz1 | Problem:
An equiangular hexagon has side lengths $1, 1, a, 1, 1, a$ in that order. Given that there exists a circle that intersects the hexagon at 12 distinct points, we have $M < a < N$ for some real numbers $M$ and $N$. Determine the minimum possible value of the ratio $\frac{N}{M}$. | [
"Solution:\n\nWe claim that the greatest possible value of $M$ is $\\sqrt{3}-1$, whereas the least possible value of $N$ is $3$.\n\nTo begin, note that the condition requires the circle to intersect each side of the hexagon at two points on its interior. This implies that the center must be inside the hexagon as it... | United States | HMMT November 2017 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Geometric Inequalit... | null | proof and answer | (3*sqrt(3)+3)/2 | |
0g8p | 設點 $M$ 為三角形 $ABC$ 的外接圓上一點。自 $M$ 點引對三角形 $ABC$ 的內切圓相切的 (兩條) 直線, 分別交 $BC$ 於 $X_1, X_2$ 點。證明三角形 $MX_1X_2$ 的外接圓與 $ABC$ 的外接圓的第二個交點 (即不同於 $M$ 的那個交點) 就是 $ABC$ 的外接圓與角 $A$ 內的偽內切圓的切點。
(註: 角 $A$ 內的**偽內切圓**指的是與邊 $AB, AC$ 皆相切, 並且內切於 $ABC$ 的外接圓的圓。)
Let $M$ be an arbitrary point on the circumcircle of triangle $ABC$ and let the tange... | [
"Assume without loss of generality that $M$ lies on the same side of line $BC$ as the vertex $A$. In this case, we denote by $(I)$ the common incircle of triangles $ABC$, and $MX_1X_2$ with radius $r$, and let $D, E, F, Y_1, Y_2$ be the tangency points of $(I)$ with the sidelines $BC, CA, AB, MX_1, MX_2$.\n\nConsid... | Taiwan | 2014 IMO Training Camp Stage 3 Mock Competition (1) | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Transformations > Inversion",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal... | null | proof only | null | |
05zl | Problem:
Pour un réel $x$, on note $\lfloor x\rfloor$ le plus grand entier relatif inférieur ou égal à $x$ (par exemple, $\lfloor 2,7\rfloor=2$, $\lfloor\pi\rfloor=3$ et $\lfloor-1,5\rfloor=-2$). On note aussi $\{x\}=x-\lfloor x\rfloor$.
Trouver tous les réels $x$ tels que
$$
\lfloor x\rfloor\{x\}<x-1
$$ | [
"Solution:\n\nOn réécrit $x=\\lfloor x\\rfloor+\\{x\\}$ dans le terme de droite et on met tous les termes à gauche pour que l'inégalité devienne\n$$\n\\lfloor x\\rfloor\\{x\\}-\\lfloor x\\rfloor-\\{x\\}+1<0\n$$\nou encore\n$$\n(\\lfloor x\\rfloor-1)(\\{x\\}-1)<0\n$$\nRemarquons que pour tout $x$ réel, on a $0 \\leq... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES - ENVOI 2 : AlgèBre | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | x ≥ 2 | |
0fxt | Problem:
Sei $P$ die Menge der ersten 2009 Primzahlen und sei $X$ die Menge aller natürlichen Zahlen, welche nur Primfaktoren aus $P$ besitzen. Bestimme alle natürlichen Zahlen $k$, für die eine Funktion $f: X \rightarrow X$ existiert, welche für alle $m, n \in X$ die folgende Gleichung erfüllt:
$$
f(m f(n))=f(m) n^{k... | [
"Solution:\n\nEs existiert genau dann eine solche Funktion, wenn $k$ eine Quadratzahl ist.\n\nSei zuerst $k=l^{2}$ ein Quadrat, dann hat die Funktion $f(n)=n^{l}$ die gewünschte Eigenschaft, denn beide Seiten der Gleichung lauten $m^{l} n^{k}$.\n\nUmgekehrt existiere so ein $f$. Sei $a=f(1) \\in X$. Setze $n=1$, da... | Switzerland | IMO Selektion | [
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Abstract Algebra > Permutations / basic group theory"
] | null | proof and answer | k is a perfect square | |
0hfr | Prime number $p > 2$ and the polynomial $Q$ with integer coefficients are such that there do not exist two positive integers $i, j$, such that $i < j < p$ and the number $(Q(j) - Q(i))(jQ(j) - iQ(i))$ is divisible by $p$. What's the smallest possible degree of $Q$? | [
