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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
0g67 | 設 $O$ 為銳角 $\triangle ABC$ 外接圓之圓心，直線 $AO$ 交 $BC$ 於一點 $D$。分別在邊 $AB$ 及 $AC$ 上各取一點 $E, F$，使得 $ED = BD$ 且 $DF = CD$。試證：$EF \parallel BC$。 | [
"首先，作直線 $AO$ 交圓 $O$ 於 $P$，則 $\\angle ABP = 90^\\circ = \\angle ACP$。\n\n接著，自 $D$ 點作邊 $BE$ 及 $CF$ 的高，分別交 $BE$ 及 $CF$ 於點 $G$ 及 $H$。因為 $\\triangle AGD \\sim \\triangle ABP$ 且 $\\triangle AHD \\sim \\triangle ACP$，故 $AG/AB = AD/AP = AH/AC$。因此 $\\triangle AGH \\sim \\triangle ABC$，從而 $GH \\parallel BC$。\n\n最後，因為 $ED = B... | Taiwan | 二〇一二數學奧林匹亞競賽第一階段選訓營 | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof only | null | |
0fc5 | Problem:
En una recta tenemos cuatro puntos $A, B, C$ y $D$, en ese orden, de forma que $A B = C D$. $E$ es un punto fuera de la recta tal que $C E = D E$. Demuestra que $\angle C E D = 2 \angle A E B$ si y sólo si $A C = E C$. | [
"Solution:\n\nSea $F$ el punto tal que los triángulos $A B F$ y $C D E$ son iguales. Claramente un triángulo es el otro desplazado por $A C$, luego $E F = A C$ y $A F = C E = D E = B F$. Trazamos la circunferencia de centro $F$ que pasa por $A$ y $B$, y como $\\angle A F B = \\angle C E D$, por ser el ángulo centra... | Spain | Problemas Primera Sesión | [
"Geometry > Plane Geometry > Transformations > Translation",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Anal... | null | proof only | null | |
01ra | There are $n$ towns in a country. Some of them are connected with domestic flights. Each flight connects exactly two towns. If there is a flight from town $A$ to town $B$, then there is a flight from $B$ to $A$. It is known that for any two towns there exists a unique route such that one can reach the second town from ... | [
"We use the graph theory language and consider the towns as the vertices of the graph $\\Gamma$. We connect two vertices by the edge iff there exists a flight from one of them to the other town. The problem conditions are equivalent to the following conditions for the graph $\\Gamma$:\n\ni) any two different vertic... | Belarus | Final Round | [
"Discrete Mathematics > Graph Theory"
] | English | proof only | null | |
08e1 | Problem:
Determinare tutte le coppie $(a, b)$ di numeri interi positivi che verificano le seguenti tre condizioni:
- $b > a$ e $b - a$ è un numero primo,
- la cifra delle unità di $a + b$ è $3$,
- $ab$ è il quadrato di un numero intero. | [] | Italy | XXXVI Olimpiade Italiana di Matematica | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 9) | |
0dcu | Let $ABC$ be a triangle inscribed in a circle $(\omega)$ and $I$ is the incenter. Denote $D, E$ as the intersection of $AI, BI$ with $(\omega)$. And $DE$ cuts $AC, BC$ at $F, G$ respectively. Let $P$ be a point such that $PF \parallel AD$ and $PG \parallel BE$. Suppose that the tangent lines of $(\omega)$ at $A, B$ mee... | [
"Suppose that $KA$ cuts $PF$ at $M$, $KB$ cuts $PG$ at $N$. By angle chasing, we have\n$$\n\\angle IEF = \\angle BAI = \\angle FAI,\n$$\nthen $AIFE$ is cyclic. In addition, we get\n$$\n\\begin{aligned}\n\\angle AMF = \\angle KAI & = \\angle KAB + \\angle BAI \\\\\n& = \\angle AEI + \\angle FEI = \\angle AEF\n\\end{... | Saudi Arabia | SAUDI ARABIAN MATHEMATICAL COMPETITIONS | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
... | English | proof only | null | |
0ivb | Problem:
Let $P$ be a graph with one vertex $v_{n}$ for each positive integer $n$. If $a < b$, then an edge connects vertices $v_{a}$ and $v_{b}$ if and only if $\frac{b}{a}$ is a prime number. What is the chromatic number of $P$? Prove your answer. | [
"Solution:\n\nAt least two colors are needed in a good coloring of $P$. We show that two is sufficient. Write the positive integer $n$ as $p_{1}^{e_{1}} p_{2}^{e_{2}} \\ldots p_{k}^{e_{k}}$, for distinct primes $p_{1}, p_{2}, \\ldots, p_{k}$, and let $f(n) = e_{1} + e_{2} + \\ldots + e_{k}$. Notice that if $v_{a}$ ... | United States | Harvard-MIT Math Tournament | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | 2 | |
00yp | Problem:
In how many ways can the set of integers $\{1,2, \ldots, 1995\}$ be partitioned into three nonempty sets so that none of these sets contains two consecutive integers? | [
"Solution:\n\nWe construct the three subsets by adding the numbers successively, and disregard at first the condition that the sets must be non-empty. The numbers $1$ and $2$ must belong to two different subsets, say $A$ and $B$. We then have two choices for each of the numbers $3,4, \\ldots, 1995$, and different c... | Baltic Way | Baltic Way | [
"Discrete Mathematics > Combinatorics > Recursion, bijection",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 2^{1993} - 1 | |
0c6y | Let $a, b, c$ be distinct complex numbers, such that $|a| = |b| = |c| = 1$. Prove that if $|a+b-c|^2 + |b+c-a|^2 + |c+a-b|^2 = 12$, then the geometric images of $a, b, c$ are the vertices of an equilateral triangle.
Mihaela Berindeanu | [] | Romania | 2019 ROMANIAN MATHEMATICAL OLYMPIAD | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Transformations > Rotation"
] | English | proof only | null | |
002l | En el trapecio $ABCD$, la suma de las bases $AB$ y $CD$ es igual a la diagonal $BD$. Sea $M$ el punto medio de $BC$ y $E$ el simétrico de $C$ respecto de la recta $DM$. Demostrar que $\angle AEB = \angle ACD$. | [] | Argentina | XIV Olimpiada Matemática Rioplatense | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | Español | proof only | null | |
01pb | Find all integers $n$ and $x_1, \ldots, x_n$ ($1 \le x_i \le 50$) such that
$$ \sum_{i=1}^{n} x_i(100 - x_i) = 1515. $$ | [
"Answer: $n = 2$, $\\{x_1, x_2\\} = \\{4, 13\\}$.\n\nIt is clear that there are no solutions for $n = 1$. Hence consider the case $n \\ge 2$.\nLet $s = \\sum_{i} x_i$, $S = \\sum_{i} x_i^2$. Then $100s - S = 1515$ and hence $100s > 1515 > 100s - s^2$. Taking into account that $s$ is a nonnegative integer, we obtain... | Belarus | Belarusian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > Jensen / smoothing",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | English | proof and answer | n = 2; {x1, x2} = {4, 13} | |
08f0 | Problem:
Per ogni reale non negativo $x$, definiamo $\lfloor x\rfloor$ come la parte intera di $x$, ovvero il più grande intero minore o uguale di $x$, e $\{x\}=x-\lfloor x\rfloor$ come la parte frazionaria di $x$.
Sia $p$ una soluzione reale positiva non intera dell'equazione $\{z\lfloor z\rfloor\}=2021\{z\}$. Qual è... | [
"Solution:\n\nLa risposta è $\\mathbf{( E )}$. Poniamo $n=\\lfloor p\\rfloor$ e $\\alpha=\\{p\\}$. Allora l'equazione si riscrive come $\\{n(n+\\alpha)\\}=2021 \\alpha$, ovvero\n$$\n\\{n \\alpha\\}=2021 \\alpha\n$$\nPoiché a sinistra abbiamo una parte frazionaria, che è $<1$, ed inoltre $p$ non è intero, otteniamo ... | Italy | Italian Mathematical Olympiad, February Round | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | MCQ | E | |
0hfm | For which largest $k$ does there exist a permutation $(a_1, a_2, \ldots, a_{2022})$ of integers $(1, 2, \ldots, 2022)$ such that for some $k$ integers $1 \le i \le 2022$ the fraction $\frac{a_1 + a_2 + \cdots + a_i}{1 + 2 + \cdots + i}$ is an integer larger than $1$?
(Oleksii Masalitin) | [
"Denote by $s_i = a_1 + a_2 + \\dots + a_i$, $t_i = 1 + 2 + \\dots + i$. We will show that there exists at most one $i \\ge 1011$, for which $s_i \\ne t_i$ is divisible by $t_i$. Indeed, note that $s_i \\le 2022 + 2021 + \\dots + (2023 - i) = 2023i - 1011$. Also, for $i \\ge 1011$ we have $2023i - 1011 < 3t_i$, as ... | Ukraine | 62nd Ukrainian National Mathematical Olympiad, Third Round, Second Tour | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | 1011 | |
05l1 | Problem:
Combien y a-t-il de nombres à six chiffres qui ont quatre chiffres pairs, deux chiffres impairs et qui sont multiples de $5$ ?
