IR-MO-R2 2016

If $0<a\leq b\leq c$ prove that $$\frac{(c-a)^2}{6c}\leq \frac{a+b+c}{3}-\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}$$

Let $ABC$ be a triangle such that $\angle C=2\angle B$ and $\omega$ be its circumcircle. a tangent from $A$ to $\omega$ intersect $BC$ at $E$ . $\Omega$ is a circle passing throw $B$ that is tangent to $AC$ at $C$ . Let $\Omega\cap AB=F$ . $K$ is a point on $\Omega$ such that $EK$ is tangent to $\Omega$ ( $A,K$ aren't in one side of $BC$ ). Let $M$ be the midpoint of arc $BC$ of $\omega$ (not containing $A$ ). Prove that $AFMK$ is a cyclic quadrilateral.

$B$ $C$ $F$ $A$ $E$ $M$ $K$

A council has $6$ members and decisions are based on agreeing and disagreeing votes. We call a decision making method an Acceptable way to decide if it satisfies the two following conditions:

Ascending condition: If in some case, the final result is positive, it also stays positive if some one changes their disagreeing vote to agreeing vote. Symmetry condition: If all members change their votes, the result will also change.Weighted Voting for example, is an Acceptable way to decide. In which members are allotted with non-negative weights like $\omega_1,\omega_2,\cdots , \omega_6$ and the final decision is made with comparing the weight sum of the votes for, and the votes against. For instance if $\omega_1=2$ and for all $i\ge2, \omega_i=1$ , decision is based on the majority of the votes, and in case when votes are equal, the vote of the first member will be the decider. Give an example of some Acceptable way to decide method that cannot be represented as a Weighted Voting method.