I'm posting here the question because I want to see a nice synthetic solution (not using complex numbers or inversive geometry) for the 3rd problem from IMO 2015. The problem is as follows:
Let $ABC$ be an acute triangle with $AB>AC$. Let $\Gamma$ be its circumcircle, $H$ its orthocenter, and $F$ the foot of the altitude from $A$. Let $M$ be the midpoint of $BC$. Let $Q$ be the point on $\Gamma$ such that $\angle HQA=90^{\circ}$ and let $K$ be the point on $\Gamma$ such that $\angle HKQ=90^{\circ}$. Assume that the points $A,B,C,K$ and $Q$ are all different and lie on $\Gamma$ in this order.
Prove that the circumcircles of triangles $KQH$ and $FKM$ are tangent to each other.
I checked the post from AoPS but I was not satisfied. I solved the problem using complex numbers and I want to see (if possible) a full synthetic solution and the motivation using some points or constructions.
