Let $ABC$ be a triangle. $r$ and $s$ are the angle bisectors of $\angle ABC$ and $\angle BCA$, respectively. The points $E$ in $r$ and $D$ in $s$ are such that $AD \| BE$ and $AE \| CD$. The lines $BD$ and $CE$ cut each other at $F$. $I$ is the incenter of $ABC$. Show that if $A,F,I$ are collinear, then $AB=AC$.

Find the smallest number \(n\) such that any set of \(n\) ponts in a Cartesian plan, all of them with integer coordinates, contains two poitns such that the square of its mutual distance is a multiple of \(2016\).

Let it \(k\) be a fixed positive integer. Alberto and Beralto play the following game: Given an initial number \(N_0\) and starting with Alberto, they alternately do the following operation: change the number \(n\) for a number \(m\) such that \(m < n\) and \(m\) and \(n\) differ, in its base-2 representation, in exactly \(l\) consecutive digits for some \(l\) such that \(1 \leq l \leq k\). If someone can't play, he loses. We say a non-negative integer \(t\) is a winner number when the gamer who receives the number \(t\) has a winning strategy, that is, he can choose the next numbers in order to guarrantee his own victory, regardless the options of the other player. Else, we call it loser. Prove that, for every positive integer \(N\), the total of non-negative loser integers lesser than \(2^N\) is \(2^{N-\lfloor \frac{log(min\{N,k\})}{log 2} \rfloor}\)

What is the greatest number of positive integers lesser than or equal to 2016 we can choose such that it doesn't have two of them differing by 1,2, or 6?

Consider the second-degree polynomial \(P(x) = 4x^2+12x-3015\). Define the sequence of polynomials \(P_1(x)=\frac{P(x)}{2016}\) and \(P_{n+1}(x)=\frac{P(P_n(x))}{2016}\) for every integer \(n \geq 1\). Show that exists a real number \(r\) such that \(P_n(r) < 0\) for every positive integer \(n\). Find how many integers \(m\) are such that \(P_n(m)<0\) for infinite positive integers \(n\).

Lei it \(ABCD\) be a non-cyclical, convex quadrilateral, with no parallel sides. The lines \(AB\) and \(CD\) meet in \(E\). Let it \(M \not= E\) be the intersection of circumcircles of \(ADE\) and \(BCE\). The internal angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(I\). The external angle bisectors of \(ABCD\) form an convex, cyclical quadrilateral with circumcenter \(J\). Show that \(I,J,M\) are colinear.