"Example: $Q(x) = x^{p-2} - 1$. For any $i < j < p$:\n$$\nQ(j) - Q(i) = y^{p-2} - x^{p-2} \\equiv \\frac{1}{y} - \\frac{1}{x} \\pmod{p},\n$$\nwhich clearly isn't divisible by $p$, and\n$$\njQ(j) - iQ(i) = (y^{p-1} - x^{p-1}) - (y - x) \\equiv (x - y) \\pmod{p}\n$$\nalso isn't divisible by $p$.\n\nSuppose that there... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | English | proof and answer | p-2 | |
04zn | Does there exist a positive real number $C$ such that the inequality
$$
x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4 \le C(x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1)
$$
holds for arbitrary positive real numbers $x_1, x_2, x_3, x_4$? | [
"For simplicity, denote $A = x_1x_2 + x_1x_3 + x_1x_4 + x_2x_3 + x_2x_4 + x_3x_4$ and $B = x_1x_2 + x_2x_3 + x_3x_4 + x_4x_1$.\n\nChoose $x_1 = x_3 = u$ and $x_2 = x_4 = 1$. Then $A \\ge x_1x_3 = u^2$ and $B = 4u$. Hence $\\frac{A}{B} \\ge \\frac{u}{4}$. As $u$ can be arbitrarily large, no constant $C$ such that $A... | Estonia | Selected Problems from the Final Round of National Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof only | null | |
0kaw | A *snake of length* $k$ is an animal which occupies an ordered $k$-tuple $(s_1, \dots, s_k)$ of cells in an $n \times n$ grid of square unit cells. These cells must be pairwise distinct, and $s_i$ and $s_{i+1}$ must share a side for $i = 1, \dots, k-1$. If the snake is currently occupying $(s_1, \dots, s_k)$ and $s$ is... | [
"The following solution is due to Brian Lawrence. For illustration reasons, we give below a figure of a snake of length 89 turning around in an $11 \\times 11$ square (which generalizes readily to odd $n$). We will see that a snake of length $(n-1)(n-2)-1$ can turn around in an $n \\times n$ square, so this certain... | United States | USA IMO TST | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | Yes | |
0h7t | Given a stripe $1 \times n$, $n \ge 4$. One wrote a positive integer in each cell of it (the numbers are not obliged to be equal). After under each number one writes a positive integer which is equal to the quantity of such integers in the previous row. (i.e. if $10$ was written in the previous row three times then und... | [
"Obviously, it is possible to rearrange the numbers in each row so that they increase. Denote the initial row as $0^{\\text{th}}$, the next - $1^{\\text{st}}$ and so on.\n\nLet us describe the properties of the numbers inside rows beginning from the $1^{\\text{st}}$ one.\n\n**Lemma 1.** If $n$ occur in $k$ th row, ... | Ukraine | UkraineMO | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | For n = 2016, the maximum number of steps before stabilization is 11. In general, the maximum is floor(log2 n) + 1. | |
0eqw | There are twice as many girls as boys at a school. If $30\%$ of the girls and $45\%$ of the boys have already completed their holiday project, what percentage of the learners still needs to complete their project? | [
"For convenience assume there are $300$ learners, i.e. $200$ girls and $100$ boys.\n\n$30\\%$ of $200$ is $60$ and $45\\%$ of $100$ is $45$.\n\nThus, $60 + 45 = 105$ learners have completed their project.\n\nThe percentage of learners who still need to complete their project is thus\n$$\n\\frac{300 - 105}{300} \\ti... | South Africa | South African Mathematics Olympiad Second Round | [
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | English | final answer only | 65% | |
03fk | Given the triangle $ABC$ and $M$ - midpoint of $AB$. Given the angles $\angle ABC = 30^\circ$ and $\angle BCM = 105^\circ$. Prove that $CM \cdot AC = BM \cdot BC$.