Note: un nombre ne commence pas par un $0$. | [
"Solution:\n\nUn tel nombre termine par $0$ ou $5$. Nous allons compter séparément les nombres qui se terminent par $0$ et ceux qui se terminent par $5$.\n\nNombres dont le dernier chiffre est $0$ :\n- Si le premier chiffre est pair, nous avons $4$ possibilités pour le choisir ($2,4,6,8$, mais pas $0$). Il reste à ... | France | Envoi de combinatoire | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Number Theory > Divisibility / Factorization"
] | null | proof and answer | 40625 | |
051e | Let $ABC$ be a triangle with median $AK$. Let $O$ be the circumcenter of the triangle $ABK$.
a) Prove that if $O$ lies on a midline of the triangle $ABC$, but does not coincide with its endpoints, then $ABC$ is a right triangle.
b) Is the statement still true if $O$ can coincide with an endpoint of the midsegment? | [
"\n\n\nFig. 2\nFig. 3\nFig. 4\n\nSolution:\na) Let $L$ and $M$ be the midpoints of the sides $CA$ and $AB$, respectively. If $O$ lies on the segment $KM$ (Fig. 2), then the segment $KM$ and the perpendicular bisector of $AB$ have two differe... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | a) The triangle is right. b) No; for an equilateral triangle the circumcenter coincides with a midsegment endpoint but the triangle is not right. | |
056i | Is it possible to find four distinct prime numbers for which the sum of any three of them is also a prime number? | [
"Suitable prime numbers are $5$, $7$, $17$, and $19$. The sums of the corresponding triplets are prime numbers $29$, $31$, $41$, and $43$."
] | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | 5, 7, 17, 19 | |
03kh | Problem:
Let $f(x) = \frac{9^{x}}{9^{x} + 3}$. Evaluate the sum
$$
f\left(\frac{1}{1996}\right) + f\left(\frac{2}{1996}\right) + f\left(\frac{3}{1996}\right) + \cdots + f\left(\frac{1995}{1996}\right)
$$ | [] | Canada | Canadian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 1995/2 | |
08r4 | Consider the equation
$$
x^2 + xy + y^2 + 3x + 6y + 6 = 0. \quad (*)
$$
(1) Find all pair $(x, y)$ of integers with $x = 1$ which satisfy equation $(*)$.
(2) Find all pair $(x, y)$ of integers which satisfy equation $(*)$. | [
"(1) Substituting $x = 1$ into the given equation, we obtain a quadratic equation $y^2 + 7y + 10 = 0$, and solving this equation normally, we get $y = -2, -5$. Hence the solutions are $(x, y) = (1, -2), (1, -5)$.\n\n(2) Assume that $(x, y)$ is a root of the given equation. Substituting the value of $x$, we obtain a... | Japan | The 4th Japanese Junior Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | English | proof and answer | Part (1): (1, -2), (1, -5). Part (2): (-2, -2), (-1, -4), (-1, -1), (1, -5), (1, -2), (2, -4). | |
0hw2 | Problem:
On a certain island, there are knights, who always tell the truth, knaves, who always lie, and spies, who could do either. Suppose you meet three people, and you know one is a knight, one is a knave, and one is a spy, but you don't know which is which.
Find a method to ask three yes/no questions, each to one o... | [
"Solution:\nOne strategy is as follows:\n1. Ask the first person, \"Is the second person more likely to tell the truth than the third person?\"\nIf the response is \"yes\", then if the first person is the knight, the second must be the spy and the third must be the knave. If the first person is the knave, the secon... | United States | Berkeley Math Circle: Monthly Contest 7 | [
"Discrete Mathematics > Logic"
] | null | proof only | null | |
048z | Determine all $a$ such that there exists unique $(x, y) \in \mathbb{R}^2$ satisfying
$$
2^{|x|} + |x| = x^2 + y + a, \quad x^2 + y^2 = 1.
$$ | [
"If the pair $(x, y)$ is the solution of the given system, then the pair $(-x, y)$ is also the solution. We conclude that the unique solution of this system has to be of the form $(0, y)$.\n\nTaking $x = 0$ in the given system we get\n$$\n\\begin{aligned}\n1 &= y + a, \\\\\ny^2 &= 1,\n\\end{aligned}\n$$\nso $y = 1$... | Croatia | CroatianCompetitions2011 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | a = 0 | |
0jo5 | Problem:
Let $P$ be a (non-self-intersecting) polygon in the plane. Let $C_{1}, \ldots, C_{n}$ be circles in the plane whose interiors cover the interior of $P$. For $1 \leq i \leq n$, let $r_{i}$ be the radius of $C_{i}$. Prove that there is a single circle of radius $r_{1}+\cdots+r_{n}$ whose interior covers the inte... | [
"Solution:\nIf $n=1$, we are done. Suppose $n>1$. Since $P$ is connected, there must be a point $x$ on the plane which lies in the interiors of two circles, say $C_{i}, C_{j}$. Let $O_{i}, O_{j}$, respectively, be the centers of $C_{i}, C_{j}$. Since $O_{i} O_{j}<r_{i}+r_{j}$, we can choose $O$ to be a point on seg... | United States | HMMT February 2015 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof only | null | |
0czw | Let $ABC$ be a triangle with medians $m_{a}, m_{b}, m_{c}$. Prove that:
a. There is a triangle with side lengths $m_{a}, m_{b}, m_{c}$.
b. This triangle is similar to $ABC$ if and only if the squares of the side lengths of triangle $ABC$ form an arithmetical sequence. | [
"Let $A'$, $B'$, $C'$ be the midpoints of sides $BC$, $CA$, $AB$, respectively. Construct the parallelogram $AC'CC''$. The points $C'$, $B'$, $C''$ are collinear, hence $BA'C''B'$ is also a parallelogram, that is $A'C'' = BB'$.\n\n\n\nThe desired triangle is $AA'C''$, and $AA' = m_{a}$, $A'... | Saudi Arabia | Saudi Arabia Mathematical Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
03rk | Given a sequence $\{a_n\}$ of numbers satisfying $a_0 = 1$, $a_{n+1} = \frac{7a_n + \sqrt{45a_n^2 - 36}}{2}$, $n \in \mathbb{N}$.
Prove that
(1) for each $n \in \mathbb{N}$, $a_n$ is a positive integer.
(2) for each $n \in \mathbb{N}$, $a_n a_{n+1} - 1$ is a perfect square. | [
"(1) By assumption, $a_1 = 5$ and $\\{a_n\\}$ is strictly increasing with\n$$\n2a_{n+1} - 7a_n = \\sqrt{45a_n^2 - 36}.\n$$\nSquare both sides, and we get\n$$\na_{n+1}^2 - 7a_n a_{n+1} + a_n^2 + 9 = 0, \\qquad \\textcircled{1}\n$$\n$$\na_n^2 - 7a_{n-1}a_n + a_{n-1}^2 + 9 = 0, \\qquad \\textcircled{2}\n$$\n$$\n\\text... | China | China Mathematical Competition (Jiangxi) | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | English | proof only | null | |
0683 | The real numbers $x, y, z$, with $x \neq z$, are mutually different and nonzero and they satisfy the following equations:
$$
(x+y)^2 + (2-xy) = 9, \\
(y+z)^2 - (3+yz) = 4.
$$
Determine the value of the expression
$$
A = \left( \frac{x}{y} + \frac{y^2}{x^2} + \frac{z^3}{x^2y} \right) \left( \frac{y}{z} + \frac{z^2}{y^2}... | [
"The given equalities can be written:\n$$\nx^2 + y^2 + xy = 7, \\qquad (1)\n$$\n$$\ny^2 + z^2 + yz = 7, \\qquad (2)\n$$\nBy subtraction we get:\n$$\nx^2 - z^2 + xy - yz = 0 \\Leftrightarrow (x-z)(x+z) + y(x-z) = 0 \\Leftrightarrow (x-z)(x+z+y) = 0.\n$$\nSince $x - z \\neq 0$, we get:\n$$\nx + y + z = 0. \\qquad (3)... | Greece | 33rd Hellenic Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | 27 | |
028t | 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\nObserve que\n$$\n2^{8}+2^{11}+2^{n} = \\left(2^{4}\\right)^{2} + 2 \\times 2^{4} \\times 2^{6} + \\left(2^{\\frac{n}{2}}\\right)^{2}\n$$\nLogo, se $n=12$, temos\n$$\n2^{8}+2^{11}+2^{12} = \\left(2^{4}+2^{6}\\right)^{2}\n$$\nLogo $n=12$ é uma solução.\n\nSolução Geral: Se $2^{8}+2^{11}+2^{n}=k^{2}$, en... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 12 | |
0kmz | Problem:
Two circles with radii $71$ and $100$ are externally tangent. Compute the largest possible area of a right triangle whose vertices are each on at least one of the circles. | [
"Solution:\n\nIn general, let the radii of the circles be $r < R$, and let $O$ be the center of the larger circle. If both endpoints of the hypotenuse are on the same circle, the largest area occurs when the hypotenuse is a diameter of the larger circle, with $[ABC] = R^2$.\n\nIf the endpoi... | United States | HMMT Spring 2021 | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 24200 | |
0acp | Solve the system:
$$
\begin{cases}
x(y+z) = 35 \\
y(z+x) = 32 \\
z(x+y) = 27
\end{cases}
$$ | [
"The given system is equivalent to\n$$\n\\begin{cases}\nxy + xz = 35 \\\\\nyz + yx = 32 \\\\\nzx + zy = 27\n\\end{cases}\n$$\nIf $a = xy$, $b = xz$, $c = yz$ then the system is\n$$\n\\begin{cases}\na+b=35 \\quad ...(1) \\\\\nc+a=32 \\quad ...(2) \\\\\nb+c=27 \\quad ...(3)\n\\end{cases}\n$$\nSubtraction of the equat... | North Macedonia | Macedonian Mathematical Competitions | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | [(5, 4, 3), (-5, -4, -3)] | |
0ah5 | Find all integer solutions of the equation
$$
x^4 + 2y^4 + 4z^4 + 8t^4 = 16xyzt
$$ | [
"It is clear that $(0, 0, 0, 0)$ is a solution of the equation. We will show that there exists no nonzero solution of the equation. Let us suppose the contrary, i.e. let $(x_0, y_0, z_0, t_0)$ be a solution of the equation with at least one nonzero coordinate $x_0^4 + 2y_0^4 + 4z_0^4 + 8t_0^4 = 16x_0 y_0 z_0 t_0$. ... | North Macedonia | 19-th Macedonian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Infinite descent / root flipping",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | (0, 0, 0, 0) | |
082m | Problem:
Determinare il numero di quadruple di numeri interi (non necessariamente distinti) compresi fra $1$ e $12$ (estremi inclusi) che verificano tutte le seguenti condizioni:
- la somma dei primi due numeri è pari
- la somma dei primi tre numeri è multipla di $3$
- la somma dei quattro numeri è multipla di $4$.
