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Hong Kong competition problems, 2021

Geometry Difficulty 5.4 AIME, harder Prove it Hong Kong

Let $\triangle ABC$ be an acute triangle with circumcircle $\Gamma$ , and let $P$ be the midpoint of the minor arc $BC$ of $\Gamma$ . Let $AP$ meet $BC$ at $D$ , and let $M$ be the midpoint of $AB$ . Also, let $E$ be the point such that $AE \perp AB$ and $BE \perp MP$ . Prove that $AE = DE$ .

Solution

Since $P$ is the midpoint of the minor arc $BC$ , $AD$ is the internal angle bisector of $\angle BAC$ .
Let $E'$ be the intersection of the perpendicular bisector of $AD$ and the line through $A$ perpendicular to $AB$ . Using the method of false position, it suffices to show $BE' \perp MP$ , and then conclude $E' = E$ . Let $\Omega$ be the circle with centre $E'$ that passes through $A$ (and hence $D$ ).

Since $AB \perp AE'$ , $AB$ is tangent to $\Omega$ . As $MA = MB$ , the powers of $M$ with respect to $\Omega$ and $B$ are equal. Since $\angle PBD = \angle PAC = \angle PAB$ , we have $\triangle PBD \sim \triangle PAB$ , so $PB^2 = PD \times PA$ . Therefore, the powers of $P$ with respect to $\Omega$ and $B$ are equal.
From above, both $M$ and $P$ lie on the radical axis of $\Omega$ and $B$ . Hence, $MP$ is perpendicular to the line joining the centres of the two circles, i.e. $E'B$ . The result follows from above.

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