OME 2017

Find the amount of different values given by the following expression: $\frac{n^2-2}{n^2-n+2}$ where $n \in \{1,2,3,..,100\}$

A midpoint plotter is an instrument which draws the exact mid point of two point previously drawn. Starting off two points $1$ unit of distance apart and using only the midpoint plotter, you have to get two point which are strictly at a distance between $\frac{1}{2017}$ and $\frac{1}{2016}$ units, drawing the minimum amount of points. ¿Which is the minimum number of times you will need to use the midpoint plotter and what strategy should you follow to achieve it?

Let $p$ be an odd prime and $S_{q} = \frac{1}{2*3*4} + \frac{1}{5*6*7} + ... + \frac{1}{q(q+1)(q+2)}$ , where $q = \frac{3p-5}{2}$ . We write $\frac{1}{2}-2S_{q}$ in the form $\frac{m}{n}$ , where $m$ and $n$ are integers. Prove that $m \equiv n (mod p)$

You are given a row made by $2018$ squares, numbered consecutively from $0$ to $2017$ . Initially, there is a coin in the square $0$ . Two players $A$ and $B$ play alternatively, starting with $A$ , on the following way: In his turn, each player can either make his coin advance $53$ squares or make the coin go back $2$ squares. On each move the coin can never go to a number less than $0$ or greater than $2017$ . The player who puts the coin on the square $2017$ wins. ¿Who is the one with the wining strategy and how should he play to win?

Let $a,b,c$ be positive real numbers so that $a+b+c = \frac{1}{\sqrt{3}}$ . Find the maximum value of $$27abc+a\sqrt{a^2+2bc}+b\sqrt{b^2+2ca}+c\sqrt{c^2+2ab}.$$

In the triangle $ABC$ , the respective mid points of the sides $BC$ , $AB$ and $AC$ are $D$ , $E$ and $F$ . Let $M$ be the point where the internal bisector of the angle $\angle ADB$ intersects the side $AB$ , and $N$ the point where the internal bisector of the angle $\angle ADC$ intersects the side $AC$ . Also, let $O$ be the intersection point of $AD$ and $MN$ , $P$ the intersection point of $AB$ and $FO$ , and $R$ the intersection point of $AC$ and $EO$ . Prove that $PR=AD$ .