(Konstantin Delchev) | [
"Let $AA_1$ be the height from point $A$ in $\\triangle ABC$. Note that it lies on the continuation of $BC$, because this triangle is obtuse. The triangle $\\triangle AA_1B$ is right-angled with angle $30^\\circ$. Therefore, $AA_1 = AM = A_1M = BM = x$. Consequently, we directly find $\\angle CMB = 45^\\circ$, $\\a... | Bulgaria | Bulgarian Spring Tournament | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | English | proof only | null | |
0gh1 | 考慮兩互質的正整數 $p, q$，並且對於 $i = 1, 2, \dots, q-1$，定義 $f(i)$ 為 $p \cdot i$ 除以 $q$ 的餘數。若 $f(i)$ 為 $f(1), f(2), \dots, f(i)$ 之間的最大數，則稱 $i$ 為大數；但若 $f(i)$ 為 $f(1), f(2), \dots, f(i)$ 之間的最小數，則稱 $i$ 為小數。注意到 1 同時為大數及小數。
給定兩個正整數 $a, b$。已知在 $1, 2, \dots, q-1$ 之中有 $a$ 個大數及 $b$ 個小數，試求 $q$ 的最小可能值。 | [
"The least possible number for $q$ is $ab + 1$.\nLet $S_{p,q}$ be the collection of small numbers, while $T_{p,q}$ be the collection of big numbers. Write out $S_{p,q} \\cup T_{p,q}$ in increasing order, and we get a sequence $c_1, \\dots, c_{a+b-1}$ (note that clearly $S_{p,q} \\cap T_{p,q} = \\{1\\}$). We claim t... | Taiwan | 2022 數學奧林匹亞競賽第二階段選訓營, 獨立研究 (二) | [
"Number Theory > Modular Arithmetic > Inverses mod n",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | Chinese; English | proof and answer | ab + 1 | |
0anu | Problem:
Which of the following quadratic expressions in $x$ have roots $\frac{g}{h}$ and $-\frac{h}{g}$?
(a) $g^{2} h^{2} x^{2}-\frac{g^{2}}{h^{2}}$
(b) $h g x^{2}+\left(g^{2}-h^{2}\right) x-h g$
(c) $h g x^{2}+\left(h^{2}-g^{2}\right) x+h g$
(d) $h g x^{2}+\left(h^{2}-g^{2}\right) x-h g$ | [] | Philippines | QUALIFYING STAGE | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas"
] | null | MCQ | d | |
0c8o | Let $n$ be a positive integer. Show that there are functions $f : \mathbb{R} \to \mathbb{R}$ satisfying $f(f(x) + y f(y)) = x + (f(y))^n$, for any $x, y \in \mathbb{R}$, if and only if $n = 2$. Determine those functions.
Cătălin Zârnă | [
"For $y = 0$ we have $f(f(x)) = x + (f(0))^n$, for any $x \\in \\mathbb{R}$, implying that $f$ is bijective. Let $a$ such that $f(a) = 0$. For $y = a$ we deduce $f(f(x)) = x$, for all real $x$, so $f(0) = 0$.\n\nFor $x = 0$ we get $f(y f(y)) = f(y)^n$, for real $y$. Substituting $y$ with $f(y)$, we get $f(y)^n = y^... | Romania | Romanian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | Solutions exist only when n equals 2. In that case, the functions are f(x) = x for all real x or f(x) = −x for all real x. | |
065h | Let the triangle $ABC$ has barycenter $G$ and circumcenter $O$. The perpendicular bisectors of $GA$, $GB$ and $GC$ intersect at the points $A_1$, $B_1$, $C_1$. Prove that $O$ is the barycentre of the triangle $A_1B_1C_1$. | [
"Let $D$, $E$, $F$ be the middles of the sides $BC$, $AC$, $AB$, respectively.\n\nLet, also $B_1C_1$, $A_1C_1$, $A_1B_1$ be the perpendicular bisectors of the line segments $GA$, $GB$ and $GC$, respectively. Then the points $A_1$, $B_1$ and $C_1$ are the circumcenters of the triangles $GBC$... | Greece | SELECTION EXAMINATION | [
"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"
] | English | proof only | null | |
0843 | Problem:
Sia $P$ un punto interno ad un triangolo $ABC$. Le rette $AP$, $BP$ e $CP$ intersecano i lati di $ABC$ in $A'$, $B'$ e $C'$ rispettivamente. Ponendo
$$
x = \frac{AP}{PA'}, \quad y = \frac{BP}{PB'}, \quad z = \frac{CP}{PC'}
$$
dimostrare che $xyz = x + y + z + 2$.
