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 ⌀ |
|---|---|---|---|---|---|---|---|---|---|
04dj | Prove that there do not exist positive integers $k$ and $n$ such that
$$
k(k + 1)(k + 2)(k + 3) = n(n + 1).
$$ | [
"We have\n$$\nk(k + 1)(k + 2)(k + 3) = (k^2 + 3k)(k^2 + 3k + 2) = (k^2 + 3k + 1)^2 - 1.\n$$\nThat means that $n(n + 1) + 1 = n^2 + n + 1$ has to be a perfect square, but that is impossible since\n$$\nn^2 < n^2 + n + 1 < n^2 + 2n + 1 = (n + 1)^2\n$$\ni.e. $n^2 + n + 1$ is between two consecutive squares.\nTherefore,... | Croatia | Mathematica competitions in Croatia | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof only | null | |
09xf | Find all quadruples $(x_1, x_2, x_3, x_4)$ of real numbers which are solutions of the following system of six equations:
$$
\begin{align*}
x_1 + x_2 &= x_3^2 + x_4^2 + 6x_3x_4, \\
x_1 + x_3 &= x_2^2 + x_4^2 + 6x_2x_4, \\
x_1 + x_4 &= x_2^2 + x_3^2 + 6x_2x_3, \\
x_2 + x_3 &= x_1^2 + x_4^2 + 6x_1x_4, \\
x_2 + x_4 &= x_1^... | [
"Subtracting the second equation from the first yields $x_2 - x_3 = x_3^2 - x_2^2 + 6x_4(x_3 - x_2)$, which we can factor as $0 = (x_3 - x_2)(x_3 + x_2 + 1 + 6x_4)$. We see that $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_4 = 0$. Similarly, we also have either $x_2 = x_3$ or $x_2 + x_3 + 1 + 6x_1 = 0$. Hence, if $x_2 \\neq ... | Netherlands | IMO Team Selection Test 1 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof and answer | All solutions are the permutations of (0, 0, 0, 0), (1/4, 1/4, 1/4, 1/4), (-1/4, -1/4, -1/4, 3/4), and (-1/2, -1/2, -1/2, 5/2). | |
0k0h | Problem:
Find the minimum possible value of
$$
\sqrt{58-42x} + \sqrt{149-140\sqrt{1-x^{2}}}
$$
where $-1 \leq x \leq 1$. | [
"Solution:\nSubstitute $x = \\cos \\theta$ and $\\sqrt{1-x^{2}} = \\sin \\theta$, and notice that $58 = 3^{2} + 7^{2}$, $42 = 2 \\cdot 3 \\cdot 7$, $149 = 7^{2} + 10^{2}$, and $140 = 2 \\cdot 7 \\cdot 10$. Therefore the first term is an application of Law of Cosines on a triangle that has two sides $3$ and $7$ with... | United States | HMMT November 2017 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(109) | |
0ahj | Find all prime numbers $p$ and $q$ which satisfy the equation
$$
(p+q)^p = (q-p)^{2q-1}.
$$ | [
"It cannot be that $q-p=1$, since $(q+p)^p > 1$, for all prime numbers $p$ and $q$.\nLet $r$ be a prime divisor of $q-p$. Then $r$ is also a divisor of $q+p$, so it is a divisor of $2q = (q+p)+(q-p)$ and of $2p = (q+p)-(q-p)$. It follows that $p=q=r$ or $r=2$. The case $p=q$ is impossible, since the right-hand side... | North Macedonia | XVI Junior Macedonian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | p=3, q=5 | |
0dk7 | Let $ABCD$ be a quadrilateral inscribed in a circle ($O$) and some point $I$ lies inside $ABCD$. Consider the lines $d_1, d_2, d_3, d_4$ pass through the midpoints of segments $IA, IB, IC, ID$ respectively and perpendicular to lines $OA, OB, OC, OD$. Line $d_1$ cuts $d_2$ at $P$, line $d_2$ cuts $d_3$ at $Q$, line $d_3... | [
"Let $R$ be the radius of the circle ($O$) and let $M, K$ be the midpoints of $IO, AI$ respectively. Then, according to the property of the midline in triangle $AIO$, we have $MK = \\frac{AO}{2}$ and $MK \\parallel AO$. Thus, we immediately have $MK \\perp SP$ and $MK = \\frac{R}{2}$. Similarly for the other sides.... | Saudi Arabia | Saudi Arabia booklet 2024 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Quadrilaterals > Inscribed/circumscribed quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
05tj | Problem:
Trouver toutes les fonctions $f: \mathbb{R} \rightarrow \mathbb{R}$ telles que
$$
f(x+f(y))+x=f(f(2 x)+y)
$$
pour tous les réels $x$ et $y$. | [
"Solution:\n\nTout d'abord, on constate aisément que, pour tout réel $a$, la fonction $f: x \\rightarrow x+a$ est une solution du problème. Réciproquement, on va montrer qu'il n'existe pas d'autre solution.\n\nDans ce qui suit, on considère une fonction solution $f$, on pose $a=f(0)$, et on notera $\\mathbf{E}(x, y... | France | Préparation Olympique Française de Mathématiques | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | All functions are f(x) = x + a for an arbitrary real constant a. | |
0k27 | Problem:
Let $z$ be a complex number. In the complex plane, the distance from $z$ to $1$ is $2$, and the distance from $z^{2}$ to $1$ is $6$. What is the real part of $z$? | [
"Solution:\nAnswer: $\\frac{5}{4}$\nNote that we must have $|z-1|=2$ and $|z^{2}-1|=6$, so $|z+1|=\\frac{|z^{2}-1|}{|z-1|}=3$. Thus, the distance from $z$ to $1$ in the complex plane is $2$ and the distance from $z$ to $-1$ in the complex plane is $3$. Thus, $z$, $1$, $-1$ form a triangle with side lengths $2$, $3$... | United States | HMMT November 2018 | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Complex numbers in geometry",
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Algebra > Intermediate Algebra > Complex numbers"
] | null | proof and answer | 5/4 | |
0gt7 | There are $100$ empty red and $k$ empty white baskets. At each step we choose one red and one white basket and add equal amounts of water into two chosen baskets. It was observed that after finite number of steps all baskets are non empty and any two baskets simultaneously chosen at some step contain equal amounts of w... | [
"Answer: $k=100$.\n\n$k=100$ is obviously possible: partition all baskets into pairs of red and white baskets and in each of $100$ steps choose baskets from some new pair and add any amount of water into chosen baskets.\n\nNow let us show that $k=100$ is the only possible value. Define a graph on $100+k$ vertices w... | Turkey | Team Selection Test | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | 100 | |
0fr5 | Problem:
Sea $O$ un punto interior del triángulo $A B C$ y sean $M, N$ y $P$ las intersecciones de $A O$ con $B C$, $B O$ con $C A$ y $C O$ con $A B$, respectivamente. Demostrar que de entre los seis triángulos que se forman, hay al menos dos cuya área es menor o igual que $[A B C] / 6$. | [
"\n\nSolution:\n\nSea $S = [A B C]$, $A_{1} = [B O M]$, $A_{2} = [M O C]$, $A_{3} = [N O C]$, $A_{4} = [A O N]$, $A_{5} = [A O P]$ y $A_{6} = [B O P]$. Usando el teorema de Ceva obtenemos que\n$$\n\\frac{B M}{M C} \\cdot \\frac{C N}{N A} \\cdot \\frac{A P}{P B} = \\frac{A_{1}}{A_{2}} \\cdot... | Spain | FASE LOCAL DE LA OLIMPIADA MATEMÁTICA ESPAÑOLA. | [
"Geometry > Plane Geometry > Concurrency and Collinearity > Ceva's theorem",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
04fe | The angle at the vertex $B$ in a triangle $ABC$ is $120^\circ$. Let $A_1$, $B_1$ and $C_1$ be the points on the segments $\overline{BC}$, $\overline{CA}$ and $\overline{AB}$ respectively, such that $AA_1$, $BB_1$ and $CC_1$ are angle bisectors of the triangle $ABC$. Determine the angle $\angle A_1B_1C_1$. (Serbia 1997) | [
"Let $X$ be any point on the extension of the segment $\\overline{AB}$ over the vertex $B$. Note that $\\angle ABB_1 = \\angle B_1BC = \\angle CBX = 60^\\circ$.\n\n\n\nThis implies that the point $A_1$ lies on the angle bisector of the angle $\\angle B_1BX$, and it also lies on the angle bi... | Croatia | Mathematica competitions in Croatia | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 90° | |
0ben | Consider a circle centered at $O$ with radius $r$ and a line $\ell$ not passing through $O$. A grasshopper is jumping to and fro between the points of the circle and the line, the length of each jump being $r$. Prove that there are at most 8 points for the grasshopper to reach.
 | [
"We assume that, when having the choice between only two places to jump to, the grasshopper never jumps back to the point from which he got to that place. Let us denote by $P_1$ the starting point of the grasshopper, with $P_2$ the point on the line on which he has jumped from $P_1$, and so on. As the length of the... | Romania | 64th NMO Selection Tests for the Junior Balkan Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 8 | |
055v | Points $P$ and $Q$ are chosen on the side $BC$ of triangle $ABC$ in such a way that $P$ lies between $B$ and $Q$, and rays $AP$ and $AQ$ trisect the angle $BAC$. The line parallel to $AQ$ and passing through $P$ meets the side $AB$ of the triangle at point $D$, and the line parallel to $AP$ and passing through $Q$ meet... | [
"Assume that $DE$ is a midsegment of $ABC$, then $D$ is the midpoint of $AB$ (Fig. 24). As $DP \\parallel AQ$, $DP$ is a midsegment of triangle $ABQ$. Hence, $P$ is a midpoint of $BQ$ and $AP$ is a median of triangle $ABQ$. As rays $AP$ and $AQ$ trisect the angle $BAC$, $AP$ is a bisector of angle $QAB$. Thus, $ABQ... | Estonia | Estonian Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles"
] | English | proof only | null | |
0bjv | The medians $AD$, $BE$ and $CF$ of triangle $ABC$ intersect at $G$. Let $P$ be a point lying in the interior of the triangle, not belonging to any of its medians. The line through $P$ parallel to $AD$ intersects the side $BC$ at $A_1$. Similarly one defines the points $B_1$ and $C_1$. Prove that
$$
\overrightarrow{A_1D... | [
"Let us draw through point $P$ parallels to the triangle's sides. If the parallels to $AB$ and $AC$ intersect the side $BC$ at points $A_B$ and $A_C$, then the triangles $ABC$ and $PA_B A_C$ are similar, and since $PA_1$ is parallel to the median $AD$, it follows that $A_1$ is the midpoint of the line segment $A_B ... | Romania | 65th Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Vectors"
] | null | proof only | null | |
0j5w | Problem:
Let $ABC$ be a triangle, and let $D$, $E$, and $F$ be the midpoints of sides $BC$, $CA$, and $AB$, respectively. Let the angle bisectors of $\angle FDE$ and $\angle FBD$ meet at $P$. Given that $\angle BAC = 37^{\circ}$ and $\angle CBA = 85^{\circ}$, determine the degree measure of $\angle BPD$. | [
"Solution:\n\nAnswer: $61^{\\circ}$\n\nBecause $D$, $E$, $F$ are midpoints, we have $ABC \\sim DEF$. Furthermore, we know that $FD \\parallel AC$ and $DE \\parallel AB$, so we have\n$$\n\\angle BDF = \\angle BCA = 180^{\\circ} - 37^{\\circ} - 85^{\\circ} = 58^{\\circ}\n$$\nAlso, $\\angle FDE = \\angle BAC = 37^{\\c... | United States | Harvard-MIT November Tournament | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof and answer | 61° | |
0cax | Let $m$ and $n$ be positive integers, where $m < 2^n$. Determine the smallest possible number of not necessarily pairwise distinct powers of 2 that add up to $m \cdot (2^n - 1)$. | [
"The required minimum is $n$.\nTo prove this, notice that the sum of two like powers of 2 is again a power of 2, so the number of powers of 2 that add up to a positive integer $k$ can successively be decreased while keeping the sum constant. The process then ends up with $k$ being expressed as a sum of pairwise dis... | Romania | THE 73rd NMO SELECTION TESTS FOR THE BALKAN AND INTERNATIONAL MATHEMATICAL OLYMPIADS - FIRST SELECTION TEST | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Invariants / monovariants"
] | null | proof and answer | n | |
0g5f | 試求所有滿足下式的非負整數數對 $(m, n)$:
$$
m^2 + 2 \cdot 3^n = m(2^{n+1} - 1).
