, 2010

Geometry Difficulty 5.4 AIME, harder Prove it Ukraine

The point $P$ lies inside triangle $ABC$ . Denote by $O_A, O_B, O_C$ the circumcenters of triangles $PBC, PAC, PAB$ respectively. Let $O_P$ be the circumcenter of triangle $O_A O_B O_C$ . Prove that the point $P$ satisfies the condition $O_P = P$ if $P$ is the orthocenter of triangle $ABC$ .

Solution

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 $O_P = P$ .

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