This website contains problems from math contests. Problems and corresponding tags were obtained from the Art of Problem Solving website.

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Found problems: 5802

The sequence $a_i$ is defined as $a_1 = 2, a_2 = 3$, and $a_{n+1} = 2a_{n-1}$ or $a_{n+1} = 3a_n - 2a_{n-1}$ for all integers $n \ge 2$. Prove that no term in $a_i$ is in the range $[1612, 2012]$.
For a positive integer $n\geq 3$ plot $n$ equally spaced points around a circle. Label one of them $A$, and place a marker at $A$. One may move the marker forward in a clockwise direction to either the next point or the point after that. Hence there are a total of $2n$ distinct moves available; two from each point. Let $a_n$ count the number of ways to advance around the circle exactly twice, beginning and ending at $A$, without repeating a move. Prove that $a_{n-1}+a_n=2^n$ for all $n\geq 4$.
Find all the functions $f: \mathbb R \rightarrow \mathbb R$ satisfying : $(x+y)(f(x)-f(y))=(x-y)f(x+y)$ for all $x,y \in \mathbb R$
Prove that for every non-negative integer $n$ there exist integers $x, y, z$ with $gcd(x, y, z) = 1$, such that $x^2 + y^2 + z^2 = 3^{2^n}$.(Poland)
Consider a convex polygon having $n$ vertices, $n\geq 4$. We arbitrarily decompose the polygon into triangles having all the vertices among the vertices of the polygon, such that no two of the triangles have interior points in common. We paint in black the triangles that have two sides that are also sides of the polygon, in red if only one side of the triangle is also a side of the polygon and in white those triangles that have no sides that are sides of the polygon. Prove that there are two more black triangles that white ones.
Consider the configurations of integers $a_{1,1}$ $a_{2,1} \quad a_{2,2}$ $a_{3,1} \quad a_{3,2} \quad a_{3,3}$ $\dots \quad \dots \quad \dots$ $a_{2017,1} \quad a_{2017,2} \quad a_{2017,3} \quad \dots \quad a_{2017,2017}$ Where $a_{i,j} = a_{i+1,j} + a_{i+1,j+1}$ for all $i,j$ such that $1 \leq j \leq i \leq 2016$. Determine the maximum amount of odd integers that such configuration can contain.
Let $a,b,c$ be distinct positive real numbers, and let $k$ be a positive integer greater than $3$. Show that \[\left\lvert\frac{a^{k+1}(b-c)+b^{k+1}(c-a)+c^{k+1}(a-b)}{a^k(b-c)+b^k(c-a)+c^k(a-b)}\right\rvert\ge \frac{k+1}{3(k-1)}(a+b+c)\] and \[\left\lvert\frac{a^{k+2}(b-c)+b^{k+2}(c-a)+c^{k+2}(a-b)}{a^k(b-c)+b^k(c-a)+c^k(a-b)}\right\rvert\ge \frac{(k+1)(k+2)}{3k(k-1)}(a^2+b^2+c^2).\] [i]Calvin Deng.[/i]
For real numbers $x$ and $y$, define \[\nabla(x,y)=x-\dfrac1y.\] If \[\underbrace{\nabla(2, \nabla(2, \nabla(2, \ldots \nabla(2,\nabla(2, 2)) \ldots)))}_{2016 \,\nabla\text{s}} = \dfrac{m}{n}\] for relatively prime positive integers $m$, $n$, compute $100m + n$. [i] Proposed by David Altizio [/i]
Let $k$ be a nonnegative integer. Evaluate \[ \sum_{j=0}^k 2^{k-j} \binom{k+j}{j}. \]
Suppose $P(n) $ is a nonconstant polynomial where all of its coefficients are nonnegative integers such that \[ \sum_{i=1}^n P(i) | nP(n+1) \] for every $n \in \mathbb{N}$. Prove that there exists an integer $k \ge 0$ such that \[ P(n) = \binom{n+k}{n-1} P(1) \] for every $n \in \mathbb{N}$.
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals. Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded? [i]Proposed by Mihai Bălună, Romania[/i]
Find all functions $f:\mathbb{R^{+}}\to\mathbb{R^+}$ such that for all $x,y\in\mathbb{R^+}$ it holds that $$f\left(xy\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x+y}\right)\right)=f\left(xy\left(\frac{1}{x}+\frac{1}{y}\right)\right)+f(x)f\left(\frac{y}{x+y}\right).$$
For a positive integer $n$ we denote by $s(n)$ the sum of the digits of $n$. Let $P(x)=x^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0$ be a polynomial, where $n \geqslant 2$ and $a_i$ is a positive integer for all $0 \leqslant i \leqslant n-1$. Could it be the case that, for all positive integers $k$, $s(k)$ and $s(P(k))$ have the same parity?