$$ | [
"(6, 3), (9, 3), (9, 5) 和 (54, 5).\n\n注意到當 $n$ 固定時, 這是一個 $m$ 的二次方程; $m$ 有整數解的條件是判別式是完全平方數。解 $n = 0, 1, 2, 3, 4, 5$ 的二次方程可知 $n \\le 5$ 時恰有如前所述的四組解。以下我們證明 $n \\ge 6$ 時無解。\n\n假設 $(m, n)$ 滿足本題的方程式且 $n \\ge 6$, 則 $m|2 \\cdot 3^n = m(2^{n+1}-m-1)$,\n故 $m$ 可以寫成 $m = 3^p$ 或 $m = 2 \\cdot 3^q$.\n\n在第一種情形, 令 $q = n - p$, 有\n... | Taiwan | 二〇一一數學奧林匹亞競賽第三階段選訓營 | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Modular Arithmetic > Fermat / Euler / Wilson theorems",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Intermediate Algebra > Quadratic functions"
... | null | proof and answer | [(6, 3), (9, 3), (9, 5), (54, 5)] | |
0i0x | Problem:
If $a, b, c, d, e, f$ are real numbers such that $a+b+c+d+e+f=0$ and $a^{3}+b^{3}+c^{3}+d^{3}+e^{3}+f^{3}=0$, prove that $(a+c)(a+d)(a+e)(a+f)=(b+c)(b+d)(b+e)(b+f)$. | [
"Solution:\nFirst, let $S$ be the sum of all possible products of three distinct variables from the set $\\{a, b, c, d, e, f\\}$, i.e. $S=abc+abd+abe+\\cdots+def$. Now, we assert that\n$$\n\\begin{aligned}\n& (a+b+c+d+e+f)^3 + 2\\left(a^3+b^3+c^3+d^3+e^3+f^3\\right) \\\\\n& \\quad - 3(a+b+c+d+e+f)\\left(a^2+b^2+c^2... | United States | Berkeley Math Circle | [
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions",
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
0aoy | Problem:
Let $ABC$ be an equilateral triangle. Let $\overrightarrow{AB}$ be extended to a point $D$ such that $B$ is the midpoint of $\overline{AD}$. A variable point $E$ is taken on the same plane such that $DE = AB$. If the distance between $C$ and $E$ is as large as possible, what is $\angle BED$? | [
"Solution:\n\n$15^\\circ$\n\nTo make $C$ and $E$ as far as possible, $C$, $D$, $E$ must be collinear in that order.\n\nWith $\\angle ABC = 60^\\circ$, we have $\\angle CBD = 120^\\circ$. Since $BC = BD$, we then have $\\angle CDB = \\frac{1}{2}\\left(180^\\circ - 120^\\circ\\right) = 30^\\circ$. Finally, since $BD ... | Philippines | Tenth Philippine Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 15° | |
0fqu | Problem:
Considera el conjunto de números enteros positivos $n$ cumpliendo que $1 \leq n \leq 1000000$. En ese conjunto, indica si es mayor la cantidad de números que pueden expresarse de la forma $a^{3}+m b^{2}$, con $a, b \in \mathbb{N}$ y $m \in \{0,2,4,6,8\}$, o la cantidad de números que no pueden expresarse de e... | [
"Solution:\n\nComo $0 \\leq a^{3}, b^{2} \\leq a^{3}+m b^{2} \\leq 1000000$, tendremos que $0 \\leq a \\leq 100$ y $0 \\leq b \\leq 1000$. Para $m=0$, tenemos que $a^{3}+m b^{2}=a^{3}$ y la cantidad de números de esa forma será $100$. Para cada $m=2,4,6,8$, la cantidad de números será menor o igual que $100 \\cdot ... | Spain | OME fase local | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | proof and answer | More numbers cannot be expressed in that form. | |
08p4 | Problem:
If $x, y, z$ are non-negative real numbers such that $x^{2}+y^{2}+z^{2}=x+y+z$, then show that:
$$
\frac{x+1}{\sqrt{x^{5}+x+1}}+\frac{y+1}{\sqrt{y^{5}+y+1}}+\frac{z+1}{\sqrt{z^{5}+z+1}} \geq 3
$$
When does the equality hold? | [
"Solution:\nFirst we factor $x^{5}+x+1$ as follows:\n$$\n\\begin{aligned}\nx^{5}+x+1 & =x^{5}-x^{2}+x^{2}+x+1=x^{2}\\left(x^{3}-1\\right)+x^{2}+x+1=x^{2}(x-1)\\left(x^{2}+x+1\\right)+x^{2}+x+1 \\\\\n& =\\left(x^{2}+x+1\\right)\\left(x^{2}(x-1)+1\\right)=\\left(x^{2}+x+1\\right)\\left(x^{3}-x^{2}+1\\right)\n\\end{al... | JBMO | Junior Balkan Mathematics Olympiad | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof and answer | Equality holds only for x = y = z = 0. | |
0hdz | Vika chose a 20-letter word that consists only of letters $A$ and $B$. Oleksii wants to know what Vika's word is. He can ask Vika if there are more $A$'s or $B$'s among several (possibly one) consecutive letters of her word. If there are as many $A$'s as there are $B$'s, Vika's answer may be any of the two letters. Wha... | [
"It is clear how Oleksii can determine the word in 20 questions – it suffices to ask about each letter separately.\n\nWe want to show that smaller number of questions would not be enough. Suppose Oleksii determined the word in no more than 19 questions. Suppose Vika chose a word that consists of 20 letters $A$. Cle... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Algorithms",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | null | proof and answer | 20 | |
02vr | Problem:
Seja $A$ um conjunto infinito de inteiros positivos. Sabe-se que se tomarmos qualquer subconjunto finito $B$ do conjunto $A$ existe um inteiro positivo $b$ maior que 1 tal que $b$ divide todos os elementos do conjunto $B$. Prove que existe um inteiro positivo $d$ maior que 1 que divide todos os elementos do c... | [
"Solution:\n\nConsidere um elemento $a$ do conjunto $A$ e sejam $p_{1}, p_{2}, \\ldots, p_{k-1}, p_{k}$ os divisores primos de $a$. Se algum primo $p_{i}$ divide todos os elementos do conjunto $A$, então $d = p_{i}$ satisfaz o enunciado. Vamos provar que isso obrigatoriamente acontece. Suponha, por absurdo, que par... | Brazil | Brazilian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof only | null | |
0juc | Problem:
Let $P$ be an odd-degree integer-coefficient polynomial. Suppose that $x P(x) = y P(y)$ for infinitely many pairs $x, y$ of integers with $x \neq y$. Prove that the equation $P(x) = 0$ has an integer root. | [
"Solution:\n\nLet $n$ be the (odd) degree of $P$. Suppose, for contradiction, that $P$ has no integer roots, and let $Q(x) = x P(x) = a_{n} x^{n+1} + a_{n-1} x^{n} + \\cdots + a_{0} x^{1}$. WLOG $a_{n} > 0$, so there exists $M > 0$ such that $Q(M) < Q(M+1) < \\cdots$ and $Q(-M) < Q(-M-1) < \\cdots$ (as $n+1$ is eve... | United States | HMMT Invitational Competition | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | null | proof only | null | |
09kk | Consider all positive integers with at least 4 digits that can be formed using the digits $0$, $1$, $2$, and $3$. How many of these integers contain at least four occurrences of the digit $1$? | [] | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Other"
] | English | final answer only | infinitely many | |
06co | Determine all functions $f : \mathbb{R} \to \mathbb{R}$ such that $f(x + y f(x)) = f(x) + x f(y)$ for all $x, y \in \mathbb{R}$. | [
"The solutions are $f(x) = 0$ for all $x \\in \\mathbb{R}$ and $f(x) = x$ for all $x \\in \\mathbb{R}$.\n\nIt is easy to check that both are solutions. Label the equation as follows.\n$$\nf(x + y f(x)) = f(x) + x f(y) \\tag{1}\n$$\nPutting $x = 1$ in (1), we have\n$$\nf(1 + y f(1)) = f(1) + f(y). \\tag{2}\n$$\nIf $... | Hong Kong | CHKMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity"
] | null | proof and answer | f(x) = 0 for all real x; f(x) = x for all real x | |
076w | Let $m$ and $n$ be two integers and define $a_0 = m$, $a_1 = n$ and $a_{k+1} = 4a_k - 5a_{k-1}$ for $k \ge 1$. If $p > 5$ is a prime such that $p-1$ is divisible by $4$, then show that there are integers $m$ and $n$ such that $p$ does not divide $a_k$ for any $k \ge 0$. | [
"Let $t$ be an integer such that $p$ divides $t^2 + 1$. There exists such an integer since $p-1$ is divisible by $4$. Let $m = 1$ and $n = t + 2$. Then\n$$\nn^2 = t^2 + 4t + 4 \\equiv 4t + 3 \\equiv 4n - 5m \\pmod{p}.\n$$\nTherefore, if $a_0 = 1$ and $a_1 = n$ then $a_2 \\equiv n^2 \\pmod{p}$. By induction, it is e... | India | India_2017 | [
"Number Theory > Residues and Primitive Roots > Quadratic residues",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations"
] | English | proof only | null | |
09yo | Find all functions $f: \mathbb{Z}_{>0} \to \mathbb{Z}_{>0}$ for which $f(n) \mid f(m) - n$ if and only if $n \mid m$ for all natural numbers $m$ and $n$. | [
"Substituting $m = n$ gives $f(n) \\mid f(n) - n$, so for all natural numbers $n$ we have $f(n) \\mid n$. Applying this to the original condition, it follows that $f(n) \\mid f(m)$ if and only if $n \\mid m$.\n\nWe show that $f(n) = n$ by induction on the number of prime factors of $n$. The base of the induction is... | Netherlands | BxMO Team Selection Test | [
"Algebra > Algebraic Expressions > Functional Equations",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | English | proof and answer | f(n) = n for all positive integers n | |
0he5 | Consider a white square $ABCD$ of size $8 \times 8$, that consists of $64$ unit squares of size $1 \times 1$. In a turn one can choose any
a) square;
b) rectangle,
that consists of a whole number of unit squares and contains at least one of the vertices of the square $ABCD$, and change a color of every unit square in... | [
"a) Clearly, choosing the same square and acting in it twice is equivalent to not acting at all. Thus, we can assume that every square can be chosen not more than once. There are $32$ squares for which one can make a turn. Thus, there are not more than $2^{32}$ different colorings, which is less than $2^{64}$ – the... | Ukraine | 60th Ukrainian National Mathematical Olympiad | [
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | a) No. b) Yes. | |
0481 | Prove that there exist integers $x_1, x_2, \dots, x_{10}, y_1, y_2, \dots, y_{10}$ satisfying:
(1) For $i = 1, 2, \dots, 10$, $|x_i| \le 10^{10}$ and $|y_i| \le 10^{10}$;
(2) The point set in the plane
$$
X = \left\{ \left( \sum_{i=1}^{10} a_i x_i, \sum_{i=1}^{10} a_i y_i \right) \middle| a_1, a_2, \dots, a_{10} \in \{... | [
"**Proof 1:** Take $x_i = 3^i$, $y_i = 9^i$ ($i = 1, 2, \\dots, 10$). We show these satisfy the conditions. (1) and (2) are obvious. For (3), we use:\n**Lemma:** If real numbers $z_1, \\dots, z_m$ satisfy $|z_{i+1}| \\ge 2|z_i|$ for $i = 1, 2, \\dots, m-1$, then for any $r_1, \\dots, r_m \\in \\{-1, 0, 1\\}$ not al... | China | 2025 International Mathematical Olympiad China National Team Selection Test | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Discrete Mathematics > Combinatorics > Expected values"
] | English | proof only | null | |
0kpq | Problem:
Rectangle $R_{0}$ has sides of lengths $3$ and $4$. Rectangles $R_{1}$, $R_{2}$, and $R_{3}$ are formed such that:
- all four rectangles share a common vertex $P$,
- for each $n=1,2,3$, one side of $R_{n}$ is a diagonal of $R_{n-1}$,
- for each $n=1,2,3$, the opposite side of $R_{n}$ passes through a vertex o... | [
"Solution:\n\nLet $ABCD$ be $R_{0}$ such that $\\overline{AB}=3$ and $\\overline{BC}=4$. Then, let $\\overline{AC}$ be a side length of $R_{1}$ and let the other two vertices be $E$ and $F$ such that $B$ lies on segment $EF$. Notice that the area of $\\triangle ABC$ is both half of the area of $R_{0}$ and half of t... | United States | HMMT February 2022 | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof and answer | 30 | |
09le | Let $n$ be a positive integer. Consider an $n \times n$ chessboard with certain cells colored green. A rook can be placed on any green cell and moves only to other green cells, changing its direction horizontally and vertically with each subsequent move. It is important to note that remaining in the same cell is not co... | [
"Let $N$ be the number of green cells. Let's construct a bipartite graph $G$ on $n$ rows and $n$ columns that $i$-th row connects $j$-th column iff the cell at the intersection of $i$-th row and $j$-th column is colored green. This graph has $N$ edges and contains no cycles of length greater than six. Now let's pro... | Mongolia | Mongolian Mathematical Olympiad | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | English | proof only | null | |
044v | Given rational number $r = \frac{p}{q} \in (0, 1)$, $p$, $q$ are coprime positive integers, and $pq$ divides $3600$. The number of such rational numbers $r$ is ______. | [
"Suppose set $\\Omega = \\{ r \\mid r = \\frac{p}{q},\\ p, q \\in \\mathbb{N}_+,\\ (p, q) = 1,\\ pq \\mid 3600 \\}$.\nWe consider the reduced fractional form $\\frac{p}{q}$ of any element $r$ of $\\Omega$. Since the standard factorization of $3600$ is $2^4 \\times 3^2 \\times 5^2$, we can set $p = 2^A \\times 3^B \... | China | China Mathematical Competition | [
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Discrete Mathematics > Combinatorics > Enumeration with symmetry"
] | null | final answer only | 112 | |
0k4s | Problem:
Let $G$ be a simple graph with $k$ connected components, which have $a_{1}, \ldots, a_{k}$ vertices, respectively. Determine the number of ways to add $k-1$ edges to $G$ to form a connected graph, in terms of the numbers $a_{i}$. | [
"Solution:\n\nThe answer is\n$$\na_{1} \\ldots a_{k}\\left(a_{1}+\\cdots+a_{k}\\right)^{k-2}\n$$\n\nWe will show that\n$$\nf\\left(a_{1}, \\ldots, a_{k}\\right)=k !\\left(a_{1} \\ldots a_{k}\\right)\\left(a_{1}+\\cdots+a_{k}\\right)^{k-2}\n$$\ncounts the number of ways to pick $k-1$ edges, in order. The proof is by... | United States | Berkeley Math Circle: Monthly Contest 1 | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | a1⋯ak · (a1 + ⋯ + ak)^{k−2} | |
0c8y | Let $\triangle ABC$ be a triangle in which $\angle ABC = 75^\circ$ and $\angle BAC = 45^\circ$. One considers the points $F$, $X$ and $Y$ such that $F$ is the projection of $B$ on $AC$, $\overline{CX} = \frac{1}{2}\overline{BC}$ and $\overline{BY} = \overline{FX} + \overline{FA}$. Prove that the centroid of triangle $A... | [
"We will prove that $YF$ is the Euler line of the triangle $ABX$, hence it contains the centroid of that triangle.\n\nBecause $BFA$ is a right isosceles triangle, it follows that $BF = AF$. In the right triangle $BFC$ we have $\\angle CBF = 30^\\circ$, hence $FC = \\frac{BC}{2}$, thus $FC = CX$, and a short computa... | Romania | Romanian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Analytic / Coordi... | English | proof only | null | |
0g00 | Problem:
Finde alle Funktionen $f: \mathbb{R} \rightarrow \mathbb{R}$, sodass für alle $x, y \in \mathbb{R}$ gilt:
$$
(y+1) f(x)+f(x f(y)+f(x+y))=y
$$ | [
"Solution:\nSoit $(y+1) f(x)+f(x f(y)+f(x+y))=y$. On remarque que $f(x) \\equiv 1$ n'est pas une solution. Donc il existe $x_{0}$ tel que $f\\left(x_{0}\\right) \\neq 1$. En substituant $x=x_{0}$, on obtient :\n$$\nf\\left(x_{0} f(y)+f\\left(x_{0}+y\\right)\\right)=y\\left(1-f\\left(x_{0}\\right)\\right)-f\\left(x_... | Switzerland | SMO - Finalrunde | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | null | proof and answer | f(x) = -x | |
0e56 | In the triangle $ABC$ the bisector of the angle $\angle BAC$ meets the segment $BC$ at $D$. The triangle $ADC$ is isosceles with the apex at $D$ and the lengths of the segments $CD$ and $BD$ are $|CD| = 36$ and $|BD| = 64$. Find the lengths of the sides of the triangle $ABC$. | [
"Obviously $|BC| = 100$. Let us find the lengths of the other sides. We have $\\angle BAD = \\angle DAC = \\angle ACB$, so the triangles $ABD$ and $CBA$ are congruent and $\\frac{|AC|}{|AD|} = \\frac{|AB|}{|BD|} = \\frac{|BC|}{|AB|}$. The second equality implies $|AB| = \\sqrt{|BD|^2 \\cdot |BC|} = \\sqrt{64 \\cdot... | Slovenia | National Math Olympiad | [
"Geometry > Plane Geometry > Triangles",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | AB = 80, AC = 45, BC = 100 | |
06ih | Find the sum of all positive integers $m$ such that for each such $m$, there exists a positive integer $n$ satisfying $\left\lfloor \frac{(2m+1)^n}{m(3m+1)} \right\rfloor = 2013$, where $\lfloor x \rfloor$ denotes the greatest integer not exceeding $x$. | [
"The answer is $759$.\n\nFor $m = 1, 2, 3$, the left-hand side becomes $\\left[\\frac{3^n}{4}\\right]$, $\\left[\\frac{5^n}{14}\\right]$, $\\left[\\frac{7^n}{30}\\right]$. Since\n$$\n\\left[\\frac{3^8}{4}\\right] < 2013 < \\left[\\frac{3^9}{4}\\right], \\quad \\left[\\frac{5^6}{14}\\right] < 2013 < \\left[\\frac{5^... | Hong Kong | IMO HK TST | [
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof and answer | 759 | |
0jdl | Problem:
Wesyu is a farmer, and she's building a cao (a relative of the cow) pasture. She starts with a triangle $A_{0} A_{1} A_{2}$ where angle $A_{0}$ is $90^{\circ}$, angle $A_{1}$ is $60^{\circ}$, and $A_{0} A_{1}$ is $1$. She then extends the pasture. First, she extends $A_{2} A_{0}$ to $A_{3}$ such that $A_{3} A... | [
"Solution:\n\nAnswer: $\\sqrt{3}$\n\nFirst, note that for any $i$, after performing the operation on triangle $A_{i} A_{i+1} A_{i+2}$, the resulting pasture is triangle $A_{i+1} A_{i+2} A_{i+3}$. Let $K_{i}$ be the area of triangle $A_{i} A_{i+1} A_{i+2}$. From $A_{n+1} A_{n-2} = \\frac{1}{2^{n} - 2} A_{n} A_{n-2}$... | United States | HMMT 2013 | [
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry"
] | null | proof and answer | sqrt(3) | |
036s | Problem:
In the cells of a square table the numbers $1$, $0$ or $-1$ are written in such a way that there is exactly one $1$ and exactly one $-1$ in every row and in every column. Is it always possible to obtain the opposite table by rearranging the rows and the columns of the initial table? (Two tables are called opp... | [
"Solution:\n\nWe shall prove that one can obtain the opposite table by rearranging the rows and columns of the initial table. Denote the columns from left to right and the rows from up to down by $1,2, \\ldots, n$. Denote by $a_{ij}$ the number written in the $i$-th row and $j$-th column.\n\nExchanging rows and col... | Bulgaria | Team selection test for 47. IMO | [
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Algebra > Linear Algebra > Matrices"
] | null | proof only | null | |
07ja | Given the triangle $ABC$ with orthocenter $H$. The point $S$ is on the circumcircle of $AHC$ such that $\angle ASB = 90^\circ$. The point $P$ on the ray $AC$ is such that $\angle APS = \angle BAS$. Prove that the circumcircle of $BPC$, the line $CS$, and the circle with the diameter $AC$, has another common point diffe... | [
"Let $D$ be the foot of altitude from $A$. According to the statement of the problem, $ASDB$ is cyclic. Then $\\angle SDC = \\angle SAB = \\angle APS$. Therefore, $CSDP$ would also be cyclic. Let $BC$ meet the circumcircle of $AHD$ for the second time at $E$. Since $\\angle AEC = \\angle AHC = 180^\\circ - \\angle ... | Iran | 41th Iranian Mathematical Olympiad | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneou... | null | proof only | null | |
05kt | Problem:
Pour tout entier strictement positif $x$, on note $S(x)$ la somme des chiffres de son écriture décimale.