(D... | [
"Solution:\n\nLa risposta è $864$. Per ogni scelta del primo numero, il secondo numero può essere scelto in $6$ modi per soddisfare la prima condizione; ancora, per ogni possibile somma dei primi $2$ numeri esistono $4$ scelte possibili per il terzo affinché la seconda condizione sia soddisfatta; infine, per ogni p... | Italy | Progetto Olimpiadi di Matematica 2003 GARA di SECONDO LIVELLO TRIENNIO | [
"Number Theory > Other"
] | null | final answer only | 864 | |
009t | The following operation is allowed on several given nonnegative integers. A positive number $a$ is chosen among them, and each number $b \ge a$ is replaced by $b-a$, including the choice $a$ itself. Starting with $1, 2, \ldots, 2013$, after several operations numbers with sum $10$ are obtained. What can these numbers b... | [
"Call the set $S_k = \\{1, 2, \\dots, k\\}$ a block, for $k=1, 2, \\dots$; for consistency assume that $S_0$ is the empty block. Suppose that several numbers can be partitioned into blocks. The key observation is that the same holds true after any operation is applied. Indeed let $S_k$ be one of the blocks, and let... | Argentina | NATIONAL XXX OMA | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | {1,2,3,4} or {1,1,1,2,2,3} or {1,1,1,1,1,2,3} or {1,1,1,1,2,2,2} or {1,1,1,1,1,1,2,2} or {1,1,1,1,1,1,1,1,2} or {1,1,1,1,1,1,1,1,1,1} | |
04na | Let $S = \{0, 95\}$. In each step, Lucija is extending the set $S$ in the following way. She chooses a polynomial with coefficients in $S$, distinct from zero, and extends $S$ with all integer roots of a chosen polynomial. She repeats the procedure by choosing another polynomial with coefficients from the extended set ... | [
"If the coefficients of a polynomial are integers, then its roots must divide the constant term. Without loss of generality, we may assume that the constant term of the chosen polynomial is non-zero, so we conclude that we can extend $S$ only with the divisors of $95$. Since $95$ has only finitely many divisors, Lu... | Croatia | Croatia_2018 | [
"Algebra > Algebraic Expressions > Polynomials > Irreducibility: Rational Root Theorem, Gauss's Lemma, Eisenstein",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | 9 | |
0k5v | Problem:
What is the smallest positive integer that cannot be written as the sum of two nonnegative palindromic integers? (An integer is palindromic if the sequence of decimal digits are the same when read backwards.) | [
"Solution:\n\nWe need to first prove that every positive integer $N$ less than $21$ can be written as sum of two nonnegative palindromic integers. If $N$ is in the interval $[1,9]$, then it can be written as $0+N$. If $N$ is in the interval $[10,18]$, it can be written as $9+(N-9)$. In addition, $19$ and $20$ can b... | United States | HMMT February 2019 | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof and answer | 21 | |
0bql | Problem:
Fie $z_{A}$, $z_{B}$, $z_{C}$ afixele vârfurilor triunghiului $ABC$. Fie $J$, de afix $z_{J}$ un punct interior triunghiului $ABC$ şi $S_{a}$, $S_{b}$, $S_{c}$ ariile triunghiurilor $JBC$, $JAC$, respectiv $JAB$.
a) Demonstraţi că $z_{J} = \frac{S_{a} z_{A} + S_{b} z_{B} + S_{c} z_{C}}{S}$, unde $S$ este ari... | [] | Romania | Olimpiada Națională de Matematică - Etapa Locală | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Advanced Configurations > Isogonal/isotomic conjugates... | null | proof only | null | |
0254 | Problem:
Vivi, Tânia e Rosa estão em fila, não necessariamente nessa ordem, e gritam sucessivamente, cada uma, um múltiplo de $3$.
| 3 | 6 | 9 |
| :---: | :---: | :---: |
| 12 | 15 | 18 |
| $\vdots$ | $\vdots$ | $\vdots$ |
Vivi foi a primeira a gritar um número maior que $2003$ e Rosa a primeira a gritar um número d... | [
"Solution:\n\nObserve que aquela que gritou os números $9$, $18$, etc, sempre gritou múltiplos de $9$. O primeiro múltiplo de $3$ com quatro algarismos é $1002$ e o primeiro múltiplo de $3$ maior do que $2003$ é $2004$. Logo, Vivi gritou $2004$ e Rosa $1002$. Nenhum desses números é múltiplo de $9$, portanto, foi T... | Brazil | Nível 2 | [
"Number Theory > Modular Arithmetic"
] | null | proof and answer | 666: Tânia; 888: Vivi | |
08p3 | Problem:
Determine the number of pairs of integers $(m, n)$ such that
$$
\sqrt{n+\sqrt{2016}}+\sqrt{m-\sqrt{2016}} \in \mathbb{Q}
$$ | [
"Solution:\nLet $r=\\sqrt{n+\\sqrt{2016}}+\\sqrt{m-\\sqrt{2016}}$. Then\n$$\nn+m+2 \\sqrt{n+\\sqrt{2016}} \\cdot \\sqrt{m-\\sqrt{2016}}=r^{2}\n$$\nand\n$$\n(m-n) \\sqrt{2016}=\\frac{1}{4}\\left(r^{2}-m-n\\right)^{2}-m n+2016 \\in \\mathbb{Q}\n$$\nSince $\\sqrt{2016} \\notin \\mathbb{Q}$, it follows that $m=n$. Then... | JBMO | Junior Balkan Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | 1 | |
07px | Determine all triples $(p, m, n)$ where $p$ is prime and $m$ and $n$ are non-negative integers satisfying the equation
$$
p^m - n^3 = 27.
$$ | [
"Because $p^n = n^3 + 27 = (n+3)(n^2 - 3n + 9)$, there are positive integers $x, y$ such that $p^x = n+3$ and $p^y = n^2 - 3n + 9$. As $n^2 - 3n + 9 - (n+3) = n^2 - 4n + 6 = (n-2)^2 + 2 > 0$ we have $p^x < p^y$ and so $p^x|p^y$. Hence $(n+3)|(n^2 - 3n + 9)$. But $n^2 - 3n + 9 = (n+3)(n-6) + 27$, thus $(n+3)|27$, wh... | Ireland | Ireland | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | (3, 3, 0) and (3, 5, 6) | |
0ko9 | Problem:
Aerith bakes some cookies. On the first day, she gives away 1 cookie and then $1/8$ of the remaining cookies; on the second day, she gives away 2 cookies and then $1/8$ of the remaining cookies, and so on. On the 7th day, she gives away 7 cookies and then there are none left. How many cookies did she bake? | [
"Solution:\n\nThe number of cookies is $49$.\n\nWorking backwards, on the 6th day, $1/8$ of the remaining cookies must have been $1$ cookie (since there were $7$ left after that), so there were $6+1+7=14$ cookies at the start of the day. Similarly, since on the 5th day there were $14$ cookies left, the $1/7$ of the... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Prealgebra / Basic Algebra > Fractions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | final answer only | 49 | |
0fpm | Consider a $25 \times 25$ chessboard with cells $C(i, j)$ for $1 \le i, j \le 25$. Find the smallest possible number $n$ of colors with which these cells can be colored subject to the following condition: For $1 \le i < j \le 25$ and for $1 \le s < t \le 25$, the three cells $C(i, s)$, $C(j, s)$, $C(j, t)$ carry at lea... | [
"The forbidden configuration is given by\n\n\nFor a $3 \\times 3$ chessboard, the minimum number is given by 2. Indeed:\n\n\nIf we deal with a $5 \\times 5$ chessboard, it is sufficient to consider 3 colours:\n\n\nIt seems that $m_n = \\frac... | Spain | MEDITERRANEAN MATHEMATICAL COMPETITION | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Counting two ways",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | Spanish | proof and answer | 13 | |
05gp | Problem:
Soient $\left(a_{n}\right)_{n \geqslant 1}$ une suite croissante d'entiers strictement positifs et $k$ un entier strictement positif. Supposons que pour un certain $r \geqslant 1$, on $\frac{r}{a_{r}}=k+1$. Montrer qu'il existe un entier $s \geqslant 1$ tel que $\frac{s}{a_{s}}=k$. | [
"Solution:\n\nSoit $v_{n}=n-k a_{n}$ pour tout $n \\geqslant 1$. Le problème revient à montrer qu'il existe $s$ tel que $v_{s}=0$.\n\nOn observe que $v_{1}=1-k a_{1} \\leqslant 0$, que $v_{r}=r-k a_{r}=a_{r}>0$ et qu'enfin $v_{n+1}= (n+1)-k a_{n+1}=n-k a_{n+1}+1 \\leqslant n-k a_{n}+1=v_{n}+1$, car la suite $\\left... | France | Préparation Olympique Française de Mathématiques | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof only | null | |
011x | Problem:
Let $n$ be a positive integer. Prove that at least $2^{n-1} + n$ numbers can be chosen from the set $\{1, 2, 3, \ldots, 2^n\}$ such that for any two different chosen numbers $x$ and $y$, $x + y$ is not a divisor of $x \cdot y$. | [
"Solution:\nWe choose the numbers $1, 3, 5, \\ldots, 2^n - 1$ and $2, 4, 8, 16, \\ldots, 2^n$, i.e., all odd numbers and all powers of $2$.\n\nConsider the three possible cases.\n\n(1) If $x = 2a - 1$ and $y = 2b - 1$, then $x + y = (2a - 1) + (2b - 1) = 2(a + b - 1)$ is even and does not divide $xy = (2a - 1)(2b -... | Baltic Way | Baltic Way | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
02hx | Problem:
O quadrado abaixo é chamado quadrado mágico, porque a soma dos números de cada linha, de cada coluna e de cada diagonal é sempre a mesma. Neste caso essa soma é $15$.
