A set $ S$ of points from the space will be called completely symmetric if it has at least three elements and fulfills the condition that for every two distinct points $ A$ and $ B$ from $ S$, the perpendicular bisector plane of the segment $ AB$ is a plane of symmetry for $ S$. Prove that if a completely symmetric set is finite, then it consists of the vertices of either a regular polygon, or a regular tetrahedron or a regular octahedron.

Let $n \geq 2$ be a fixed integer. Find the least constant $C$ such the inequality \[\sum_{i<j} x_{i}x_{j} \left(x^{2}_{i}+x^{2}_{j} \right) \leq C \left(\sum_{i}x_{i} \right)^4\] holds for any $x_{1}, \ldots ,x_{n} \geq 0$ (the sum on the left consists of $\binom{n}{2}$ summands). For this constant $C$, characterize the instances of equality.

Let $n$ be an even positive integer. We say that two different cells of a $n \times n$ board are neighboring if they have a common side. Find the minimal number of cells on the $n \times n$ board that must be marked so that any cell (marked or not marked) has a marked neighboring cell.

Find all the pairs of positive integers $(x,p)$ such that p is a prime, $x \leq 2p$ and $x^{p-1}$ is a divisor of $ (p-1)^{x}+1$.

Click for solution First of all, if $p=2$, then $x$ is $1$ or $2$. Assume now $p\ge 3$. Clearly, $x$ must be odd (let's say that $x=2k+1$), and let's assume $x>1$. Let $q$ be the smallest prime divisor of $x$. We have $q|((p-1)^2x-1,(p-1)^{q-1}-1)=(p-1)^2-1=p(p-2)$. Assume $q\ne p$. We find that $q|p-2$. At the same time, we have $q|(p-1)^x+1=(p-1)^{2k+1}+1=[(p-1)^{2k}-1](p-1)+p$, so $q|p$, which contradicts the fact that $q|p-2$ and $q$ is odd. This means that $q=p$, so $x\in\{p,2p\}$, and since it's odd, we have $x=p$. We now have to find those odd primes $p$ s.t. $p^{p-1}|(p-1)^p+1$. After expanding the binomial on the right, we find that the expression on the right is $p^2+kp^3$, so $p-1\le 2$, meaning that the only odd prime satisfying this is $p=3$ (it it does satisfy the relation, since $3^2|2^3+1$). The solutions are then $(1,p)$ for any prime $p$, $(2,2)$ and $(3,3)$.

Two circles $\Omega_{1}$ and $\Omega_{2}$ touch internally the circle $\Omega$ in M and N and the center of $\Omega_{2}$ is on $\Omega_{1}$. The common chord of the circles $\Omega_{1}$ and $\Omega_{2}$ intersects $\Omega$ in $A$ and $B$. $MA$ and $MB$ intersects $\Omega_{1}$ in $C$ and $D$. Prove that $\Omega_{2}$ is tangent to $CD$.

Click for solution {U,V} = o1^o2; E=AN^o2; F=BN^o2;p=DF^CE ; X = centre(o1); Y = centre(o2). AC*AM = AV*AU = AE*AN ==> MNEC cyclic; <ACE = <ANM = <ABM = <CDM ==> CE tangent to o1; <AEC = <AMN = <ABN = <EFN ==> EC tangent to o2; G= o2^XY, H = foot(Y) on CX; YH//CE ==> XH = r1 – r2 = XG ==> CXG == YXH ==> <CGX = 90° XY symmetry axes of CD and EF ==> <DGX = 90°.

Find all the functions $f: \mathbb{R} \to\mathbb{R}$ such that \[f(x-f(y))=f(f(y))+xf(y)+f(x)-1\]for all $x,y \in \mathbb{R} $.