Soit $k>0$ un entier. On définit la suite $\left(x_{n}\right)$ par $x_{1}=1$ et $x_{n+1}=S\left(k x_{n}\right)$ pour tout $n>0$.
Prouver que $x_{n}<27 \sqrt{k}$, pour tout $n>0$. | [
"Solution:\n\nLemme. Pour tout entier $s \\geqslant 0$, on a $10^{s} \\geqslant (s+1)^{3}$.\nPreuve du lemme. On raisonne par récurrence sur $s$.\nPour $s=0$, on a $10^{0}=1=(0+1)^{3}$.\nSupposons l'inégalité vraie pour la valeur $s$. Alors :\n$$\n10^{s+1} = 10 \\times 10^{s} \\geqslant 10(s+1)^{3}\n$$\nd'après l'h... | France | Olympiades Françaises de Mathématiques | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Number Theory > Other",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Discrete Mathematics > Combinatorics > Induction / smoothing"
] | null | proof only | null | |
018t | Let $AB$ and $CD$ be two diameters of the circle $C$. For an arbitrary point $P$ on $C$, let $R$ and $S$ be the feet of the perpendiculars from $P$ to $AB$ and $CD$, respectively. Show that the length of $RS$ is independent from the choice of $P$. | [
"Let $O$ be the centre of $C$. Then $P$, $R$, $S$, and $O$ are points on a circle $C'$ with diameter $OP$, equal to the radius of $C$. The segment $RS$ is a chord in this circle subtending the angle $AOC$ or its supplementary angle. Since the angle as well as radius of $C'$ are independent of $P$, so is $RS$.\n\n![... | Baltic Way | Baltic Way 2011 Problem Shortlist | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0inp | Problem:
A convex quadrilateral is determined by the points of intersection of the curves $x^{4}+y^{4}=100$ and $x y=4$; determine its area. | [
"Solution:\nAnswer: $4 \\sqrt{17}$. By symmetry, the quadrilateral is a rectangle having $x=y$ and $x=-y$ as axes of symmetry. Let $(a, b)$ with $a>b>0$ be one of the vertices. Then the desired area is\n$$\n(\\sqrt{2}(a-b)) \\cdot (\\sqrt{2}(a+b)) = 2\\left(a^{2}-b^{2}\\right) = 2 \\sqrt{a^{4}-2 a^{2} b^{2}+b^{4}} ... | United States | Harvard-MIT Mathematics Tournament | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Cartesian coordinates",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 4 sqrt(17) | |
0c4r | A triangle $ABC$ has $AB = AC$ and $\angle BAC = 110^\circ$. A point $D$ is taken on the perpendicular bisector of the segment $[AC]$, inside triangle $ABC$, so that $\angle DAC = 25^\circ$, and a point $E$ is taken on $BC$, so that $BE = AD$.
a) Find the measure of $\angle AED$.
b) Prove that $D$ is the circumcenter... | [] | Romania | SHORTLISTED PROBLEMS FOR THE 2019 ROMANIAN NMO | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Circles > Tangents"
] | English | proof and answer | ∠AED = 55°. Moreover, D is the circumcenter of triangle AEC. | |
057v | Find all pairs $(a, b)$ of positive integers such that $a \ge b$ and
$$
\frac{1}{a} + \frac{1}{b} = \frac{1}{2021}.
$$ | [
"*Solution 1:* Let $d = \\gcd(a,b)$, $a = da'$, $b = db'$. The given equation reduces to $\\frac{1}{da'} + \\frac{1}{db'} = \\frac{1}{2021}$ which is equivalent to\n$$\n2021(a' + b') = da'b'. \\qquad (4)\n$$\nAs $a'$ and $b'$ are relatively prime to each other, they both are relatively prime to $a' + b'$, implying ... | Estonia | Estonian Math Competitions | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis... | English | proof and answer | The pairs are (4086462, 2022), (97008, 2064), (88924, 2068), (4230, 3870), and (4042, 4042). | |
0gkn | Let $\triangle ABC$ be a triangle with $AB \leq AC$ and let $P$ be an interior point on the angle bisector of $\angle BAC$. Let $D, E$ be points on the segments $PC, PB$ respectively such that $\angle PBD = \angle PCE$. The line $BD$ meets $AC$ at $X$, and $CE$ meets $AB$ at $Y$.
Prove that $BX \leq CY$. | [
"We use Kelly's lemma\n\n**Lemma 1** (Kelly). Given a triangle $ABC$. Suppose the cevians $BE$ and $CF$ are such that $\\angle CBE \\geq \\angle BCF$ and $\\angle ABE \\geq \\angle ACF$. Then $BE \\leq CF$.\n\n*Proof* (From Crux). Choose $Q$ on the segment $AE$ so that $\\angle QBE = \\angle QCF$. Let $CF$ meets $B... | Thailand | Thailand Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | English | proof only | null | |
07zk | Problem:
Sia $f$ una funzione definita nell'insieme degli interi positivi a valori interi positivi.
Diciamo che:
- $f$ è crescente se $n < m$ implica $f(n) < f(m)$
- $f$ è moltiplicativa se $\operatorname{MCD}(m, n) = 1$ implica $f(n m) = f(n) \cdot f(m)$
- $f$ è completamente moltiplicativa se $f(n m) = f(n) \cdot f(... | [
"Solution:\n\na.\nSe $m$ è il minimo intero positivo tale che $f(m) < m$, gli $m-1$ numeri che precedono $m$ devono avere tutti immagine minore di $f(m)$ e dunque non possono essere tutti diversi perché si hanno a disposizione al più $m-2$ valori, assurdo.\n\nb.\nSe $m$ è il minimo intero positivo tale che $f(m) > ... | Italy | GARA NAZIONALE di MATEMATICA | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Number Theory > Other"
] | null | proof only | null | |
06vo | The incircle $\omega$ of acute-angled scalene triangle $A B C$ has centre $I$ and meets sides $B C$, $C A$, and $A B$ at $D, E$, and $F$, respectively. The line through $D$ perpendicular to $E F$ meets $\omega$ again at $R$. Line $A R$ meets $\omega$ again at $P$. The circumcircles of triangles $P C E$ and $P B F$ meet... | [
"Step 1. The external bisector of $\\angle B A C$ is the line through $A$ perpendicular to $I A$. Let $D I$ meet this line at $L$ and let $D I$ meet $\\omega$ at $K$. Let $N$ be the midpoint of $E F$, which lies on $I A$ and is the pole of line $A L$ with respect to $\\omega$. Since $A N \\cdot A I = A E^{2} = A R ... | IMO | IMO 2019 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 > Circles > Radical axis theorem",
"Geometry > Plane Geometry > Advanced Configurations > Polar... | English | proof only | null | |
0jm0 | Problem:
In the octagon $COMPUTER$ exhibited below, all interior angles are either $90^{\circ}$ or $270^{\circ}$ and we have $CO = OM = MP = PU = UT = TE = 1$.
Point $D$ (not to scale in the diagram) is selected on segment $RE$ so that polygons $COMPUTED$ and $CDR$ have the same area. Find ... | [
"Solution:\n\nThe area of the octagon $COMPUTER$ is equal to $6$. So, the area of $CDR$ must be $3$. So, we have the equation\n$$\n\\frac{1}{2} \\times CD \\times DR = [CDR] = 3.\n$$\nAnd from $CD = 3$, we have $DR = 2$."
] | United States | HMMT November 2014 | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | proof and answer | 2 | |
072w | Let $ABC$ be a triangle with sides $a$, $b$, $c$, circumradius $R$ and inradius $r$.