| 4 | 9 | 2 |
| :--- | :--- | :--- |
| 3 | 5 | 7 |
| 8 | 1 | 6 |
Complete os cinco números que faltam no quadrado abaixo para que ele seja um... | [
"Solution:\n\nComo a soma dos números de uma diagonal é $4+0+(-4)=0$, este deve ser o valor da soma dos números de cada linha, coluna e diagonal.\n\nAssim, obtemos de imediato os números que faltam nas casas em cinza no primeiro tabuleiro: $16$, $8$ e $12$, pois $(-12)+16+(-4)=0$ (na primeira linha), $(-12)+8+4=0$ ... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | -12 16 -4
8 0 -8
4 -16 12 | |
08ko | Problem:
Let $a$, $b$, $c$ and $m_a$, $m_b$, $m_c$ be respectively the lengths of the sides and the medians of an acute-angled triangle $ABC$. Prove that
$$
\frac{m_a^2}{b^2 + c^2 - a^2} + \frac{m_b^2}{c^2 + a^2 - b^2} + \frac{m_c^2}{a^2 + b^2 - c^2} \geq \frac{9}{4}
$$ | [
"Solution:\n\nTaking into consideration that the triangle $ABC$ is acute-angled, using the formulae\n$$\n4 m_a^2 = 2 b^2 + 2 c^2 - a^2, \\quad 4 m_b^2 = 2 a^2 + 2 c^2 - b^2, \\quad 4 m_c^2 = 2 a^2 + 2 b^2 - c^2\n$$\nand using the notations $x = b^2 + c^2 - a^2 > 0$, $y = a^2 + c^2 - b^2 > 0$, $z = a^2 + b^2 - c^2 >... | JBMO | JBMO Shortlist | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
0f86 | Problem:
What is the minimal value of $b / (c + d) + c / (a + b)$ for positive real numbers $b$ and $c$ and nonnegative real numbers $a$ and $d$ such that $b + c \geq a + d$? | [
"Solution:\n\nAnswer: $\\sqrt{2} - 1 / 2$.\n\nObviously $a + d = b + c$ at the minimum value, because increasing $a$ or $d$ reduces the value. So we may take $d = b + c - a$. We also take $b \\gg c$ (interchanging $b$ and $c$ if necessary). Dividing through by $b / 2$ shows that there is no loss of generality in ta... | Soviet Union | 22nd ASU | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | sqrt(2) - 1/2 | |
01uz | Points $C_1$ and $B_1$ are marked on the sides $AB$ and $AC$ of the triangle $ABC$ respectively. Segments $BB_1$ and $CC_1$ intersect at point $X$, and segments $B_1C_1$ and $AX$ intersect at point $A_1$. The circumcircles of the triangles $BXC_1$ and $CXB_1$ intersect the side $BC$ at points $D$ and $E$ respectively. ... | [
"First we prove the equality\n$$\n\\frac{B_1 E}{C_1 D} = \\frac{B_1 A_1}{C_1 A_1}.\n$$\nSince the quadrilateral $XB_1CE$ is cyclic, $\\angle XB_1E = \\angle XCE$. Therefore, the triangle $BB_1E$ is similar to the triangle $BCX$. Hence,\n$$\n\\frac{B_1 E}{X C} = \\frac{B B_1}{B C}. \\qquad (1)\n$$\nSimilarly one can... | Belarus | Selection and Training Session | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Menelaus' theorem",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
068x | A company consisting of $n$ friends play a table game according to the following rules:
(a) At each round play exactly 3 players.
(b) The game stops after $n$ rounds.
(c) Each couple of players have played together at least at one round.
Determine the maximal possible value of $n$. | [
"Since in each round play exactly 3 players, the number of couples playing at each round is $\\binom{3}{2} = 3$. Therefore, at the end of the game after $n$ rounds, the total number of couples played the game will be $3n$. According to the last rule:\n$$\n\\binom{n}{2} \\le 3n \\Leftrightarrow \\frac{n(n-1)}{2} \\l... | Greece | 34th Hellenic Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 7 | |
01kz | Let $\tau(n)$ be the number of the divisors of positive integer $n$ (including $1$ and $n$) and $\sigma(n)$ be their sum. Prove that $\sqrt{n} \le \frac{\sigma(n)}{\tau(n)} \le \frac{n+1}{2}$. | [
"Note that if $d$ takes on the values of all divisors of positive integer $n$, then $n/d$ takes on the values of all divisors of $n$, too. Let $d_1, \\dots, d_{\\tau(n)}$ denote the divisors of $n$. We have\n$$\n\\sigma(n) = \\frac{1}{2} \\left( \\left( d_1 + \\frac{n}{d_1} \\right) + \\left( d_2 + \\frac{n}{d_2} \... | Belarus | 60th Belarusian Mathematical Olympiad | [
"Number Theory > Number-Theoretic Functions > τ (number of divisors)",
"Number Theory > Number-Theoretic Functions > σ (sum of divisors)",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | English | proof only | null | |
0d1b | Find all the finite sets $A$ of real positive numbers having at least two elements, with the property that $a^2 + b^2 \in A$ for every $a, b \in A$ with $a \neq b$. | [
"If $a_1 < a_2 < \\dots < a_n$ are the elements of $A$, then $a_1^2 + a_2^2, a_1^2 + a_3^2, \\dots, a_1^2 + a_n^2, a_2^2 + a_n^2, \\dots, a_{n-1}^2 + a_n^2$ belong to $A$, and we have\n$a_1^2 + a_2^2 < a_1^2 + a_3^2 < \\dots < a_1^2 + a_n^2 < a_2^2 + a_n^2 < \\dots < a_{n-1}^2 + a_n^2$,\nwhich implies $2n - 3 \\le ... | Saudi Arabia | Saudi Arabia Mathematical Competitions 2012 | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | English | proof and answer | All such sets are exactly the two-element sets {t, sqrt(t(1 − t))} with t in (0, 1) and t ≠ 1/2; equivalently, two distinct positive reals x, y with x^2 + y^2 equal to one of x or y. No set with three or more elements exists. | |
0f4n | Problem:
An integer is put in each cell of an $n \times n$ array. The difference between the integers in cells which share a side is $0$ or $1$. Show that some integer occurs at least $n$ times. | [
"Solution:\n\nLet the integers in the array be $a_{i,j}$ for $1 \\leq i, j \\leq n$. For any two adjacent cells (sharing a side), $|a_{i,j} - a_{k,l}| = 0$ or $1$.\n\nLet $m$ be the minimum integer in the array. Consider the set of cells where the value is $m$. Any cell adjacent to such a cell can have value $m$ or... | Soviet Union | 16th ASU | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof only | null | |
0j3l | Problem:
How many functions $f$ from $\{-1005, \ldots, 1005\}$ to $\{-2010, \ldots, 2010\}$ are there such that the following two conditions are satisfied?
- If $a<b$ then $f(a)<f(b)$.
- There is no $n$ in $\{-1005, \ldots, 1005\}$ such that $|f(n)|=|n|$. | [
"Solution:\n\n11733467826666773000724417738143880005531795870067107864012250438426995524609421666308605302966355504513409792805200762540756742811158611534813828022157596601875355477425764387233393584166695775000921640409535245687759455481741935349426766583008743635349407582844600705064877936286986176650915007126065... | United States | Harvard-MIT Mathematics Tournament | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | final answer only | 11733467826666773000724417738143880005531795870067107864012250438426995524609421666308605302966355504513409792805200762540756742811158611534813828022157596601875355477425764387233393584166695775000921640409535245687759455481741935349426766583008743635349407582844600705064877936286986176650915007126065996533696012706527... | |
05y3 | Problem:
Soit $n \geqslant 6$. Prouvez que chaque carré peut être découpé en exactement $n$ carrés (pas nécessairement de même taille). | [
"Solution:\n\nOn peut le montrer par récurrence. Si on a un découpage du carré en $n$ carrés, on peut obtenir un découpage en $n+3$ carrés en subdivisant l'un des carrés en $4$. Il suffit donc de montrer que l'on peut découper un carré en $6$, $7$ et $8$ carrés.\n\nPour $6$, on subdivise le carré en une grille $3 \... | France | PRÉPARATION OLYMPIQUE FRANÇAISE DE MATHÉMATIQUES | [
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
0kck | Problem:
Suppose a sequence $s_{1}, s_{2}, \ldots$, of positive integers satisfies $s_{n+2} = s_{n+1} + s_{n}$ for all positive integers $n$ (but not necessarily $s_{1} = s_{2} = 1$). Prove that there exists an integer $r$ such that $s_{n} - r$ is not divisible by $8$ for any integer $n$. | [
"Solution:\n\nWe start by observing that the \"classic\" Fibonacci sequence goes $1, 1, 2, 3, 5, 0, 5, 5, 2, 7, 1, 0$ modulo $8$ (repeating forever with period $12$), and hence takes on only $6$ distinct values mod $8$ (the values $4$ and $6$ are omitted). So the statement is true there.\n\nNow let $s_{1} = a$ and ... | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other"
] | null | proof only | null | |
0ep9 | How many positive factors of $128$ are not factors of $120$? | [
"$128$ has factors that are all powers of $2$, i.e. $2^0, 2^1, \\ldots, 2^7$.\n\nThe largest power of $2$ that divides $120$ is $2^3$.\n\nThe powers of $2$ dividing $128$ and exceeding $8$ are $2^4, 2^5, 2^6$ and $2^7$.\n\nSo, there are $4$ positive factors of $128$ that are not factors of $120$."
] | South Africa | South African Mathematics Olympiad | [
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | final answer only | 4 | |
0hnr | Problem:
Let $ABC$ be a triangle. A line is drawn not passing through any vertex of $ABC$. Prove that some side of $ABC$ is not cut by the line. | [
"Solution:\nConsider the two sides of the line $\\ell$. By the pigeonhole principle on the three vertices of $ABC$, two of these vertices, say $A$ and $B$, lie on the same side of $\\ell$. Then segment $AB$ does not intersect $\\ell$."
