Let real coefficient polynomial $f(x) = x^n + a_1 \cdot x^{n-1} + \ldots + a_n$ has real roots $b_1, b_2, \ldots, b_n$, $n \geq 2,$ prove that $\forall x \geq max\{b_1, b_2, \ldots, b_n\}$, we have \[f(x+1) \geq \frac{2 \cdot n^2}{\frac{1}{x-b_1} + \frac{1}{x-b_2} + \ldots + \frac{1}{x-b_n}}.\]

Click for solution Put $x-b_i=a_i$. We must prove that for positive numbers $a_i$ we have $ (1+a_1)...(1+a_n)(1/a_1+...+1/a_n)\geq 2n^2$. USe Huygens inequality to find that the LHS is greater than or equal to $n(1+G)^n/G$ where $G$ is the geometric mean of $a_i$. A study of the function $\frac{(1+x)^n}{x}$ shows that it has minimal value in $\frac{1}{n-1}$ and this one is $ (n-1)(n/(n-1))^n$ obviously greater than $2n$.

For $i = 1,2, \ldots, 1991$, we choose $n_i$ points and write number $i$ on them (each point has only written one number on it). A set of chords are drawn such that: (i) They are pairwise non-intersecting. (ii) The endpoints of each chord have distinct numbers. If for all possible assignments of numbers the operation can always be done, find the necessary and sufficient condition the numbers $n_1, n_2, \ldots, n_{1991}$ must satisfy for this to be possible.

$5$ points are given in the plane, any three non-collinear and any four non-concyclic. If three points determine a circle that has one of the remaining points inside it and the other one outside it, then the circle is said to be good. Let the number of good circles be $n$; find all possible values of $n$.

Click for solution Yeah! Federico is our Team leader! he has shown us many times the proof of this beautiful result! anyway it is to hard to write it down, i just give some hints: Draw all circles and show that moving a point to another region which is a neighboor of it starting region doesnt change the number of halving circles! (the point may only pass trough one boundary) Conclude this number is invariant Use induction in $n$, to prove for $2n+1$ points it is $n^2$.

We choose 5 points $A_1, A_2, \ldots, A_5$ on a circumference of radius 1 and centre $O.$ $P$ is a point inside the circle. Denote $Q_i$ as the intersection of $A_iA_{i+2}$ and $A_{i+1}P$, where $A_7 = A_2$ and $A_6 = A_1.$ Let $OQ_i = d_i, i = 1,2, \ldots, 5.$ Find the product $\prod^5_{i=1} A_iQ_i$ in terms of $d_i.$

Let $f$ be a function $f: \mathbb{N} \cup \{0\} \mapsto \mathbb{N},$ and satisfies the following conditions: (1) $f(0) = 0, f(1) = 1,$ (2) $f(n+2) = 23 \cdot f(n+1) + f(n), n = 0,1, \ldots.$ Prove that for any $m \in \mathbb{N}$, there exist a $d \in \mathbb{N}$ such that $m | f(f(n)) \Leftrightarrow d | n.$

Click for solution Let's note that our sequence $f(n)$ satisfies: $f(n)=f(n-k-1)f(k)+f(n-k)f(k+1)$ (it can be checked by induction). We know that there doesn't exist such $n$ that $m|f(n)$ and $gcd(m,f(n+1))>1$ for any $m \in N-\{1\}$, because going backwards recurentially we would obtain $1<gcd(m,f(1))=gcd(m,1)$ what's false. For every $m \in N$ there exists the smallest $k_0 \in N$ such that $m|f(k_0)$. Let's proove that $m|f(n)$ if and only if $k_0|n$. Let's assume that $m>1$ because for $m=1$ there is nothing to proove. Suppose there exists such $t \in N$ that $m|f(t)$ and $k_0$ doesn't divide $t$. Thus we have: $m|f(t)=>m|f(t-k_0-1)f(k_0)+f(t-k_0)f(k_0+1)=>m|f(t-k_0)$, cause $m$ satisfies $gcd(m,f(k_0+1))=1$-what we have already mentioned. So if $m|f(t)$ then $m|f(t-k_0)$. Because $t$ is not divisible by $k_0$ then there exists such $i$ : $i \geq 0$ that $0<t-ik_0<k_0$, so we received a number $t-ik_0$, which is smaller than $k_0$ and satisfies $m|f(t-ik_0)$ - we have a contradiction. Thus we proved that for every $m \in N$ there exists such $k_0 \in N$ that: $m|f(n)$ if and only if $k_0|n$. So $m|f(f(n))$ if and only if $k_0|f(n)$ if and only if $k_1|n$. So our $d$ satisfies: $d=k_1$.

All edges of a polyhedron are painted with red or yellow. For an angle of a facet, if the edges determining it are of different colors, then the angle is called excentric. The excentricity of a vertex $A$, namely $S_A$, is defined as the number of excentric angles it has. Prove that there exist two vertices $B$ and $C$ such that $S_B + S_C \leq 4$.