Prove that
$$
\frac{R}{2r} \ge \left( \frac{64a^2 b^2 c^2}{(4a^2 - (b-c)^2)(4b^2 - (c-a)^2)(4c^2 - (a-b)^2)} \right)^2
$$ | [
"(by Riddhipratim Basu) We introduce $s - a = x$, $s - b = y$, $s - c = z$, where $s = (a + b + c)/2$. Then $x$, $y$, $z$ are positive, and $a = y + z$, $b = z + x$, $c = x + y$. We may express\n$$\n\\frac{R}{2r} = \\frac{abcs}{4\\Delta^2} = \\frac{(x+y)(y+z)(z+x)}{8xyz}.\n$$\nSimilarly\n$$\n\\begin{align*}\n4a^2 -... | India | Indija TS 2006 | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, Euler line, nine-point circle",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
00zu | Problem:
On a $5 \times 5$ chessboard, two players play the following game. The first player places a knight on some square. Then the players alternately move the knight according to the rules of chess, starting with the second player. It is not allowed to move the knight to a square that has been visited previously. ... | [
"Solution:\n\nThe first player has a winning strategy.\n\nDivide all the squares of the board except one in pairs so that the squares of each pair are accessible from each other by one move of the knight (see Figure 10 where the squares of each pair are marked with the same number, and the remaining square is marke... | Baltic Way | Baltic Way 1997 | [
"Discrete Mathematics > Graph Theory > Matchings, Marriage Lemma, Tutte's theorem",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof and answer | The first player has a winning strategy. | |
09n5 | Let us consider the equation
$$
(a^n - b^n)^2 = a^{n+m} - b^{n+m}
$$
for positive integers $n$ and $m$ such that $n \ge m \ge 1$. Show that this equation has no integer solution $(a, b)$ satisfying $\gcd(a, b) = 1$ and $|a| > |b| > 1$. | [
"From the contrary, suppose there exist integers $(a, b)$ with $\\gcd(a, b) = 1$ and $|a| > |b| > 1$ that satisfy the equation\n$$\n(a^n - b^n)^2 = a^{n+m} - b^{n+m}\n$$\nfor $n \\ge m \\ge 1$.\nWe can rewrite the given equation as follows:\n$$\n(a^n - b^n)^2 = a^n(a^m - b^m) + b^m(a^n - b^n).\n$$\nThis implies tha... | Mongolia | Mongolian Mathematical Olympiad | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)"
] | English | proof only | null | |
0e17 | Prove that the inequality $x^2 + y^2 + 1 \ge 2(xy - x + y)$ holds for any two real $x$ and $y$. When does the equality hold? | [
"Rewrite the inequality $x^2 + y^2 + 1 \\ge 2(xy - x + y)$ as $x^2 - 2xy + y^2 + 2x - 2y + 1 \\ge 0$. If we further rearrange the left-hand side into $(x - y)^2 + 2(x - y) + 1 \\ge 0$, we notice that it is a perfect square, and the inequality becomes $((x - y) + 1)^2 \\ge 0$. Hence, the inequality holds for all rea... | Slovenia | National Math Olympiad | [
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | null | proof and answer | Equality holds exactly when y = x + 1. | |
06bs | Let the angle bisectors of $\angle A$, $\angle B$, $\angle C$ of $\triangle ABC$ intersect the circumcircle of $\triangle ABC$ at $P$, $Q$, $R$ respectively. Prove that $AP + BQ + CR > BC + CA + AB$. | [
"Let $I$ be the incentre of $\\triangle ABC$. It is well-known that $PB = PI = PC$, etc. By the triangle inequality, we have\n$$\n\\begin{aligned}\n& BI + CI > BC, \\\\\n& CI + AI > CA, \\\\\n& AI + BI > AB, \\\\\n& 2IP = BP + CP > BC, \\\\\n& 2IQ = CQ + AQ > CA, \\\\\n& 2IR = AR + BR > AB.\n\\end{aligned}\n$$\nAdd... | Hong Kong | IMO HK TST | [
"Geometry > Plane Geometry > Triangles > Triangle centers: centroid, incenter, circumcenter, orthocenter, Euler line, nine-point circle",
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities"
] | null | proof only | null | |
00d9 | Sea $n \ge 3$ un entero. Lucas y Matías juegan un juego en un polígono regular de $n$ lados con un vértice marcado como *trampa*. Inicialmente Matías ubica una ficha en un vértice del polígono. En cada paso, Lucas dice un entero positivo y Matías mueve la ficha ese número de vértices en sentido horario o en sentido ant... | [
"Numeramos los vértices con $0, 1, 2, \\ldots, n$ en sentido horario y suponemos que la trampa está ubicada en el $0$. Si la ficha está en $x$ y Lucas dice $y$, entonces Matías puede mover la ficha a los vértices con número $x-y$ o $x+y$ módulo $n$.\n\nVeamos que si $n$ tiene un divisor impar $d \\neq 1$, entonces ... | Argentina | Nacional OMA | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Number Theory > Modular Arithmetic"
] | Spanish | proof and answer | Matías can avoid the trap for all n that are not powers of two, by always staying on vertices not divisible by some fixed odd divisor of n. Lucas can force a fall into the trap exactly when n is a power of two, by repeatedly calling the power of two equal to the current two-adic valuation to strictly increase it until ... | |
0jal | Problem:
Find the largest integer less than $2012$ all of whose divisors have at most two $1$'s in their binary representations. | [
"Solution:\nCall a number good if all of its positive divisors have at most two $1$'s in their binary representations. Then, if $p$ is an odd prime divisor of a good number, $p$ must be of the form $2^{k}+1$. The only such primes less than $2012$ are $3, 5, 17$, and $257$, so the only possible prime divisors of $n$... | United States | HMMT November | [
"Number Theory > Divisibility / Factorization > Prime numbers",
"Number Theory > Divisibility / Factorization > Factorization techniques",
"Algebra > Prealgebra / Basic Algebra > Integers"
] | null | proof and answer | 1536 | |
0jiv | Problem:
Find the number of nonempty sets $\mathcal{F}$ of subsets of the set $\{1, \ldots, 2014\}$ such that:
a. For any subsets $S_{1}, S_{2} \in \mathcal{F}$, $S_{1} \cap S_{2} \in \mathcal{F}$.
b. If $S \in \mathcal{F}$, $T \subseteq \{1, \ldots, 2014\}$, and $S \subseteq T$, then $T \in \mathcal{F}$. | [
"Solution:\nAnswer: $2^{2014}$\n\nFor a subset $S$ of $\\{1, \\ldots, 2014\\}$, let $\\mathcal{F}_{S}$ be the set of all sets $T$ such that $S \\subseteq T \\subseteq \\{1, \\ldots, 2014\\}$. It can be checked that the sets $\\mathcal{F}_{S}$ satisfy the conditions 1 and 2. We claim that the $\\mathcal{F}_{S}$ are ... | United States | HMMT 2014 | [
"Discrete Mathematics > Combinatorics > Recursion, bijection"
] | null | proof and answer | 2^2014 | |
08k7 | Problem:
Find the maximum value of the area of a triangle having side lengths $a$, $b$, $c$ with
$$
a^{2}+b^{2}+c^{2}=a^{3}+b^{3}+c^{3}
$$ | [
"Solution:\nWithout any loss of generality, we may assume that $a \\leq b \\leq c$.\nOn the one hand, Tchebyshev's inequality gives\n$$\n(a+b+c)\\left(a^{2}+b^{2}+c^{2}\\right) \\leq 3\\left(a^{3}+b^{3}+c^{3}\\right)\n$$\nTherefore using the given equation we get\n$$\na+b+c \\leq 3 \\text{ or } p \\leq \\frac{3}{2}... | JBMO | OJBM | [
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Geometric Inequalities > Optimization in geometry",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof and answer | sqrt(3)/4 | |
025l | Problem:
Área de triângulo - Se $AC = 1\ \mathrm{cm}$ e $AD = 4\ \mathrm{cm}$, qual é a relação entre as áreas dos triângulos $\triangle ABC$ e $\triangle CBD$?
 | [
"Solution:\n\nOs triângulos $\\triangle ABC$ e $\\triangle CBD$ têm bases $AC$ e $CD$, respectivamente, e a mesma altura $h$ em relação a essas bases.\n\n\n\nAssim, temos\n$$\n\\text{área } \\triangle ABC = \\frac{AC \\times h}{2} \\text{ e área } \\triangle CBD = \\frac{CD \\times h}{2}\n$... | Brazil | Nível 2 | [
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 3/5 | |
0g2k | Problem:
Sei $A$ ein Punkt und sei $k$ ein Kreis durch $A$. Seien $B$ und $C$ zwei weitere Punkte auf $k$. Weiter seien $X$ der Schnittpunkt der Winkelhalbierenden von $\angle A B C$ mit $k$ und $Y$ die Spiegelung von $A$ am Punkt $X$. Sei $D$ der Schnittpunkt der Geraden $Y C$ mit $k$. Zeige, dass der Punkt $D$ nicht... | [
"\n\nSei $\\angle A B X=\\alpha$. Da $B X$ die Winkelhalbierende von $\\angle A B C$ ist, gilt $\\angle X B C=\\alpha$. Da $A B C X$ ein Sehnenviereck ist, gilt auch $\\angle X A C=\\angle A C X=\\alpha$. Dies bedeutet, dass $A X=X C$. Zusammen mit der Bedingung, dass $Y$ die Spiegelung von... | Switzerland | SMO 2019 | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Transformations > Homothety",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0io1 | Problem:
Andrew has a fair six sided die labeled with $1$ through $6$ as usual. He tosses it repeatedly, and on every third roll writes down the number facing up as long as it is not the $6$. He stops as soon as the last two numbers he has written down are squares or one is a prime and the other is a square. What is t... | [
"Solution:\n\nAnswer: $\\frac{4}{25}$. We can safely ignore all of the rolls he doesn't record. The probability that he stops after writing two squares consecutively is the same as the probability that he never rolls a prime. For, as soon as the first prime is written, either it must have been preceded by a square ... | United States | Harvard-MIT Mathematics Tournament | [
"Statistics > Probability > Counting Methods > Other"
] | null | proof and answer | 4/25 | |
0fbw | Problem:
Determina todas las ternas de números reales $(a, b, c)$, que satisfacen el sistema de ecuaciones siguiente:
$$
\left\{
\begin{array}{l}
a^{5} = 5 b^{3} - 4 c \\
b^{5} = 5 c^{3} - 4 a \\
c^{5} = 5 a^{3} - 4 b
\end{array}
\right.
$$ | [
"Solution:\n\nSin pérdida de generalidad podemos suponer que $a = \\max\\{a, b, c\\}$.\n\nPrimer caso: $c \\geq b$. Entonces $a^{5} + 4c \\geq c^{5} + 4b$ y $b \\geq a$. De este modo $a = b = c$.\n\nSegundo caso: $b \\geq c$. Entonces $b^{5} + 4a \\geq c^{5} + 4b$ y $c \\geq a$. Por tanto $a = b = c$.\n\nAsí todo s... | Spain | null | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations",
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | proof and answer | (0, 0, 0), (1, 1, 1), (-1, -1, -1), (2, 2, 2), (-2, -2, -2) | |
027p | Problem:
Os números $10$, $11$ e $12$ podem pertencer a uma mesma progressão geométrica? | [
"Solution:\nSuponha que $10$, $11$ e $12$ sejam termos de uma mesma progressão geométrica:\n$$\n\\begin{aligned}\n10 & = a r^{p-1} \\\\\n11 & = a r^{q-1} \\\\\n12 & = a r^{k-1}\n\\end{aligned}\n$$\nDaí,\n$$\n\\begin{aligned}\n& 11 / 10 = r^{q-p} \\\\\n& 12 / 11 = r^{k-q}\n\\end{aligned}\n$$\nAlém disso, temos\n$$\n... | Brazil | null | [
"Algebra > Algebraic Expressions > Sequences and Series",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | null | proof and answer | No | |
0ie3 | Problem:
How many real numbers $x$ are solutions to the following equation?
$$
|x-1|=|x-2|+|x-3|
$$ | [
"Solution:\n\nIf $x<1$, the equation becomes $(1-x)=(2-x)+(3-x)$ which simplifies to $x=4$, contradicting the assumption $x<1$.\n\nIf $1 \\leq x \\leq 2$, we get $(x-1)=(2-x)+(3-x)$, which gives $x=2$.\n\nIf $2 \\leq x \\leq 3$, we get $(x-1)=(x-2)+(3-x)$, which again gives $x=2$.\n\nIf $x \\geq 3$, we get $(x-1)=(... | United States | Harvard-MIT Mathematics Tournament | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations"
] | null | final answer only | 2 | |
0eis | Problem:
Naj bo $K$ končna podmnožica celih števil s $k \geq 3$ elementi. Pravimo, da sta števili $a, b \in K$ povezani, če in samo če obstajajo števila $x_{1}, x_{2}, \ldots, x_{k}$, za katera velja:
- $\{x_{1}, x_{2}, \ldots, x_{k}\}=K$,
- $x_{1}=a, x_{k}=b$,
- $|x_{i}-x_{i+1}|$ je liho število za vsak $i \in\{1,2, ... | [
"Solution:\n\nTretji pogoj v nalogi pove, da sta števili $a$ in $b$ povezani, če lahko števila v množici $K$ razporedimo v vrsto, ki se začne z $a$ in konča z $b$, tako da sta vsaki dve sosednji števili v tej vrsti različne parnosti. Vrsto, v kateri sta vsaki dve sosednji števili različne parnosti, bomo imenovali a... | Slovenia | 63. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Discrete Mathematics > Combinatorics > Invariants / monovariants",
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Other"
] | null | proof only | null | |
08lo | Problem:
Let $O$ be a point inside the parallelogram $A B C D$ such that
$$
\angle A O B + \angle C O D = \angle B O C + \angle C O D
$$
Prove that there exists a circle $k$ tangent to the circumscribed circles of the triangles $\triangle A O B$, $\triangle B O C$, $\triangle C O D$ and $\triangle D O A$. | [
"Solution:\nFrom given condition it is clear that $\\varangle A O B + \\varangle C O D = \\varangle B O C + \\varangle A O D = 180^\\circ$.\nLet $E$ be a point such that $A E = D O$ and $B E = C E$. Clearly, $\\triangle A E B \\equiv \\triangle D O C$ and from that $A E \\parallel D O$ and $B E \\parallel C O$. Als... | JBMO | 2008 Shortlist JBMO | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Circles > Tangents",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Constructions and loci"
] | null | proof only | null | |
0d52 | Let $a$, $b$, $c$ be positive real numbers such that $a + b + c = 1$. Prove that
$$
2\left(\frac{a b}{a + b} + \frac{b c}{b + c} + \frac{c a}{c + a}\right) + 1 \geq 6(a b + b c + c a).