] | United States | Berkeley Math Circle: Monthly Contest 4 | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
025f | Problem:
Florinda foi dar um passeio. Ela saiu do ponto $A$ e caminhou $1~\mathrm{m}$. Nesse ponto ela virou para a esquerda um ângulo de $90^\circ$ e caminhou $2~\mathrm{m}$. No último ponto ela virou para esquerda, e caminhou $3~\mathrm{m}$. Ela continuou andando desta maneira até que no último trecho ela caminhou $3... | [
"Solution:\nNesse exercício todas as distâncias estão dadas em metros. Observe na seguinte figura que os segmentos retilíneos horizontais possuem comprimentos ímpares enquanto que os verticais possuem comprimentos pares.\n\nSe consideramos somente os segmentos retilíneos horizontais, obtemo... | Brazil | null | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | sqrt(481) | |
032k | Problem:
Consider the function
$$
f(x) = (a^{2} + 4a + 2)x^{3} + (a^{3} + 4a^{2} + a + 1)x^{2} + (2a - a^{2})x + a^{2}
$$
where $a$ is a real parameter.
a) Prove that $f(-a) = 0$.
b) Find all values of $a$ such that the equation $f(x) = 0$ has three different positive roots. | [
"Solution:\n\na) It follows by a direct verification.\n\nb) Writing the equation in the form\n$$\n(x + a)\\left((a^{2} + 4a + 2)x^{2} + (1 - a)x + a\\right) = 0\n$$\nwe get that $a < 0$. Moreover, the quadratic polynomial in (3) must have two distinct real zeros, i.e.\n$$\nD = (1 - a)^{2} - 4a(a^{2} + 4a + 2) > 0 \... | Bulgaria | 53. Bulgarian Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a ∈ (-2 - √2, -3) ∪ (-3, (-11 - √137)/8) ∪ (-1, -2 + √2) | |
0ehc | Problem:
Dana sta dva enako dolga vektorja $\frac{1}{3} \vec{a}+\vec{b}$ in $\frac{2}{3} \vec{a}+\vec{b}$, kjer sta $\vec{a}$ in $\vec{b}$ neničelna vektorja, za katera velja $|\vec{a}|=\sqrt{3}|\vec{b}|$.
a) Izračunaj velikost kota med vektorjema $\vec{a}$ in $\vec{b}$.
b) Izračunaj velikost kota med vektorjema $\f... | [
"Solution:\n\n(a) Vektorja $\\frac{1}{3} \\vec{a}+\\vec{b}$ in $\\frac{2}{3} \\vec{a}+\\vec{b}$ sta enako dolga, zato je $\\left|\\frac{1}{3} \\vec{a}+\\vec{b}\\right|^{2}=\\left|\\frac{2}{3} \\vec{a}+\\vec{b}\\right|^{2}$. Kot med vektorjema $\\vec{a}$ in $\\vec{b}$ označimo s $\\varphi$ in poračunamo\n$$\n\\left|... | Slovenia | 62. matematično tekmovanje srednješolcev Slovenije | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Algebra > Linear Algebra > Vectors"
] | null | proof and answer | Angle between a and b: 5π/6; angle between (1/3)a + b and (2/3)a + b: π/3 | |
07fa | Prove that for any positive integers $m > n$, there are infinitely many positive integers $a, b$ such that set of prime divisors of $a^m + b^n$ is equal to set of prime divisors of $a^{2019} + b^{1398}$. | [
"We are going to construct infinite pairs like $(a, b)$ that suit the problem's condition. Consider $c$ be an arbitrary natural number then put $a = c^{n+1398}$ and $b = c^{m+2019}$. Then\n$$\na^m + b^n = c^{mn} (c^{1398m} + c^{2019n}),\n$$\nand\n$$\na^{2019} + b^{1398} = c^{1398 \\times 2019} (c^{1398m} + c^{2019n... | Iran | 37th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof only | null | |
02sf | Problem:
Nos lados $AB$ e $BC$ de um triângulo equilátero $ABC$, fixam-se dois pontos $D$ e $E$, respectivamente, de modo que $\overline{AD} = \overline{BE}$.
Se os segmentos $AE$ e $CD$ se cortam no ponto $P$, determine $\measuredangle APC$. | [
"Solution:\n\nObserve os triângulos $DAC$ e $EBA$. Sabe-se que $\\overline{DA} = \\overline{EB}$. Além disso, como o triângulo $ABC$ é equilátero, então $\\overline{AC} = \\overline{BA}$. Mais ainda, em um triângulo equilátero todos os ângulos internos medem $60^\\circ$. Logo $\\measuredangle DAC = \\measuredangle ... | Brazil | Brazilian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 120° | |
0d62 | Find the number of binary sequences $S$ of length $2015$ such that for any two segments $I_{1}, I_{2}$ of $S$ of the same length, we have
- The sum of digits of $I_{1}$ differs from the sum of digits of $I_{2}$ by at most $1$;
- If $I_{1}$ begins on the left end of $S$ then the sum of digits of $I_{1}$ is not greater t... | [
"Note that if the sequence $S$ ends with zero then all digits of $S$ will be zero.\n\nNow suppose that the sequence $S$ ends with one. Let $k$ be the number of $1$'s in $S$ then $1 \\leq k \\leq 2015$. Let $\\alpha = \\frac{2015}{k} \\geq 1$. We consider the sequence $S_{k} = a_{1} a_{2} \\ldots a_{2015}$ where $a_... | Saudi Arabia | SAMC 2015 | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English, Arabic | proof and answer | 2016 | |
0fv3 | Problem:
Bestimme alle positiven reellen Lösungen des folgenden Gleichungssystems:
$$
\begin{aligned}
& a=\max \left\{\frac{1}{b}, \frac{1}{c}\right\} \\
& b=\max \left\{\frac{1}{c}, \frac{1}{d}\right\} \\
& c=\max \left\{\frac{1}{d}, \frac{1}{e}\right\} \\
& d=\max \left\{\frac{1}{e}, \frac{1}{f}\right\} \\
& e=\max ... | [
"Solution:\n\nWegen der zyklischen Symmetrie des Systems können wir annehmen, dass $a$ maximal ist unter allen sechs Zahlen. Daraus folgt direkt\n$$\nf=\\max \\left\\{\\frac{1}{a}, \\frac{1}{b}\\right\\}=\\frac{1}{b}, \\quad e=\\max \\left\\{\\frac{1}{f}, \\frac{1}{a}\\right\\}=\\frac{1}{f}=b, \\quad d=\\max \\left... | Switzerland | SMO Finalrunde | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | All solutions are the cyclic permutations of (x, x, 1/x, x, x, 1/x) with x ≥ 1; equivalently, the cyclic permutations of (x, 1/x, 1/x, x, 1/x, 1/x) with x ≤ 1. | |
0l90 | Let $a$, $b$, $c$ be real numbers such that the polynomial $P(x) = x^3 + a x^2 + b x + c$ has three real roots (not necessarily distinct).
Prove that:
$$
12ab + 27c \leq 6a^3 + 10(a^2 - 2b)^{3/2}.
$$
When does equality occur? | [
"The demanded inequality can be written in the form:\n$$\n-6a(a^2 - 2b) \\leq -27c + 10(a^2 - 2b)^{3/2} \\quad (1)\n$$\nLet $\\alpha, \\beta, \\gamma$ be three real roots of the polynomial $P(x)$. By the theorem of Viet, (1) is equivalent to\n$$\n6(\\alpha + \\beta + \\gamma)(\\alpha^2 + \\beta^2 + \\gamma^2) \\leq... | Vietnam | THE 2002 VIETNAMESE MATHEMATICAL OLYMPIAD | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > Cauchy-Schwarz"
] | English | proof and answer | Equality holds if and only if the roots are a permutation of minus t, two t, and two t with t nonnegative, equivalently a equals minus three t, b equals zero, and c equals four times t cubed for some nonnegative t. | |
0as4 | Problem:
If $\frac{x-a-b}{c}+\frac{x-b-c}{a}+\frac{x-c-a}{b}=3$, where $a, b, c$ are positive constants, find $x$ in terms of $a, b$ and $c$. | [
"Solution:\n\n(ans. $x=a+b+c$.\nSubstituting this into the equation, one gets $\\frac{c}{c}+\\frac{a}{a}+\\frac{b}{b}=3$.)"
] | Philippines | 13th Philippine Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Fractions"
] | null | final answer only | a+b+c | |
00sg | 100 couples are invited to a traditional Moldovan dance. The 200 people stand in a line, and then in a step, two of them (not necessarily adjacent) may swap positions. Find the least $C$ such that whatever the initial order, they can arrive at an ordering where everyone is dancing next to their partner in at most $C$ s... | [
"With 100 replaced by $N$, the answer is $C = C(N) = N - 1$. Throughout, we will say that the members of a couple have the same.\n\n$N=2$: We use this as a base case for induction for both bounds. Up to labelling, there is one trivial initial order, and two non-trivial ones, namely\n$$\n1, 1, 2, 2; \\quad 1, 2, 2, ... | Balkan Mathematical Olympiad | BMO 2019 Shortlist | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof and answer | 99 | |
0kwm | Problem:
The number $5.6$ may be expressed uniquely (ignoring order) as a product $\underline{a} \cdot \underline{b} \times \underline{c} . \underline{d}$ for digits $a, b, c, d$ all nonzero. Compute $\underline{a} \cdot \underline{b}+\underline{c} \cdot \underline{d}$. | [
"Solution:\n\nWe want $\\overline{a b} \\times \\overline{c d} = 560 = 2^{4} \\times 5 \\times 7$. To avoid a zero digit, we need to group the $5$ with the $7$ to get $3.5$ and $1.6$, and our answer is $3.5 + 1.6 = 5.1$."
