Mathematical Olympiad in Ukraine, Fourth Round (March 23, ), 2010

Let $A' = PA \cap O_B O_C$, $B' = PB \cap O_A O_C$, $C' = PC \cap O_B O_A$. $O_A O_C$ is a perpendicular bisector of $BP$, thus $B'$ is a midpoint of $BP$. By analogy, $A', C'$ are midpoints of $PA, PC$ (Fig.07). If $P$ is a circumcenter of $\triangle O_A O_B O_C$, then the perpendicular from $P$ to $O_A O_C$ passes through the midpoint of $O_A O_C$, hence $B'$ is also a midpoint of $O_A O_C$. Following the same lines, we get that $A', C'$ are midpoints of $O_B O_C, O_A O_B$.

We also have $PB \perp O_A O_C \parallel A'C' \parallel AC$, because $A'C'$ is a midline of triangles $O_A O_B O_C$ and $APC$. By analogy, $PA \perp BC, PC \perp AB$, thus $P$ is an orthocenter of $\triangle ABC$. Obviously, if $P$ is an orthocenter then $O_P = P$.