Team selection test for the 54th IMO

Denote by $A_1$, $B_1$ and $C_1$ the projections of $A$, $B$ and $C$ on $NP$, $PM$ and $MN$, respectively. It follows from the condition of the problem (Carnot's theorem) that $(N{C}_1^2-M{C}_1^2)+(M{B}_1^2-P{B}_1^2)+(P{A}_1^2-N{A}_1^2)=0$. Hence $0=(C{N}^2-C{M}^2)+(B{M}^2-B{P}^2)+(A{P}^2-A{N}^2)=(C{N}^2-A{N}^2)+(A{P}^2-B{P}^2)+(B{M}^2-C{M}^2)$ and the same theorem implies that the perpendiculars from $M$, $N$ and $P$ to $BC$, $AC$ and $AB$ are concurrent at a point $O$.

Since $OM{H}_{C}N$ and $OM{H}_{B}P$ are parallelograms we conclude that $P{H}_B{H}_{C}N$ is also a parallelogram. Therefore the segments $P{H}_C$ and $N{H}_B$ bisect each other at some point. The segment $M{H}_A$ passes through the same point.