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Junior Balkan Mathematical Olympiad

Geometry Difficulty 5.6 AIME, harder Prove it North Macedonia

Let $ABC$ be an acute triangle, $A'$ , $B'$ and $C'$ be the reflections of the vertices $A$ , $B$ and $C$ with respect to $BC$ , $CA$ and $AB$ , respectively, and let the circumcircles of triangles $ABB'$ and $ACC'$ meet again at $A_1$ . Points $B_1$ and $C_1$ are defined similarly. Prove that the lines $AA_1$ , $BB_1$ and $CC_1$ have a common point.

Solution

Let $O_1$ , $O_2$ and $O$ be the circumcenters of triangles $ABB'$ , $ACC'$ and $ABC$ respectively. As $AB$ is the perpendicular bisector of the line segment $CC'$ , $O_2$ is the intersection of the perpendicular bisector of $AC$ with $AB$ . Similarly, $O_1$ is the intersection of the perpendicular bisector of $AB$ with $AC$ . It follows that $O$ is the orthocenter of triangle $AO_1O_2$ . This means that $AO$ is perpendicular to $O_1O_2$ . On the other hand, the segment $AA_1$ is the common chord of the two circles, thus it is perpendicular to $O_1O_2$ . As a result, $AA_1$ passes through $O$ . Similarly, $BB_1$ and $CC_1$ pass through $O$ , so the three lines are concurrent at $O$ .

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