] | United States | HMMT November 2023 | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Decimals"
] | null | final answer only | 5.1 | |
03wo | On a coordinate plane there are two regions, $M$ and $N$:
$M$ is confined by
$$
\begin{cases}
y \ge 0, \\
y \le x, \\
y \le 2-x
\end{cases}
$$
and $N$ is determined by the inequalities $t \le x \le t+1$, $0 \le t \le 1$. Then the size of the common area of $M$ and $N$ is given by $f(t) = \underline{\hspace{2cm}}$. | [
"As shown in the figure, we have\n$$\n\\begin{aligned}\nf(t) &= S_{\\text{shaded area}} \\\\\n&= S_{\\triangle AOB} - S_{\\triangle OCD} - S_{\\triangle BEF} \\\\\n&= 1 - \\frac{1}{2}t^2 - \\frac{1}{2}(1-t)^2 \\\\\n&= -t^2 + t + \\frac{1}{2}, \\quad 0 \\le t \\le 1.\n\\end{aligned}\n$$\n"
] | China | China Mathematical Competition | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Triangles"
] | English | final answer only | f(t) = -t^2 + t + 1/2 for 0 ≤ t ≤ 1 | |
04cs | One box contains three blue balls and one red ball, and a second box contains three red and one blue ball. First we randomly choose one ball from the first box and put it in the second one, and then we randomly choose one ball from the second box. What is the probability that the chosen ball is red? | [] | Croatia | Mathematica competitions in Croatia | [
"Statistics > Probability > Counting Methods > Other"
] | English | proof and answer | 13/20 | |
027w | Problem:
Pedrinho escreveu dois números inteiros e positivos num pedaço de papel e mostrou para Joãozinho. Depois disso, Pedrinho calculou o dobro do produto destes dois números. Joãozinho somou 21 com o dobro do primeiro número e depois o resultado com o segundo número. Para surpresa dos dois, o resultado foi o mesmo... | [
"Solution:\n\nSejam $x$ e $y$ os números escritos por Pedrinho. Então o número que Pedrinho calculou foi $2 x y$, enquanto que Joãozinho calculou $21+2 x+y$. Como eles encontraram o mesmo resultado, vale que:\n$$\n2 x y=2 x+y+21\n$$\nVamos agora manipular a equação acima:\n$$\n\\begin{aligned}\n2 x y-2 x & =y+21 \\... | Brazil | null | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | (1, 23) and (6, 3) | |
0iun | Problem:
The curves $x^{2} + y^{2} = 36$ and $y = x^{2} - 7$ intersect at four points. Find the sum of the squares of the $x$-coordinates of these points. | [
"Solution:\n\nIf we use the system of equations to solve for $y$, we get $y^{2} + y - 29 = 0$ (since $x^{2} = y + 7$). The sum of the roots of this equation is $-1$. Combine this with $x^{2} = y + 7$ to see that the sum of the square of the possible values of $x$ is $2 \\cdot (-1 + 7 \\cdot 2) = 26$."
] | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | final answer only | 26 | |
05d1 | Kati writes the numbers
$$
2^0, 2^1, 2^2, \dots, 2^{100}, 3^0, 3^1, 3^2, \dots, 3^{100}, 6^0, 6^1, 6^2, \dots, 6^{100}
$$
on the board. In each step, she performs one of the following operations:
(1) She can pick two numbers and replace them with their greatest common divisor and least common multiple; or
(2) She can p... | [
"At all points throughout the process, every number on the board can be written as $2^\\alpha 3^\\beta$ for some $0 \\le \\alpha, \\beta \\le 100$. Moreover, the list of exponents of both $2$ and $3$ does not change throughout the process. Indeed, in a step of the first kind, two numbers $2^{\\alpha_1} 3^{\\beta_1}... | Estonia | Estonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Algebra > Equations and Inequalities > Combinatorial optimization",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | English | proof and answer | Greatest sum: 101 + 2*(6^101 - 1)/5. Least sum: 2^102 - 3*2^51 + 2*3^50 + 3^101. | |
0ehw | Problem:
Kolikšna je velikost središčnega kota, ki pripada krožnemu loku, ki ima enako dolžino kot polmer krožnice?
(A) $\frac{45^{\circ}}{\pi}$
(B) $\frac{90^{\circ}}{\pi}$
(C) $\frac{135^{\circ}}{\pi}$
(D) $\frac{180^{\circ}}{\pi}$
(E) $\frac{270^{\circ}}{\pi}$ | [
"Solution:\nDolžino krožnega loka izračunamo po formuli $l = \\frac{\\pi r \\alpha}{180^{\\circ}}$. Ker je dolžina krožnega loka enaka polmeru kroga, enačbo preuredimo v $r = \\frac{\\pi r \\alpha}{180^{\\circ}}$. Iz enačbe izrazimo $\\alpha$ in dobimo rezultat $\\frac{180^{\\circ}}{\\pi}$."
] | Slovenia | 19. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol Državno tekmovanje | [
"Geometry > Plane Geometry > Circles",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | MCQ | D | |
0k3q | Problem:
Let $a$ and $b$ be real numbers greater than $1$ such that $a b = 100$. The maximum possible value of $a^{\left(\log_{10} b\right)^2}$ can be written in the form $10^{x}$ for some real number $x$. Find $x$. | [
"Solution:\nLet $p = \\log_{10} a$, $q = \\log_{10} b$. Since $a, b > 1$, $p$ and $q$ are positive. The condition $a b = 100$ translates to $p + q = 2$. We wish to maximize\n$$\nx = \\log_{10} a^{\\left(\\log_{10} b\\right)^2} = \\left(\\log_{10} a\\right)\\left(\\log_{10} b\\right)^2 = p q^2\n$$\nBy AM-GM,\n$$\n\\... | United States | HMMT November 2018 | [
"Algebra > Intermediate Algebra > Logarithmic functions",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | 32/27 | |
01d9 | Let $k$ and $t$ be integers with $1 \le k/2 < t < k$. Each square of a $k \times k$ checkerboard is coloured either red or blue. A move consists of choosing a row or a column with at most $t$ red squares and switching the colour of these red squares to blue. Assume that it is possible to make all squares of the checker... | [
"Assume that $m \\ge k+1$. Let $T$ be the set of all initially red squares, and assume that $|T|$ is as small as possible. It follows that $m = k+1$ because otherwise after the first $m - (k+1)$ moves the number of red squares has decreased and still there are $k+1$ moves required, a contradiction.\nIf each row con... | Baltic Way | Baltic Way 2016 | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
02ox | Problem:
(a) A soma de quatro inteiros positivos consecutivos pode ser um número primo? Justifique sua resposta.
(b) A soma de três inteiros positivos consecutivos pode ser um número primo? Justifique sua resposta. | [
"Solution:\n\n(a) Seja $x$ o menor dos números. Então, a soma em questão é\n$$\nx + (x+1) + (x+2) + (x+3) = 4x + 6 = 2(x+3)\n$$\nEste número é par maior que $2$, portanto não pode ser um número primo.\n\n(b) Seja $y$ o menor dos números. Então, a soma em questão é\n$$\ny + (y+1) + (y+2) = 3y + 3 = 3(y+1)\n$$\nEste ... | Brazil | Brazilian Mathematical Olympiad, Nível 2 | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | a) no; b) no | |
00du | There are some cards on the table. Each card has an integer number written on it. Beto performs the following operation many times: he picks two cards from the table, computes the difference between the numbers that are written on them, he writes this difference on his notebook and then removes those two cards from the... | [
"a. To begin, let us note that, as long as there are more than 7 cards on the table, we can always find two cards with the same remainder upon division by 7 (because there are only 7 possible remainders). In that situation, Beto can pick those two cards and write in his notebook the difference, which will be divisi... | Argentina | XXIX Rioplatense Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Modular Arithmetic"
] | English | proof and answer | Yes | |
02f4 | A real number with absolute value less than $1$ is written in each cell of an $n \times n$ array, so that the sum of the numbers in each $2 \times 2$ square is zero. Show that for $n$ odd the sum of all the numbers is less than $n$. | [
"Consider a $L$-shaped piece formed by a square and two neighbouring squares.\nThe sum of its numbers is the opposite of the number that completes the $L$\nto a $2 \\times 2$ square, so the sum of numbers in any $L$ is less than $1$.\nWe proceed by induction on $n$. For $n=3$, divide the array in four regions: a $2... | Brazil | XV OBM | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof only | null | |
063b | Problem:
Die Zentralbank von Sikinien prägt Münzen im Wert von 11 und 12 Kulotnik. Bei einem Einbruch haben 11 sikinische Ganoven einen Tresor geknackt und Münzen im Gesamtwert von 5940 Kulotnik erbeutet. Sie versuchen für eine Weile, die Beute gerecht unter sich aufzuteilen - also so, dass jeder gleich viel erhält - ... | [
"Solution:\n\nDie Ganoven mögen $a$ Münzen im Wert von 11 Kulotnik und $b$ Münzen im Wert von 12 Kulotnik erbeutet haben. Dabei sind $a$ und $b$ zwei nichtnegative ganze Zahlen mit\n$$\n11 a + 12 b = 5940 = 11 \\cdot 540 = 12 \\cdot 495\n$$\nWir nehmen nun $b > 0$ an und versuchen, eine gerechte Aufteilung der Beut... | Germany | Germany TST | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof only | null | |
