SAUDI ARABIAN MATHEMATICAL COMPETITIONS

Let $I$ be the midpoint of $BC$. We have $\angle ICM=\angle MAC=\angle MCP$ then $CM$ is the bisector of $\angle ICP$. This follows that $M$ is the insimilicenter of $(I)$ and $(P)$. However, $\angle MCN=90^{\circ }$ hence $N$ is the exsimilicenter of $(I)$ and $(P)$.

Let $L$, $T$ be the intersections of $FM$ and $(P)$ ($L$ is between $F$ and $M$). Consider the homothety center at $M$, ratio $\frac{-PC}{IC}$, ${\mathcal{H}}_M^{\frac{-PC}{IC}}:F\mapsto T$, we get $IF\parallel PT$.

By the same way, consider the homothety ${\mathcal{H}}_N^{\frac{PC}{IC}}:F\mapsto E$, we get $IF\parallel PE$.

Therefore $E$, $P$, $T$ are collinear or $ET$ is the diameter of $(P)$. This means $\angle ELP=90^{\circ }$.

Denote $J$ the second intersection of $FN$ and $(O)$.

Since 3 circles $(O)$, $(P)$, $(ELMJ)$ have $EL$, $BC$, $JM$ as their radical axes then $EL$ cuts $JM$ at point $K$ lying on $BC$.

In conclusion, orthocenter $K$ of triangle $EFM$ lies on $BC$. $\square$