$$ | [
"Using the condition $a + b + c = 1$, we have\n$$\n\\begin{aligned}\n\\left(\\frac{a b}{a + b} + \\frac{b c}{b + c} + \\frac{c a}{c + a}\\right) &= (a + b + c)\\left(\\frac{a b}{a + b} + \\frac{b c}{b + c} + \\frac{c a}{c + a}\\right) \\\\\n&= (a b + b c + c a) + a b c\\left(\\frac{1}{a + b} + \\frac{1}{b + c} + \\... | Saudi Arabia | SAMC 2015 | [
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English, Arabic | proof only | null | |
0el1 | Problem:
Dano je dvomestno naravno število $n$. Če seštejemo naslednja tri zaporedna naravna števila, ki sledijo številu $n$, dobimo dvomestno število, ki ima enaki števki kot število $n$. Koliko pozitivnih deliteljev ima število $n$ ?
(A) 3
(B) 4
(C) 5
(D) 6
(E) 8 | [
"Solution:\n\nOznačimo $n=\\overline{ab}=10a+b$. Tedaj je $(n+1)+(n+2)+(n+3)=\\overline{ba}=10b+a$. Od tod sledi $10b+a=3n+6=30a+3b+6$ oziroma $29a=7b-6$. Ker pa sta $a$ in $b$ števki, je $29a \\leq 7 \\cdot 9-6=57$, od koder sledi $a=1$. Torej je $7b=35$ oziroma $b=5$ in zato $n=15$. Število $n$ ima 4 pozitivne de... | Slovenia | 67. matematično tekmovanje srednješolcev Slovenije, Državno tekmovanje | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Number-Theoretic Functions > τ (number of divisors)"
] | null | MCQ | B | |
0hau | Two cyclists left towns $A$ and $B$, went towards one another with the speeds $v_1$ and $v_2$, where $v_1 \ge v_2$, and met for the first time after 1 hour. After having met, they both continued their journeys without stopping until their destination town. If one of them reached his/her final point, he/she turned aroun... | [
"Until the first meeting point, the first cyclist travelled the distance $S_1 = v_1$, while the second cyclist travelled $S_2 = v_2$. Therefore, the distance between towns $A$ and $B$ equals $v_1 + v_2$. There are two possible cases.\n\nCase 1. Before their second meeting, both cyclists reached their destination to... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Algebra > Prealgebra / Basic Algebra > Simple Equations",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | t2 = 2 hours if v1 < 2*v2; otherwise t2 = 2*v2/(v1 - v2) hours | |
0404 | Let $P(x) = x^{2012} + a_{2011}x^{2011} + a_{2010}x^{2010} + \dots + a_1x + a_0$ be a polynomial of degree $2012$ of real coefficients with $1$ as its leading coefficient. Find the minimum of real number $c$ such that $|\operatorname{Im} z| \le c |\operatorname{Re} z|$, where $\operatorname{Re} z$ and $\operatorname{Im... | [
"First, we point out that $c \\ge \\cot \\frac{\\pi}{4022}$. Consider the polynomial $P(x) = x^{2012} - x$. Changing the sign of coefficients of $P(x)$, we obtain four polynomials $P(x)$, $-P(x)$, $Q(x) = x^{2012} + x$ and $-Q(x)$. Note that $P(x)$ and $-P(x)$ have the same roots; one of the roots is $z_1 = \\cos \... | China | China National Team Selection Test | [
"Algebra > Algebraic Expressions > Polynomials > Roots of unity",
"Algebra > Intermediate Algebra > Complex numbers"
] | English | proof and answer | cot(pi/4022) | |
0hbp | Three numbers $2^{100}$, $3^{100}$ and $5^{100}$ are written on a long paper strip without any space in-between, thus, creating one big number $N$. Arsenii claims that he can change the last digit of number $N$ so that the new number is a power of $13$. Is he right? | [
"Suppose Arsenii's claim is correct and by changing the last digit of number $N$ he obtained $13^k$, where $k$ is a positive integer. Clearly, he had to change the last digit, because it is $5$. Since $2^{100} \\equiv 1 \\pmod{3}$, $3^{100} \\equiv 0 \\pmod{3}$ and $5^{100} \\equiv 1 \\pmod{3}$, the sums of their d... | Ukraine | 59th Ukrainian National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof and answer | No | |
0d2c | For positive integers $a$ and $b$, $\operatorname{gcd}(a, b)$ denotes their greatest common divisor and $\operatorname{lcm}(a, b)$ their least common multiple. Determine the number of ordered pairs $(a, b)$ of positive integers satisfying the equation
$$
a b+63=20 \operatorname{lcm}(a, b)+12 \operatorname{gcd}(a, b).
$... | [
"Let $d=\\operatorname{gcd}(a, b)$ and $a=d a'$, $b=d b'$ with $a', b'$ two relatively prime positive integers. The equation becomes\n$$\na' b' d^2 + 63 = 20 a' b' d + 12 d\n$$\nTherefore, $d$ divides $63$ and we have\n$$\na' b' d + \\frac{63}{d} = 20 a' b' + 12\n$$\nIf $5 < d < 20$, then $a' b' d + \\frac{63}{d} <... | Saudi Arabia | Selection tests for the Balkan Mathematical Olympiad 2013 | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | English | proof and answer | 2 | |
0bvj | Let $x, y, z$ be three positive real numbers such that $x^2 + y^2 + z^2 + 3 = 2(xy + yz + zx)$. Prove that
$$
\sqrt{xy} + \sqrt{yz} + \sqrt{zx} \geq 3.
$$ | [
"*First solution.* Using the given condition, the inequality can be written equivalently $\\sqrt{xy} + \\sqrt{yz} + \\sqrt{zx} \\ge \\sqrt{3(2xy + 2yz + 2zx - x^2 - y^2 - z^2)}$ or, denoting $\\sqrt{x} = a, \\sqrt{y} = b, \\sqrt{z} = c$, we have $(ab + bc + ca)^2 \\ge 3(2a^2b^2 + 2b^2c^2 + 2c^2a^2 - a^4 - b^4 - c^4... | Romania | Eleventh STARS OF MATHEMATICS Competition | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Muirhead / majorization",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | null | proof only | null | |
069q | Let the set $X = \{1,2,...,8\}$ and two nonempty disjoint subsets $A, B$ of $X$ with union the set $X$. Let $P_A$ be the product of the elements of the set $A$ and $P_B$ be the product of the elements of the set $B$. Determine the least possible value of the sum $P_A + P_B$. | [
"We observe that: $P_A \\cdot P_B = 8! = c$. By symmetry, without loss of generality, we suppose that $P_A \\leq P_B$, and hence $P_A \\leq \\sqrt{c}$. We write\n$$\nP_A + P_B = P_A + \\frac{c}{P_A}\n$$\nand $P_A = x$. We consider the function $f(x) = x + \\frac{c}{x}$, with $1 \\leq x \\leq \\sqrt{c}$. The functio... | Greece | SELECTION EXAMINATION 2019 | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Combinatorial optimization"
] | English | proof and answer | 402 | |
03cn | Find all real numbers $a$ with the following property: for every infinite sequence $a_1, a_2, a_3, \dots$ of mutually distinct positive integers, such that the inequality $a_n \le a n$ is satisfied for every positive integer $n$, there are infinitely many terms of the sequence which has sum of their digits in numerical... | [
"Answer: $1 \\le a < 2019$.\n\nIt is clear that $a \\ge 1$, since $a_1$ is a positive integer less than or equal to $a$.\n\nDenote by $\\sigma(a_n)$ the remainder of the division of $2019$ by the sum of the digits of $a_n$ in base $4038$.\n\nWe consider the sequence $\\{b_n : \\sigma(b_n) = 0\\}_{n=0}^{\\infty}$ fo... | Bulgaria | 68. National Mathematical Olympiad | [
"Number Theory > Modular Arithmetic",
"Number Theory > Divisibility / Factorization",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | 1 ≤ a < 2019 | |
0l2n | Problem:
Let $f(x)$ be a quotient of two quadratic polynomials. Given that $f(n) = n^{3}$ for all $n \in \{1,2,3,4,5\}$, compute $f(0)$. | [
"Solution:\nLet $f(x) = \\frac{p(x)}{q(x)}$. Then, $x^{3} q(x) - p(x)$ has $1,2,3,4,5$ as roots. Therefore, WLOG, let\n\n$$\nx^{3} q(x) - p(x) = (x-1)(x-2)(x-3)(x-4)(x-5) = x^{5} - 15 x^{4} + 85 x^{3} - \\ldots\n$$\n\nThus, $q(x) = x^{2} - 15 x + 85$, so $q(0) = 85$. Plugging $x = 0$ in the above equation also give... | United States | HMMT February 2024 | [
"Algebra > Algebraic Expressions > Polynomials > Polynomial interpolation: Newton, Lagrange",
"Algebra > Intermediate Algebra > Quadratic functions"
] | null | proof and answer | 24/17 | |
08kt | Problem:
The real numbers $x$, $y$, $z$, $m$, $n$ are positive, such that $m+n \geq 2$. Prove that
$$
\begin{gathered}
x \sqrt{y z(x+m y)(x+n z)}+y \sqrt{x z(y+m x)(y+n z)}+z \sqrt{x y(z+m x)(z+n y)}
\\
\leq \frac{3(m+n)}{8}(x+y)(y+z)(z+x)
\end{gathered}
$$ | [
"Solution:\n\nUsing the AM-GM inequality we have\n$$\n\\begin{aligned}\n& \\sqrt{y z(x+m y)(x+n z)}=\\sqrt{(x z+m y z)(x y+n y z)} \\leq \\frac{x y+x z+(m+n) y z}{2} \\\\\n& \\sqrt{x z(y+m x)(y+n z)}=\\sqrt{(y z+m x z)(x y+n x z)} \\leq \\frac{x y+y z+(m+n) x z}{2} \\\\\n& \\sqrt{x y(z+m x)(z+n y)}=\\sqrt{(y z+m x ... | JBMO | 2007 Shortlist JBMO | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean"
] | null | proof only | null | |
08ts | For a convex hexagon $ABCDEF$, angles formed by any pair chosen from the 3 diagonals $AD, BE, CF$ are $60^\circ$. Prove that the inequality
$$
AB + BC + CD + DE + EF + FA \ge AD + BE + CF
$$
must hold.
