Solve the equation $3^x=2^xy+1$ in positive integers.

Let $n\geq 1$ be an integer and let $X$ be a set of $n^2+1$ positive integers such that in any subset of $X$ with $n+1$ elements there exist two elements $x\neq y$ such that $x\mid y$. Prove that there exists a subset $\{x_1,x_2,\ldots, x_{n+1} \} \in X$ such that $x_i \mid x_{i+1}$ for all $i=1,2,\ldots, n$.

Prove that if the distance from a point inside a convex polyhedra with $n$ faces to the vertices of the polyhedra is at most 1, then the sum of the distances from this point to the faces of the polyhedra is smaller than $n-2$. Calin Popescu

Prove that in any convex polygon with $4n+2$ sides ($n\geq 1$) there exist two consecutive sides which form a triangle of area at most $\frac 1{6n}$ of the area of the polygon.

Let $m,n$ be co-prime integers, such that $m$ is even and $n$ is odd. Prove that the following expression does not depend on the values of $m$ and $n$: \[ \frac 1{2n} + \sum^{n-1}_{k=1} (-1)^{\left[ \frac{mk}n \right]} \left\{ \frac {mk}n \right\} . \] Bogdan Enescu

A sequence of real numbers $\{a_n\}_n$ is called a bs sequence if $a_n = |a_{n+1} - a_{n+2}|$, for all $n\geq 0$. Prove that a bs sequence is bounded if and only if the function $f$ given by $f(n,k)=a_na_k(a_n-a_k)$, for all $n,k\geq 0$ is the null function. Mihai Baluna - ISL 2004

Let $A_0A_1A_2A_3A_4A_5$ be a convex hexagon inscribed in a circle. Define the points $A_0'$, $A_2'$, $A_4'$ on the circle, such that \[ A_0A_0' \parallel A_2A_4, \quad A_2A_2' \parallel A_4A_0, \quad A_4A_4' \parallel A_2A_0 . \] Let the lines $A_0'A_3$ and $A_2A_4$ intersect in $A_3'$, the lines $A_2'A_5$ and $A_0A_4$ intersect in $A_5'$ and the lines $A_4'A_1$ and $A_0A_2$ intersect in $A_1'$. Prove that if the lines $A_0A_3$, $A_1A_4$ and $A_2A_5$ are concurrent then the lines $A_0A_3'$, $A_4A_1'$ and $A_2A_5'$ are also concurrent.

Let $ABC$ be a triangle, and let $D$, $E$, $F$ be 3 points on the sides $BC$, $CA$ and $AB$ respectively, such that the inradii of the triangles $AEF$, $BDF$ and $CDE$ are equal with half of the inradius of the triangle $ABC$. Prove that $D$, $E$, $F$ are the midpoints of the sides of the triangle $ABC$.

Let $P$ be a polygon (not necessarily convex) with $n$ vertices, such that all its sides and diagonals are less or equal with 1 in length. Prove that the area of the polygon is less than $\dfrac {\sqrt 3} 2$.

Let $a\in\mathbb{R}-\{0\}$. Find all functions $f: \mathbb{R}\to\mathbb{R}$ such that $f(a+x) = f(x) - x$ for all $x\in\mathbb{R}$. Dan Schwartz

On the edges of a convex polyhedra we draw arrows such that from each vertex at least an arrow is pointing in and at least one is pointing out. Prove that there exists a face of the polyhedra such that the arrows on its edges form a circuit. Dan Schwartz

Let $n\geq 0$ be an integer and let $p \equiv 7 \pmod 8$ be a prime number. Prove that \[ \sum^{p-1}_{k=1} \left \{ \frac {k^{2^n}}p - \frac 12 \right\} = \frac {p-1}2 . \] Călin Popescu

Click for solution This is one nice problem (in fact, all of them seem to be very nice, except maybe for the functional equation ). Let's solve it for $n=1$, and then we'll see why this is enough. We must compute $\sum_{k=1}^{p-1}\left(\frac{k^2}p-\frac 12-\left\lfloor\frac{k^2}p-\frac 12\right\rfloor\right)$, which is equal to $\sum_{k=1}^{p-1}\left(\frac{2k^2}p-\frac 12-\left\lfloor\frac{2k^2}p\right\rfloor+1\right)-\sum_{k=1}^{p-1}\left(\frac{k^2}p-\left\lfloor\frac{k^2}p\right\rfloor\right)\ (*)$, because of the identity $\left\lfloor x\right\rfloor+\left\lfloor x+\frac 12\right\rfloor=\left\lfloor 2x\right\rfloor$ applied to $x=\frac{k^2}p-\frac 12$. Now, $(*)$ is equal to $\sum_{k=1}^{p-1}\left(\left\{\frac{2k^2}p\right\}-\left\{\frac{k^2}p\right\}\right)+\frac{p-1}2$, which equals $\frac{p-1}2$ because $2$ is a quadratic residue $\pmod p$, so the residues $2k^2\pmod p$ are the same as the residues $k^2\pmod p$. The identity is proven for $n=1$. This is enough, because the residues $k^{2^n}\pmod p$ are the same as the residues $k^2\pmod p$, since $-1$ is not a quadratic residue $\pmod p$, meaning that the equation $x^4=1$ has $2$ solutions $\pmod p$, so there are $\frac{p-1}2$ fourth powers $\pmod p$, and we can apply induction: there are also $\frac{p-1}2$ $8$'th powers, and so on, and these are precisely the quadratic residues. It says $n\ge 0$ in the text, but if it's also true for $n=0$, it's certainly not interesting enough to work out . Edit: edited in response to prowler's post below

a) Prove that there exists a sequence of digits $\{c_n\}_{n\geq 1}$ such that or each $n\geq 1$ no matter how we interlace $k_n$ digits, $1\leq k_n\leq 9$, between $c_n$ and $c_{n+1}$, the infinite sequence thus obtained does not represent the fractional part of a rational number. b) Prove that for $1\leq k_n\leq 10$ there is no such sequence $\{c_n\}_{n\geq 1}$. Dan Schwartz

On a $2004 \times 2004$ chess table there are 2004 queens such that no two are attacking each other\footnote[1]{two queens attack each other if they lie on the same row, column or direction parallel with on of the main diagonals of the table}. Prove that there exist two queens such that in the rectangle in which the center of the squares on which the queens lie are two opposite corners, has a semiperimeter of 2004.

Let $n\geq 2$ be an integer. Find the smallest real value $\rho (n)$ such that for any $x_i>0$, $i=1,2,\ldots,n$ with $x_1 x_2 \cdots x_n = 1$, the inequality \[ \sum_{i=1}^n \frac 1{x_i} \leq \sum_{i=1}^n x_i^r \] is true for all $r\geq \rho (n)$.

Let $\mathbb{N}=\{1,2,\ldots\}$. Find all functions $f: \mathbb{N}\to\mathbb{N}$ such that for all $m,n\in \mathbb{N}$ the number $f^2(m)+f(n)$ is a divisor of $(m^2+n)^2$.

We consider a polyhedra which has exactly two vertices adjacent with an odd number of edges, and these two vertices are lying on the same edge. Prove that for all integers $n\geq 3$ there exists a face of the polyhedra with a number of sides not divisible by $n$.