0i8t | Problem:
A room is built in the shape of the region between two semicircles with the same center and parallel diameters. The farthest distance between two points with a clear line of sight is $12$ m. What is the area (in $\mathrm{m}^{2}$) of the room? | [
"Solution:\n$18 \\pi$\nThe maximal distance is as shown in the figure. Call the radii $R$ and $r$, $R > r$. Then $R^{2} - r^{2} = 6^{2}$ by the Pythagorean theorem, so the area is $(\\pi / 2) \\cdot (R^{2} - r^{2}) = 18 \\pi$.\n\n"
] | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 18π | |
0dl9 | Determine all triples $(a, b, p)$ of positive integers $a, b$ and a prime number $p \ge 3$, such that $2^a + p^{2b}$ is a $(p-1)$-st power of a positive integer. | [] | Saudi Arabia | Saudi Booklet | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | (4, 1, 3) | |
063j | Problem:
Man bestimme alle Paare $(x, y)$ ganzer Zahlen, welche die Gleichung
$$
\sqrt[3]{7 x^{2}-13 x y+7 y^{2}} = |x-y| + 1
$$
erfüllen. | [
"Solution:\nDie Gleichung (1) ist symmetrisch in $x$ und $y$, so dass wir zunächst $x \\geq y$ annehmen können und für $x \\neq y$ zu jeder Lösung $(x, y)$ auch $(y, x)$ als Lösung erhalten. Mit\n$d = x - y \\geq 0$ folgt $\\sqrt[3]{7 d^{2} + x y} = d + 1$. Potenzieren liefert $x^{2} - d x + \\left(-d^{3} + 4 d^{2}... | Germany | 2. Auswahlklausur 2014/2015 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | All integer solutions are given by the following parametrizations with m ∈ {0, 1, 2, ...}:
- (x, y) = (m^3 + 2m^2 − m − 1, m^3 + m^2 − 2m − 1) and its swap (y, x),
- (x, y) = (−m^3 − m^2 + 2m + 1, −m^3 − 2m^2 + m + 1) and its swap (y, x), with the understanding that for m = 1 the second family coincides with the first ... | |
038d | Problem:
The positive integers $l, m, n$ are such that $m-n$ is a prime number and $8\left(l^{2}-m n\right)=2\left(m^{2}+n^{2}\right)+5(m+n) l$. Prove that $11 l+3$ is a perfect square. | [
"Solution:\n\nSetting $p = m - n$ and $q = m + n$ gives\n$$\nm n = \\frac{1}{4}\\left(q^{2} - p^{2}\\right) \\quad \\text{and} \\quad m^{2} + n^{2} = \\frac{1}{2}\\left(q^{2} + p^{2}\\right)\n$$\nHence the given conditions can be written as\n$$\n8 l^{2} - 2 q^{2} + 2 p^{2} = q^{2} + p^{2} + 5 q l\n$$\ni.e.\n$$\np^{... | Bulgaria | 55. Bulgarian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0kc8 | Problem:
Suppose Harvard Yard is a $17 \times 17$ square. There are 14 dorms located on the perimeter of the Yard. If $s$ is the minimum distance between two dorms, the maximum possible value of $s$ can be expressed as $a-\sqrt{b}$ where $a, b$ are positive integers. Compute $100 a+b$. | [
"Solution:\n\nIf two neighboring dorms are separated by a distance of more than $s$, we can move them slightly closer together and adjust the other dorms, increasing $s$. Therefore, in an optimal arrangement, the dorms form an equilateral 14-gon with side length $s$.\n\nBy scaling, the problem is now equivalent to ... | United States | HMMO 2020 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | final answer only | 602 | |
0e44 | A sequence of positive real numbers contains at least 5 different terms. For any two terms of this sequence we can find two more terms with the same product. At least how many terms does the progression have? | [] | Slovenia | Selection Examinations for the IMO | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | 9 | |
0jjz | Problem:
Let $\omega$ be a circle, and let $ABCD$ be a quadrilateral inscribed in $\omega$. Suppose that $BD$ and $AC$ intersect at a point $E$. The tangent to $\omega$ at $B$ meets line $AC$ at a point $F$, so that $C$ lies between $E$ and $F$. Given that $AE = 6$, $EC = 4$, $BE = 2$, and $BF = 12$, find $DA$. | [
"Solution:\n\nAnswer: $2 \\sqrt{42}$\n\nBy power of a point, we have $ED \\cdot EB = EA \\cdot EC$, whence $ED = 12$.\n\nAdditionally, by power of a point, we have $144 = FB^2 = FC \\cdot FA = FC(FC + 10)$, so $FC = 8$.\n\nNote that $\\angle FBC = \\angle FAB$ and $\\angle CFB = \\angle AFB$, so $\\triangle FBC \\s... | United States | HMMT 2014 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | null | proof and answer | 2*sqrt(42) | |
077j | Problem:
In a convex quadrilateral $ABCD$, $\angle ABD = 30^{\circ}$, $\angle BCA = 75^{\circ}$, $\angle ACD = 25^{\circ}$ and $CD = CB$. Extend $CB$ to meet the circumcircle of triangle $DAC$ at $E$. Prove that $CE = BD$. | [
"Solution:\nFirst we show that $\\angle DEC = 30^{\\circ}$. Choose a point $F$ on $AB$ such that $CF = CB$. Join $FC$ and $FD$. Observe that $\\angle DCB = 75^{\\circ} + 25^{\\circ} = 100^{\\circ}$. Since $CD = CB$, we have $\\angle CDB = \\angle CBD = 40^{\\circ}$. Therefore $\\angle CBF = 40^{\\circ} + 30^{\\circ... | India | INMO | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle"
] | null | proof only | null | |
050y | Let $ABC$ be a triangle on the plane. The angle bisector from the vertex $A$ meets the side $BC$ at $P$, and the median from the vertex $B$ meets the side $AC$ at $M$. The lines $AB$ and $MP$ meet at the point $K$. Prove that if $\frac{|PC|}{|BP|} = 2$, then $AP$ and $CK$ are perpendicular. | [
"Let $K'$ be a point on the ray $AB$, such that $B$ is the midpoint of the line segment $AK'$ (Fig. 10). Then $CB$ is the median of the triangle $ACK'$. As $P$ divides this line segment in the ratio $2:1$, $P$ must be the centroid of the triangle $ACK'$. So, $K'M$, which is also a median of the triangle $ACK'$, mus... | Estonia | Estonian Math Competitions | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
09bw | $a + b^2 + (a, b)^3 = ab \cdot (a, b)$ байх бүх $a, b$ натурал тоон хосыг ол. Энд $(a, b)$ нь $a, b$ тоонуудын ХИЕХ. | [
"$a = zx$, $b = zy$ ба $(x, y) = 1$ гэе. Тэгвэл $x + zy^2 + z^2 = z^2 x y$ ба болох тул $x = zt$, $t \\in \\mathbb{Z}$ байна. Иймд $t + y^2 + z = z^2 t y$ буюу $t = \\frac{y^2 + z}{z^2 y - 1}$.\n\n1. $z = 1$ бол $t = \\frac{y^2 + 1}{y - 1} = y + 1 + \\frac{2}{y - 1}$ тул $y = 2$ юмуу $y = 3$ байна. Эндээс $(x, y) =... | Mongolia | Mongolian Mathematical Olympiad 46 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | Mongolian | proof and answer | (a, b) = (5, 2), (5, 3), (4, 2), (4, 6) | |
06ii | In $\triangle ABC$, $\tan \angle CAB = \frac{22}{7}$ and the altitude from $A$ to $BC$ divides $BC$ into segments of lengths $3$ and $17$. Find the area of $\triangle ABC$. (1 mark)
在 $\triangle ABC$ 中, $\tan \angle CAB = \frac{22}{7}$, 且從 $A$ 到 $BC$ 的高把 $BC$ 分成長度 3 和 17 的兩段。求 $\triangle ABC$ 的面積。 (1分) | [] | Hong Kong | HONG KONG PRELIMINARY SELECTION CONTEST | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | English; Chinese | final answer only | 110 | |
0cvp | Each of the trinomials $x^2 + a x + b$ and $x^2 + a x + b + 1$ has at least one real root; moreover, all real roots of those trinomials are integers. Prove that the trinomial $x^2 + a x + b + 2$ has no real roots. | [
"Пусть $D_1, D_2, D_3$ — соответственно дискриминанты этих трёхчленов. Первые два уравнения имеют только целые корни, поэтому $D_1 = m^2, D_2 = n^2$, где числа $m$ и $n$ можно считать целыми неотрицательными. Вычитая из первого равенства второе, получаем, что $4 = m^2 - n^2$, то есть $4 = (m - n)(m + n)$. Числа $m-... | Russia | Regional round | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Intermediate Algebra > Quadratic functions",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | English; Russian | proof only | null | |
012p | Problem:
We draw $n$ convex quadrilaterals in the plane. They divide the plane into regions (one of the regions is infinite). Determine the maximal possible number of these regions. | [
"Solution:\n\nOne quadrilateral produces two regions. Suppose we have drawn $k$ quadrilaterals $Q_{1}, \\ldots, Q_{k}$ and produced $a_{k}$ regions. We draw another quadrilateral $Q_{k+1}$ and try to evaluate the number of regions $a_{k+1}$ now produced. Our task is to make $a_{k+1}$ as large as possible. Note that... | Baltic Way | Baltic Way 2002 mathematical team contest | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Combinatorial Geometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | 4n^2 - 4n + 2 | |
0dlv | Bank is planning to produce and grant a special credit card to its loyal costumers. Each of these credit cards has a unique ID-number that has 16 digits and satisfies the following properties:
(i) The first four digits of the ID-numbers is fixed and is equal to 2024.