Here, we say that a polygon is convex if all of its interior angles are less than $180^\circ$. We also denote by $XY$ ... | [
"First, we prove the following Lemma:\n**Lemma:** Suppose the angle $\\angle XYZ$ of a triangle $\\triangle XYZ$ is $60^\\circ$. Let the lengths of the sides $YZ, XZ, XY$ be $x, y, z$ respectively. Then, the inequality $2x \\ge y + z$ holds.\n**Proof:** Let $\\Gamma$ be the circum-circle of the triangle $\\triangle... | Japan | Japan Junior Mathematical Olympiad | [
"Geometry > Plane Geometry > Geometric Inequalities > Triangle inequalities",
"Geometry > Plane Geometry > Triangles > Triangle inequalities",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing",
"Geometry > Plane Geometry > Miscellaneous > Angle chasing"
] | null | proof only | null | |
0aqp | Problem:
Simplify: $\sqrt{\sin^{4} 15^{\circ} + 4 \cos^{2} 15^{\circ}} - \sqrt{\cos^{4} 15^{\circ} + 4 \sin^{2} 15^{\circ}}$. | [] | Philippines | 12th Philippine Mathematical Olympiad - Area Stage | [
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry"
] | null | final answer only | sqrt(3)/2 | |
07ir | Let $n \ge 2$ be a given positive integer. Prove that there is a $n$-tuple $(a_1, \dots, a_n)$ of pair-wise coprime positive integers, each of which greater than $1402$ such that
$$
\lfloor \frac{a_1}{a_2} \rfloor + \lfloor \frac{a_2}{a_3} \rfloor + \dots + \lfloor \frac{a_n}{a_1} \rfloor = \lfloor \frac{a_2}{a_1} \rfl... | [
"Notice that if $a_1 < a_2 < \\dots < a_n$ are positive integers such that\n$a_2/a_1, a_3/a_2, \\dots, a_n/a_{n-1} \\in (1, 2)$\n\nwhile $a_n/a_1 \\in (n, n+1)$ then $(a_1, \\dots, a_n)$ would satisfy the condition of the problem. It thus suffices to find such numbers. We shall then prove the following lemma.\n\n**... | Iran | 41th Iranian Mathematical Olympiad | [
"Number Theory > Divisibility / Factorization > Greatest common divisors (gcd)",
"Number Theory > Divisibility / Factorization > Prime numbers",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | null | proof only | null | |
0c5j | Find the integer solutions of the equation
$$
x^2(x^2 + 1) = 21^y - 1.
$$ | [
"The equation can be written as $x^4 - x^2 + 1 = 21^y$, i.e., $(x^2 - x + 1)(x^2 + x + 1) = 21^y$. The number in the left hand side is an integer, therefore $y$ must be non-negative. It is easy to see that $x^2 - x + 1$ and $x^2 + x + 1$ are co-prime. As $0 < x^2 - x + 1 < x^2 + x + 1$, we can only have $x^2 - x + ... | Romania | The DANUBE Mathematical Competition | [
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities",
"Number Theory > Divisibility / Factorization > Factorization techniques"
] | English | proof and answer | (x,y) = (0,0), (2,1), (-2,1) | |
0c3t | Problem:
Fie $p$ un număr natural mai mare sau egal cu $2$ şi fie $(M, \cdot)$ un monoid finit, astfel încât $a^{p} \neq a$, oricare ar fi $a \in M \backslash\{e\}$, unde $e$ este elementul neutru al lui $M$. Arătaţi că $(M, \cdot)$ este grup. | [
"Solution:\n\nFie $a \\in M \\backslash\\{e\\}$. Cum $M$ este finit, există două numere naturale nenule $i$ şi $k$, astfel încât $a^{i}=a^{i+k}$.\n\nPrin înmulţiri succesive cu $a^{k}$, rezultă că $a^{i}=a^{i+n k}$, oricare ar fi numărul natural nenul $n$.\n\nAlegem un număr natural $m$, astfel încât $m k>i$. Prin ... | Romania | Olimpiada Naţională de Matematică Etapa Judeţeană şi a Municipiului Bucureşti | [
"Algebra > Abstract Algebra > Group Theory"
] | null | proof only | null | |
0dsj | In $\triangle ABC$, $AB = AC = 14\sqrt{2}$, $D$ is the midpoint of $CA$ and $E$ is the midpoint of $BD$. Suppose $\triangle CDE$ is similar to $\triangle ABC$. Find the length of $BD$. | [
"Let $\\ell = AB = AC = 14\\sqrt{2}$ and $BC = x$. The 4 angles marked in the figure are all equal. This implies that $\\triangle ABC$, $\\triangle BCD$, $\\triangle CDE$ are all similar. Thus $BD = BC = x$.\n\nAlso $AB/BC = BC/CD$. That is $\\ell/x = x/(\\ell/2)$. From this we get $x = \\ell/\\sqrt{2}$. Since $\\e... | Singapore | Singapore Mathematical Olympiad | [
"Geometry > Plane Geometry > Miscellaneous > Angle chasing",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | null | proof and answer | 14 | |
0ekg | Problem:
V trapezu $ABCD$ je dolžina kraka $AD$ enaka $13~\mathrm{cm}$ in dolžina kraka $BC$ enaka $9~\mathrm{cm}$. Kot $\angle BAD$ meri $37^\circ$. Kolikšna je velikost kota $\angle BCD$, zaokrožena na dve decimalni mesti, če je kot $\angle CBA$ ostri?
(A) $37^\circ$
(B) $143^\circ$
(C) $48,29^\circ$
(D) $119,62^\c... | [
"Solution:\n\nIzračunamo višino trapeza $v = d \\cdot \\sin \\alpha \\doteq 7,82~\\mathrm{cm}$. Potem iz $\\sin \\beta = \\frac{v}{b}$ dobimo $\\beta_1 \\doteq 60,38^\\circ$ in $\\beta_2 = 180^\\circ - \\beta_1 \\doteq 119,62^\\circ$. Pravilen je odgovor $D$."
] | Slovenia | 22. tekmovanje v znanju matematike za dijake srednjih tehniških in strokovnih šol | [
"Geometry > Plane Geometry > Quadrilaterals",
"Geometry > Plane Geometry > Analytic / Coordinate Methods > Trigonometry",
"Geometry > Plane Geometry > Triangles > Triangle trigonometry"
] | null | MCQ | D | |
01fh | The expressions $x+y$, $x-y$, $x^2+xy+y^2$ and $x^2-xy+y^2$ are written on the two sides of two cards in such a way that each side of each card contains exactly one of these expressions. The cards are laid on the table on top of each other in such a way that only the top side of the uppermost card is visible. Alice and... | [
"**Answer:** Yes, Alice.\n\nLet Alice choose the card where at least one of the two expressions is a trinomial. She can do it as follows: if the visible side of the topmost card contains a trinomial then she can pick that card, otherwise the bottommost card definitely contains a trinomial and she can pick that one.... | Baltic Way | Baltic Way 2019 | [
"Discrete Mathematics > Combinatorics > Games / greedy algorithms",
"Algebra > Algebraic Expressions > Polynomials > Polynomial operations"
] | English | proof and answer | Alice | |
07f3 | a, b and c are positive real numbers such that
$$
\sum_{cyc} (a+b)^2 = 2 \sum_{cyc} a + 6abc.
$$
Prove that
$$
\sum_{cyc} (a-b)^2 \le \left| 2 \sum_{cyc} a - 6abc \right|.