Prove that for arbitary positive integer $ n\geq 4$, there exists a permutation of the subsets that contain at least two elements of the set $ G_{n} \equal{} \{1,2,3,\cdots,n\}$: $ P_{1},P_{2},\cdots,P_{2^n \minus{} n \minus{} 1}$ such that $ |P_{i}\cap P_{i \plus{} 1}| \equal{} 2,i \equal{} 1,2,\cdots,2^n \minus{} n \minus{} 2.$
Let $ S$ be a finite set of points in the plane such that no three of them are on a line. For each convex polygon $ P$ whose vertices are in $ S$, let $ a(P)$ be the number of vertices of $ P$, and let $ b(P)$ be the number of points of $ S$ which are outside $ P$. A line segment, a point, and the empty set are considered as convex polygons of $ 2$, $ 1$, and $ 0$ vertices respectively. Prove that for every real number $ x$ \[\sum_{P}{x^{a(P)}(1 \minus{} x)^{b(P)}} \equal{} 1,\] where the sum is taken over all convex polygons with vertices in $ S$. [i]Alternative formulation[/i]: Let $ M$ be a finite point set in the plane and no three points are collinear. A subset $ A$ of $ M$ will be called round if its elements is the set of vertices of a convex $ A \minus{}$gon $ V(A).$ For each round subset let $ r(A)$ be the number of points from $ M$ which are exterior from the convex $ A \minus{}$gon $ V(A).$ Subsets with $ 0,1$ and 2 elements are always round, its corresponding polygons are the empty set, a point or a segment, respectively (for which all other points that are not vertices of the polygon are exterior). For each round subset $ A$ of $ M$ construct the polynomial \[ P_A(x) \equal{} x^{|A|}(1 \minus{} x)^{r(A)}. \] Show that the sum of polynomials for all round subsets is exactly the polynomial $ P(x) \equal{} 1.$ [i]Proposed by Federico Ardila, Colombia[/i]
A polygon is decomposed into triangles by drawing some non-intersecting interior diagonals in such a way that for every pair of triangles of the triangulation sharing a common side, the sum of the angles opposite to this common side is greater than $180^o$. a) Prove that this polygon is convex. b) Prove that the circumcircle of every triangle used in the decomposition contains the entire polygon. [i]Proposed by Morteza Saghafian - Iran[/i]
The total mass of $100$ given weights with positive masses equals $2S$. A natural number $k$ is called [i]middle[/i] if some $k$ of the given weights have the total mass $S$. Find the maximum possible number of middle numbers.
Consider the segment $[0; 1]$. At each step we may split one of the available segments into two new segments and write the product of lengths of these two new segments onto a blackboard. Prove that the sum of the numbers on the blackboard never will exceed $1/2$. [i]Mikhail Lukin[/i]
Determine which positive integers $n$ have the following property: For all integers $m$ that are relatively prime to $n$, there exists a permutation $\pi:\{1,2, \ldots, n\} \rightarrow\{1,2, \ldots, n\}$ such that $\pi(\pi(k)) \equiv m k(\bmod n)$ for all $k \in\{1,2, \ldots, n\}$.
Fix an integer $n \geq 3$. Determine the smallest positive integer $k$ satisfying the following condition: For any tree $T$ with vertices $v_1, v_2, \dots, v_n$ and any pairwise distinct complex numbers $z_1, z_2, \dots, z_n$, there is a polynomial $P(X, Y)$ with complex coefficients of total degree at most $k$ such that for all $i \neq j$ satisfying $1 \leq i, j \leq n$, we have $P(z_i, z_j) = 0$ if and only if there is an edge in $T$ joining $v_i$ to $v_j$. Note, for example, that the total degree of the polynomial $$ 9X^3Y^4 + XY^5 + X^6 - 2 $$ is 7 because $7 = 3 + 4$. [i]Proposed by Andrei Chiriță, Romania[/i]
Given positive integer $n (n \geq 2)$, find the largest positive integer $\lambda$ satisfying : For $n$ bags, if every bag contains some balls whose weights are all integer powers of $2$ (the weights of balls in a bag may not be distinct), and the total weights of balls in every bag are equal, then there exists a weight among these balls such that the total number of balls with this weight is at least $\lambda$.
There are $ n$ markers, each with one side white and the other side black. In the beginning, these $ n$ markers are aligned in a row so that their white sides are all up. In each step, if possible, we choose a marker whose white side is up (but not one of the outermost markers), remove it, and reverse the closest marker to the left of it and also reverse the closest marker to the right of it. Prove that, by a finite sequence of such steps, one can achieve a state with only two markers remaining if and only if $ n \minus{} 1$ is not divisible by $ 3$. [i]Proposed by Dusan Dukic, Serbia[/i]
Let $a_0,a_1,a_2,\dots $ be a sequence of real numbers such that $a_0=0, a_1=1,$ and for every $n\geq 2$ there exists $1 \leq k \leq n$ satisfying \[ a_n=\frac{a_{n-1}+\dots + a_{n-k}}{k}. \]Find the maximum possible value of $a_{2018}-a_{2017}$.
For fixed positive integers $s, t$, define $a_n$ as the following. $a_1 = s, a_2 = t$, and $\forall n \ge 1$, $a_{n+2} = \lfloor \sqrt{a_n+(n+2)a_{n+1}+2008} \rfloor$. Prove that the solution set of $a_n \not= n$, $n \in \mathbb{N}$ is finite.
Let $a$ and $b$ be positive integers. The cells of an $(a+b+1)\times (a+b+1)$ grid are colored amber and bronze such that there are at least $a^2+ab-b$ amber cells and at least $b^2+ab-a$ bronze cells. Prove that it is possible to choose $a$ amber cells and $b$ bronze cells such that no two of the $a+b$ chosen cells lie in the same row or column.