50th Mathematical Olympiad in Ukraine, Fourth Round (March 23, 2010)

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

We also have $PB\perp O_A{O}_C\parallel A^{\prime }{C}^{\prime }\parallel AC$, because $A^{\prime }{C}^{\prime }$ 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 $△ABC$. Obviously, if $P$ is an orthocenter then $O_P=P$.