 | [
"Solution:\n\nSi verifica facilmente che, comunque scelti tre numeri reali non nulli $a, b, c$, e ponendo $x = \\frac{a + b}{c}$, $y = \\frac{b + c}{a}$ e $z = \\frac{a + c}{b}$, la relazione da dimostrare diventa un'identità algebrica.\n\nPer trovare $a, b, c$ si proceda come segue: siano $D$ ed $E$ le intersezion... | Italy | Olimpiade Italiana di Matematica | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
06ot | A point $D$ is chosen on the side $AC$ of a triangle $ABC$ with $\angle C < \angle A < 90^{\circ}$ in such a way that $BD = BA$. The incircle of $ABC$ is tangent to $AB$ and $AC$ at points $K$ and $L$, respectively. Let $J$ be the incentre of triangle $BCD$. Prove that the line $KL$ intersects the line segment $AJ$ at ... | [
"Denote by $P$ the common point of $AJ$ and $KL$. Let the parallel to $KL$ through $J$ meet $AC$ at $M$. Then $P$ is the midpoint of $AJ$ if and only if $AM = 2 \\cdot AL$, which we are about to show.\n\n\n\nDenoting $\\angle BAC = 2\\alpha$, the equalities $BA = BD$ and $AK = AL$ imply $\\... | IMO | IMO 2006 Shortlisted Problems | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing... | English | proof only | null | |
022z | Problem:
Soma de potências de 2 - Determine um valor de $n$ para o qual o número $2^{8}+2^{11}+2^{n}$ seja um quadrado perfeito. | [
"Solution:\n\nSolução 1: Observe que $2^{8}+2^{11}+2^{n} = (2^{4})^{2} + 2 \\times 2^{4} \\times 2^{6} + \\left(2^{\\frac{n}{2}}\\right)^{2}$. Logo, para $n=12$, temos $2^{8}+2^{11}+2^{12} = (2^{4}+2^{6})^{2}$. Assim, $n=12$ é uma solução.\n\n\nSolução 2: Se $2^{8}+2^{11}+2^{n}=k^{2}$, então\n$$\n\\begin{aligned}\n... | Brazil | Nível 2 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 12 | |
04oz | Determine the value of the real parameter $p$ so that the solutions of
$$
(p - 3)x^2 + (p^2 + 1)x - 11p + 18 = 0
$$
are the lengths of legs of a right-angled triangle with hypotenuse of length $\sqrt{17}$. | [] | Croatia | Croatian Mathematical Society Competitions | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | 2 | |
0grd | Two distinct positive integers are called *relatively consistent* if the larger one can be written as a sum of some distinct positive divisors of the other one. Show that there exist $2018$ positive integers such that any two of them are relatively consistent. | [
"By inducting on $n$, we show that there exist relatively consistent $n$ distinct positive integers.\n\n$= 1+2$, and hence $2$ and $3$ is a consistent pair.\n\nNow let $x_1 < x_2 < \\dots < x_n$ be relatively consistent $n$ distinct positive integers. Then for any given $i < j$ there exist distinct positive divisor... | Turkey | Team Selection Test | [
"Number Theory > Divisibility / Factorization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof only | null | |
0bhf | Let triangle $ABC$ be so that $m(\angle B) = m(\angle C) = 50^\circ$. Take $D$ on the ray $AB$ so that $BC = AD\sqrt{3}$. Let $MC$ be a segment perpendicular on the plane $ABC$ so that $CD = 2MC$.
a) Prove that $m((\widehat{MAC}), (\widehat{MCD})) = 4m((\widehat{MBC}), (\widehat{MCD}))$.
b) Find the measure of the an... | [] | Romania | Shortlisted problems for the 65th Romanian NMO | [
"Geometry > Solid Geometry > Other 3D problems",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | proof and answer | arcsin(1/sqrt(3)) | |
08r6 | For $0 \le x, y \le 1$, let
$$
f(x, y) = x y^2 \sqrt{1-x^2} - x^2 y \sqrt{1-y^2}
$$
Find the minimum constant $c$ for which the following condition holds.
For any integer $n > 1$ and any real numbers $a_1, a_2, \dots, a_n$ such that $0 \le a_1 < a_2 < \dots < a_n \le 1$,
$$
f(a_1, a_2) + f(a_2, a_3) + \dots + f(a_{n-1... | [
"Draw a semicircle whose radius is $1$, and let $A$ and $B$ be the two ends of the arc. Take two points $C$ and $D$ on arc $AB$ so that $A$, $C$, $D$ and $B$ are on the arc in this order.\nLet $2a = AC$, $2b = AD$, $\\angle CAB = \\alpha$, $\\angle DAB = \\beta$. Then $a = \\cos \\alpha$, $b = \\cos \\beta$ and are... | Japan | The 16th Japanese Mathematical Olympiad - The First Round | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | English | proof and answer | pi/4 | |
08ny | Problem:
$\omega_{1}$ and $\omega_{2}$ are two circles that are externally tangent to each other at the point $M$ and internally tangent to a circle $\omega_{3}$ at the points $K$ and $L$, respectively. Let $A$ and $B$ be the two points where the common tangent line at $M$ to $\omega_{1}$ and $\omega_{2}$ intersects $... | [
"Solution:\n\nLet $C$ be the intersection point of the tangent lines to the circles $\\omega_{1}$ at $K$ and $\\omega_{2}$ at $L$. Point $C$ lies on the radical axis of circles $\\omega_{1}$ and $\\omega_{3}$, and also on the radical axis of the circles $\\omega_{2}$ and $\\omega_{3}$. Therefore $C$ lies on the rad... | JBMO | 17th Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null |
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