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China Hong Kong Mathematical Olympiad

Geometry Difficulty 5.4 AIME, harder Prove it Hong Kong

Let $O$ be the circumcentre of $\triangle ABC$ . Suppose $AB > AC > BC$ . Let $D$ be a point on the minor arc $BC$ such that $AD$ is not a diameter of the circumcircle of $\triangle ABC$ . Let $E$ and $F$ be points on $AD$ such that $AB \perp OE$ and $AC \perp OF$ . Let $P$ be the intersection of $BE$ and $CF$ . If $PB = PC + PO$ , then prove that $\angle BAC = 30^\circ$ .

Solution

Note that $OE$ is the perpendicular bisector of $AB$ . Let $\angle EBA = \angle EAB = x$ . Similarly, let $\angle FCA = \angle FAC = y$ . Then we have
$\angle BPC = 180^\circ - \angle PBC - \angle PCB = 2x + 2y = 2 \angle BAC = \angle BOC.$
This implies $O$ , $B$ , $C$ , $P$ are concyclic. Applying Ptolemy's theorem to the cyclic quadrilateral $OBCP$ , we obtain
$OB \times PC + BC \times OP = OC \times BP.$
Thus,
$BC \times OP = OC \times BP - OB \times PC = OB(BP - PC) = OB \times OP.$
This yields $BC = OB$ . Hence, $\triangle OBC$ is equilateral. Therefore,
$\angle BAC = \frac{1}{2} \angle BOC = 30^\circ.$

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