$$ | [
"We know that $\\sum_{cyc} a^2 + \\sum_{cyc} ab = \\sum_{cyc} a + 3abc$, so\n$$\n\\begin{aligned}\n\\left(\\sum_{cyc} a^2 + \\sum_{cyc} ab\\right)^2 &= \\left(\\sum_{cyc} a + 3abc\\right)^2 \\\\\n\\Rightarrow \\quad &\\left(\\sum_{cyc} a^2\\right)^2 + \\left(\\sum_{cyc} ab\\right)^2 + 2\\left(\\sum_{cyc} a^2\\right... | Iran | 37th Iranian Mathematical Olympiad | [
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Cauchy-Schwarz",
"Algebra > Algebraic Expressions > Polynomials > Symmetric functions"
] | English | proof only | null | |
00vp | There are $n$ cities in a country, where $n \geq 100$ is a positive integer. Some pairs of cities are connected by (two-way) flights. For two cities $A$ and $B$, a path is a sequence of distinct cities $C_0, C_1, C_2, \ldots, C_k, C_{k+1}$, such that there are flights between $C_i$ and $C_{i+1}$ for every $0 \leq i \le... | [
"Use the obvious graph interpretation. We show that any such graph is one of the following: the full graph $K_n$, the circular graph $C_n$, and for $n$ even, the bipartite graph $K_{\\frac{n}{2}, \\frac{n}{2}}$. First, we show that these graphs satisfy the condition.\n* For $K_n$, we can choose any long path and th... | Balkan Mathematical Olympiad | Balkan Mathematical Olympiad Shortlisted Problems | [
"Discrete Mathematics > Graph Theory",
"Discrete Mathematics > Combinatorics > Coloring schemes, extremal arguments"
] | English | proof and answer | n, n(n−1)/2, and when n is even also n^2/4 | |
0d3z | Determine all functions $f:[0, \infty) \rightarrow \mathbb{R}$ such that $f(0)=0$ and
$$
f(x)=1+5 f\left(\left\lfloor\frac{x}{2}\right\rfloor\right)-6 f\left(\left\lfloor\frac{x}{4}\right\rfloor\right)
$$
for all $x>0$. | [
"Let $x \\geq 0$. If $x \\in (0,2)$ then $f(x)=1+5 f(0)-6 f(0)=1$. \nIf $x \\in [2,4)$ then $f(x)=1+5 f(1)-6 f(0)=6=a_{1}$. \nIf $x \\in [4,8)$ then $\\left\\lfloor\\frac{x}{2}\\right\\rfloor \\in [2,4)$ and $\\left\\lfloor\\frac{x}{4}\\right\\rfloor \\in [1,2)$, and therefore $f(x)=1+5 \\cdot 6-6 \\cdot 1=25=a_{2}... | Saudi Arabia | SAMC | [
"Algebra > Algebraic Expressions > Functional Equations",
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Algebra > Algebraic Expressions > Sequences and Series > Floors and ceilings"
] | English, Arabic | proof and answer | f(0)=0; f(x)=1 for 0<x<2; and for n>=1, f(x) = -2^{n+2} + (3^{n+2}+1)/2 for x in [2^n, 2^{n+1}). | |
07kw | Suppose $A$ is an $n \times n$ array of numbers, with $n > 2$, and denote by $A(i, j)$ the number in the $i$th row and $j$th column. We say that $A$ is an *averaging array* if it has the following property: $A(i, j)$ equals the average of the three numbers $A(i, j-1)$, $A(i-1, j)$, and $A(i-1, j-1)$, whenever $i, j \in... | [
"a.\nFor part (i), first observe that if $A(i, j) = M$ with $i, j > 1$, then the three neighbouring values that have average value $A(i, j)$ must also equal $M$, since otherwise at least one would have to be larger than $M$. Then it is not hard to see that we can propagate this value to get that the subarray with m... | Ireland | Irska | [
"Algebra > Algebraic Expressions > Sequences and Series > Recurrence relations",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Algorithms"
] | English | proof only | null | |
0au8 | Problem:
In a Word Finding game, a player tries to find a word in a $12 \times 12$ array of letters by looking at blocks of adjacent letters that are arranged horizontally, arranged vertically, or arranged diagonally. How many such 3-letter blocks are there in a given $12 \times 12$ array of letters? | [] | Philippines | 17th Philippine Mathematical Olympiad Area Stage | [
"Discrete Mathematics > Other"
] | null | final answer only | 440 | |
0crd | По кругу стоят $10^{1000}$ натуральных чисел. Между каждыми двумя соседними числами записали их наименьшее общее кратное. Могут ли эти наименьшие общие кратные образовать $10^{1000}$ последовательных чисел (расположенных в каком-то порядке)? | [
"**Ответ.** Не могут.\n\nПусть $n = 10^{1000}$. Обозначим исходные числа (в порядке обхода) через $a_1, \\dots, a_n$; мы будем считать, что $a_{n+1} = a_1$. Положим $b_i = \\text{НОК}(a_i, a_{i+1})$. Предположим что числа $b_1, \\dots, b_n$ — это $n$ подряд идущих натуральных чисел.\n\nРассмотрим наибольшую степень... | Russia | XL Russian mathematical olympiad | [
"Number Theory > Divisibility / Factorization > Least common multiples (lcm)",
"Number Theory > Divisibility / Factorization > Prime numbers"
] | null | proof and answer | No | |
0gfp | 設 $C$ 為實數空間 $\mathbb{R}^3$ 中的一個體積為正的凸子集。設 $C_1, C_2, \dots, C_n$ 是 $n$ 個將 $C$ 平移(但不旋轉)所得的集合, 滿足 $C_i \cap C \ne \emptyset$ 對每一個 $i = 1, 2, \dots, n$ 均成立, 但對不同的 $i, j$, $C_i$ 與 $C_j$ 最多只在它們的邊界相交。證明 $n \le 27$, 並證明 $27$ 是 $n$ 的最大可能值。 | [] | Taiwan | 2022 數學奧林匹亞競賽第一階段培訓營, 獨立研究(一) | [
"Geometry > Plane Geometry > Combinatorial Geometry > Minkowski's theorem",
"Geometry > Solid Geometry > Other 3D problems"
] | Chinese; English | proof and answer | 27 | |
0gr3 | The diagonals of convex quadrilateral $ABCD$ meet at point $E$. Given
$$
\frac{|AB|}{|CD|} = \frac{|BC|}{|AD|} = \sqrt{\frac{|BE|}{|DE|}}
$$
show that $ABCD$ is either a parallelogram or a cyclic quadrilateral. | [
"Using Stewart's Theorem in triangles $ABD$ and $BCD$, we obtain\n$$\n\\frac{AB^2 \\cdot ED + AD^2 \\cdot BE}{BD} - \\frac{BE \\cdot ED}{AE^2} = \\frac{BC^2 \\cdot ED + CD^2 \\cdot BE}{BD} - \\frac{BE \\cdot ED}{EC^2}.\n$$\nUsing the given relations\n$$\nAB^2 \\cdot ED = CD^2 \\cdot BE \\quad \\text{and} \\quad AD^... | Turkey | 22nd Junior Turkish Mathematical Olympiad | [
"Geometry > Plane Geometry > Quadrilaterals > Cyclic quadrilaterals",
"Geometry > Plane Geometry > Miscellaneous > Distance chasing"
] | English | proof only | null | |
0419 | Let $a$, $b$ be real numbers such that the equation $x^3 - a x^2 + b x - a = 0$ has only real roots. Find the minimum of $\frac{2a^3 - 3ab + 3a}{b+1}$. | [
"Let $x_1$, $x_2$ and $x_3$ be the real roots of the equation $x^3 - a x^2 + b x - a = 0$. By Vieta's Formula, we have\n$x_1 + x_2 + x_3 = a$, $x_1 x_2 + x_2 x_3 + x_1 x_3 = b$, $x_1 x_2 x_3 = a$.\n\nBy $(x_1 + x_2 + x_3)^2 \\ge 3(x_1 x_2 + x_2 x_3 + x_1 x_3)$, we have $a^2 \\ge 3b$, and by $a = x_1 + x_2 + x_3 \\g... | China | China Southeastern Mathematical Olympiad | [
"Algebra > Algebraic Expressions > Polynomials > Vieta's formulas",
"Algebra > Equations and Inequalities > QM-AM-GM-HM / Power Mean",
"Algebra > Equations and Inequalities > Linear and quadratic inequalities"
] | English | proof and answer | 9√3 | |
0btf | The elements of the set $\{1, 2, \dots, 21\}$ are written randomly on a circle, in the order $a_1, a_2, \dots, a_{21}$ (see figure). Consider the sums
$$
\begin{align*}
S_1 &= a_1 + a_2 + a_3 + a_4 + a_5, \
S_2 &= a_2 + a_3 + a_4 + a_5 + a_6, \
\vdots \
S_{17} &= a_{17} + a_{18} + a_{19} + a_{20} + a_{21}, \
S_{18} &=... | [
"Suppose that $S_1, S_2, \\dots, S_{18}$ leave the same remainder when divided by 5. Since $S_1 = a_1 + (a_2 + a_3 + a_4 + a_5)$ and $S_2 = (a_2 + a_3 + a_4 + a_5) + a_6$, if $S_1$ and $S_2$ leave the same remainder when divided by 5, then $a_1$ and $a_6$ leave the same remainder when divided by 5. In the same way:... | Romania | 67th Romanian Mathematical Olympiad | [
"Number Theory > Modular Arithmetic"
] | English | proof only | null | |
0ecp | Which integer $n$ satisfies
$$(n - 1)(n - 3) \dots (n - 2015) = n(n + 2)(n + 4) \dots (n + 2014)$$
(A) -4028 (B) -2014 (C) 2015 (D) 4030 (E) None. | [
"If $n$ is an odd integer the left side of the equality is even and the right side is odd. Thus the equality is not satisfied for odd integers. Similarly, if $n$ is an even integer the left side of the equality is odd and the right side is even. Therefore no integer satisfies the equality."
] | Slovenia | National Math Olympiad 2015 – First Round | [
"Algebra > Prealgebra / Basic Algebra > Integers",
"Number Theory > Diophantine Equations > Techniques: modulo, size analysis, order analysis, inequalities"
] | null | MCQ | E | |
05it | Problem:
Chacun des 400 députés d'un parlement a giflé exactement un autre député. Montrer qu'on peut créer une commission parlementaire de 134 députés telle qu'aucun membre de la commission n'ait giflé aucun autre membre. | [
"Solution:\n\nAppelons deux députés ennemis si l'un d'eux a giflé l'autre. Nous allons résoudre par récurrence un exercice plus général :\nDans un parlement d'au moins $3n-2$ députés, chaque député a giflé 0 ou 1 autre député. Il est alors toujours possible de créer une commission parlementaire de $n$ députés qui n... | France | Olympiades Françaises de Mathématiques | [
"Discrete Mathematics > Combinatorics > Pigeonhole principle",
"Discrete Mathematics > Combinatorics > Induction / smoothing",
"Discrete Mathematics > Combinatorics > Games / greedy algorithms"
] | null | proof only | null | |
00h2 | Find all functions $f$ from the set $\{\mathbf{R}\}$ of real numbers into $\{\mathbf{R}\}$ which satisfy for all $x, y, z \in \{\mathbf{R}\}$ the identity
$$
f(f(x)+f(y)+f(z))=f(f(x)-f(y))+f(2 x y+f(z))+2 f(x z-y z)
$$ | [
"It is clear that if $f$ is a constant function which satisfies the given equation, then the constant must be $0$. Conversely, $f(x)=0$ clearly satisfies the given equation, so, the identically $0$ function is a solution. In the sequel, we consider the case where $f$ is not a constant function.\nLet $t \\in \\mathb... | Asia Pacific Mathematics Olympiad (APMO) | APMO | [
"Algebra > Algebraic Expressions > Functional Equations > Injectivity / surjectivity",
"Algebra > Algebraic Expressions > Functional Equations > Existential quantifiers"
] | English | proof and answer | The only solutions are f(x) = 0 for all real x and f(x) = x^2 for all real x. | |
0evj | The incircle of a triangle $ABC$ is tangent to $BC$, $AC$, $AB$ at the points $D$, $E$, $F$, respectively. Suppose the line $EF$ intersects the lines $BI$, $CI$, $BC$, $DI$ at the points $K$, $L$, $M$, $Q$, respectively, where the incenter of $\triangle ABC$ is $I$. If the line passing through both the midpoint of $CL$... | [
"Since $BD$ and $BF$ are tangent lines to the incircle of $\\triangle ABC$, $BD = BF$. But the line $BI$ bisects $\\angle DBF$, so $DF$ and $BI$ are perpendicular to each other. Similarly, $DE$ and $CI$ are perpendicular to each other. It follows that $\\angle BKD = \\angle BKF = 90^\\circ - \\angle DFK$ and $\\ang... | South Korea | Korean Mathematical Olympiad | [
"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 > Circles > Tangents",
"Geometry > Plane Geometry > Quadrilateral... | English | proof only | null | |
027b | Problem:
Alguns alunos do sétimo e oitavo ano de uma escola participam de um torneio de pingue-pongue, onde cada aluno joga contra todos os outros exatamente uma vez recebendo 1 por vitória e 0 ponto por derrota. Existem dez vezes mais alunos do oitavo ano do que do sétimo ano. A pontuação total dos alunos do oitavo a... | [
"Solution:\n\na) Cada um dos $k$ alunos irá jogar $k-1$ vezes. Somando-se a quantidade de jogos de cada um, obtemos $k(k-1)$. Entretanto, teremos contado cada jogo duas vezes, uma para cada um dos participantes da partida. Logo, o número de jogos é $\\frac{k(k-1)}{2}$.\n\nb) Seja $n$ o número de alunos do sétimo an... | Brazil | NÍVEL 3 | [
"Discrete Mathematics > Combinatorics > Counting two ways"
] | null | proof and answer | 10 |
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