(ii) For each pair of cards their corresponding ID-n... | [
"Let the ID-number be $d_1d_2\\ldots d_{16}$, where each $d_i$ is a digit ($0$ to $9$). By condition (i), $d_1d_2d_3d_4 = 2024$ is fixed. So, the remaining $12$ digits $d_5, d_6, \\ldots, d_{16}$ can be chosen.\n\nLet $S$ be the set of all possible $12$-digit strings (from $0$ to $9$ in each position), i.e., $|S| =... | Saudi Arabia | Saudi Booklet | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 10^11 | |
0eg1 | Problem:
Naj bo $\sin \alpha + \sin \beta = 1$ in $\cos \alpha + \cos \beta = -\sqrt{3}$.
a. Izračunaj vrednost izraza $\cos (\alpha - \beta)$.
b. Poišči vse pare realnih števil $\alpha$ in $\beta$, ki ustrezajo danima enačbama. | [
"Solution:\n\na. Adicijski izrek za kosinus nam da $\\cos (\\alpha - \\beta) = \\cos \\alpha \\cos \\beta + \\sin \\alpha \\sin \\beta$. Dani enačbi kvadriramo, da dobimo\n$$\n\\begin{array}{r}\n\\sin^2 \\alpha + 2 \\sin \\alpha \\sin \\beta + \\sin^2 \\beta = 1 \\\\\n\\cos^2 \\alpha + 2 \\cos \\alpha \\cos \\beta ... | Slovenia | 61. matematično tekmovanje srednješolcev Slovenije, Odbirno tekmovanje | [
"Precalculus > Trigonometric functions"
] | null | proof and answer | cos(α − β) = 1; all solutions are α = 5π/6 + 2mπ and β = 5π/6 + 2nπ for integers m, n. | |
0jey | Problem:
Find the number of subsets $S$ of $\{1,2, \ldots, 6\}$ satisfying the following conditions:
- $S$ is non-empty.
- No subset of $S$ has the property that the sum of its elements is $10$. | [
"Solution:\nAnswer: $34$\n\nWe do casework based on the largest element of $S$. Call a set $n$-free if none of its subsets have elements summing to $n$.\n\nCase 1: The largest element of $S$ is $6$. Then $4 \\notin S$. If $5 \\notin S$, then we wish to find all $4$-free subsets of $\\{1,2,3\\}$ (note that $1+2+3=6<... | United States | HMMT November 2013 | [
"Discrete Mathematics > Combinatorics > Inclusion-exclusion"
] | null | final answer only | 34 | |
0kdp | Problem:
Estimate
$$
N = \prod_{n=1}^{\infty} n^{n^{-1.25}}
$$
An estimate of $E > 0$ will receive $\lfloor 22 \min (N / E, E / N) \rfloor$ points. | [
"Solution:\nWe approximate\n$$\n\\ln N = \\sum_{n=1}^{\\infty} \\frac{\\ln n}{n^{5/4}}\n$$\nwith an integral as\n$$\n\\begin{aligned}\n\\int_{1}^{\\infty} \\frac{\\ln x}{x^{5/4}} \\, dx & = \\left.\\left(-4 x^{-1/4} \\ln x - 16 x^{-1/4}\\right)\\right|_{1}^{\\infty} \\\\\n& = 16\n\\end{aligned}\n$$\nTherefore $e^{1... | United States | HMMT February 2020 | [
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products",
"Algebra > Intermediate Algebra > Exponential functions",
"Algebra > Intermediate Algebra > Logarithmic functions"
] | null | final answer only | e^16 (approximately 8,886,111) | |
0bw3 | a) Show that, if $n$ is a positive integer and $f: \mathbb{R} \to \mathbb{R}$ is a derivative, then the function $x \mapsto (f(x))^n$, $x \in \mathbb{R}$ has the intermediate values property.
b) Give an example of a function $f: \mathbb{R} \to \mathbb{R}$ which is a derivative, but none of the functions $x \mapsto (f(... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 68th NMO | [
"Calculus > Differential Calculus > Derivatives"
] | English | proof only | null | |
088t | Problem:
Siano $ABC$ un triangolo acutangolo e $H$ il piede dell'altezza relativa al vertice $A$. Coloriamo ogni punto $P$ interno al triangolo in questo modo: di rosso, se il vertice più vicino a $P$ è $A$; di verde, se il vertice più vicino a $P$ è $B$; di blu, se il vertice più vicino a $P$ è $C$. Sapendo che $AH=3... | [
"Solution:\n\nLa risposta è $198$. Indichiamo con $L, M, N$ i punti medi dei lati $AB, BC, CA$, e con $O$ il circocentro di $ABC$ (punto di incontro degli assi dei lati del triangolo). Per una nota proprietà dell'asse di un segmento, l'insieme dei punti rossi è il quadrilatero $ALON$, di cui bisogna quindi determin... | Italy | UNIONE MATEMATICA ITALIANA SCUOLA NORMALE SUPERIORE DI PISA GARA di SECONDO LIVELLO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 198 | |
02zo | Problem:
A calculadora $MK$-2020 pode efetuar as seguintes três operações com números em sua memória:
1) Determinar se dois números escolhidos são iguais.
2) Adicionar dois números escolhidos.
3) Para os números escolhidos $a$ e $b$, determinar as raízes reais da equação $x^{2} + a x + b = 0$ ou anunciar que tal equaç... | [
"Solution:\n\na) Podemos usar a segunda operação e gerar o número $2z$. Usando a primeira operação, podemos decidir se $z$ e $2z$ são iguais, ou seja, se $z$ é ou não igual a zero.\n\nb) Pela Fórmula de Bhaskara, as raízes de $x^{2} + 2z x + z = 0$ são dadas por:\n$$\nx = -z \\pm \\sqrt{z^{2} - z}\n$$\n\nc) Suponha... | Brazil | Brazilian Mathematical Olympiad | [
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | a) Compare z and 2z using the equality test; they are equal if and only if z = 0, so otherwise z ≠ 0. b) The roots are x = −z ± sqrt(z^2 − z). c) After confirming z ≠ 0, run the quadratic solver on x^2 + 2zx + z; it returns a single real root if and only if z = 1, otherwise z ≠ 1. | |
0l2g | Problem:
The number $17^{6}$ when written out in base 10 contains 8 distinct digits from $1,2, \ldots, 9$, with no repeated digits or zeroes. Compute the missing nonzero digit. | [
"Solution:\n\nObserve that\n\n$$\n17^{6} \\equiv (-1)^{6} = 1 \\bmod 9\n$$\n\nIf $x$ is the missing digit, then the digits of $17^{6}$ sum to $(1+2+\\cdots+9)-x=45-x$, so $45-x \\equiv 1 \\bmod 9$. We conclude $x=8$."
] | United States | HMMT November 2024 | [
"Number Theory > Modular Arithmetic"
] | null | final answer only | 8 | |
0dmd | Problem:
Нека је $k$ природан број. Доказати да за позитивне реалне бројеве $x, y, z$ чији је збир једнак $1$, важи неједнакост
$$
\frac{x^{k+2}}{x^{k+1}+y^{k}+z^{k}}+\frac{y^{k+2}}{y^{k+1}+z^{k}+x^{k}}+\frac{z^{k+2}}{z^{k+1}+x^{k}+y^{k}} \geqslant \frac{1}{7}
$$
Када важи једнакост? | [
"Solution:\n\nДати израз је симетричан, па се без губљења општости може претпоставити да је $x \\geqslant y \\geqslant z$. Тада је\n$$\nx^{k+1}+y^{k}+z^{k} \\leqslant y^{k+1}+z^{k}+x^{k} \\leqslant z^{k+1}+x^{k}+y^{k}\n$$\nЗаиста, довољно је доказати прву неједнакост, тј. да је $x^{k+1}+y^{k} \\leqslant y^{k+1}+x^{... | Serbia | СРПСКА МАТЕМАТИЧКА ОЛИМПИЈАДА | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Sequences and Series > Sums and products"
] | null | proof and answer | Equality holds if and only if x = y = z = 1/3. | |
06vk | In triangle $A B C$, let $A_{1}$ and $B_{1}$ be two points on sides $B C$ and $A C$, and let $P$ and $Q$ be two points on segments $A A_{1}$ and $B B_{1}$, respectively, so that line $P Q$ is parallel to $A B$. On ray $P B_{1}$, beyond $B_{1}$, let $P_{1}$ be a point so that $\angle P P_{1} C=\angle B A C$. Similarly, ... | [
"Solution 1. Throughout the solution we use oriented angles.\nLet rays $A A_{1}$ and $B B_{1}$ intersect the circumcircle of $\\triangle A C B$ at $A_{2}$ and $B_{2}$, respectively. By\n$$\n\\angle Q P A_{2}=\\angle B A A_{2}=\\angle B B_{2} A_{2}=\\angle Q B_{2} A_{2},\n$$\npoints $P, Q, A_{2}, B_{2}$ are concycli... | IMO | IMO 2019 Shortlisted Problems | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Concurrency and Collinearity > Pappus theorem",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
0clw | Problem:
Given a finite group of boys and girls, a covering set of boys is a set of boys such that every girl knows at least one boy in that set; and a covering set of girls is a set of girls such that every boy knows at least one girl in that set. Prove that the number of covering sets of boys and the number of cover... | [
"Solution:\n\nA set $X$ of boys is separated from a set $Y$ of girls if no boy in $X$ is an acquaintance of a girl in $Y$. Similarly, a set $Y$ of girls is separated from a set $X$ of boys if no girl in $Y$ is an acquaintance of a boy in $X$. Since acquaintance is assumed mutual, separation is symmetric: $X$ is sep... | Romanian Master of Mathematics (RMM) | Romanian Master of Mathematics Competition | [
"Discrete Mathematics > Combinatorics > Counting two ways",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Number Theory > Modular Arithmetic > Polynomials mod p"
] | null | proof only | null | |
03w4 | Suppose that $f(x) = \frac{x}{\sqrt{1+x^2}}$ and $f^{(n)}(x) = f[f[f\dots f(x)]]$. Then $f^{(99)}(1) = \underline{\hspace{2cm}}$. | [
"We have\n$$\n\\begin{aligned}\nf^{(1)}(x) &= f(x) = \\frac{x}{\\sqrt{1+x^2}}, \\\\\nf^{(2)}(x) &= f[f(x)] = \\frac{x}{\\sqrt{1+2x^2}}, \\\\\n&\\vdots \\\\\nf^{(99)}(x) &= \\frac{x}{\\sqrt{1+99x^2}}.\n\\end{aligned}\n$$\nTherefore, $f^{(99)}(1) = \\frac{1}{10}$."
] | China | China Mathematical Competition | [
"Precalculus > Functions"
] | English | final answer only | 1/10 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.