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Estonian Math Competitions

Geometry Difficulty 4.9 AIME Prove it Estonia

In an acute triangle $ABC$ , a point $P$ is chosen such that all points symmetrical to $P$ with respect to the sides of $ABC$ lie on the circumcircle of $ABC$ . Prove that $P$ is the orthocenter of $ABC$ .

Solution

Let $A'$ , $B'$ , $C'$ be points symmetric to the point $P$ with respect to the sides $BC$ , $CA$ , $AB$ (Fig. 17).

Then $|C'A| = |PA| = |B'A|$ , giving that the arcs $AC'$ and $AB'$ of the circumcircle of the triangle $ABC$ are equal. Since $A$ and $C'$ are on the same half-plane from the line $BB'$ , and $C$ on the other one, we have $\angle C'CA = \angle B'CA = \angle PCA$ . Since $P$ and $C'$ are on the same side from the line $AC$ , the points $P$ , $C$ , $C'$ are collinear. Since $PC' \perp AB$ , we must also have $PC \perp AB$ , that is, the point $P$ lies on the height drawn from the vertex $C$ in the triangle $ABC$ . Analogously we see that $P$ is on the other two heights.

Fig. 17

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