Competition problem — Geometry — Olympiad Maths Prep

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Geometry Difficulty 5.6 AIME, harder Prove it Czech Republic

Let $k$ be a semicircle with diameter $PQ$ . Consider a chord $BC$ of fixed length $d$ whose endpoints are distinct from $P$ , $Q$ . A ray of light emanating from $B$ reaches point $C$ after reflecting from $PQ$ at such a point $A$ that $\angle PAB = \angle QAC$ . Prove that $\angle BAC$ doesn't depend on the position of the chord $BC$ on $k$ .

(Šárka Gergelitsová)

Solutions — 2

Solution 1

Reflect $k$ and $C$ about $PQ$ to get $l$ and $C'$ , respectively (Fig. 1). Then $C'$ lies on $l$ and since $\angle QAC' = \angle QAC = \angle PAB$ it also lies on $BA$ . Triangle $C'CA$ is isosceles, hence
$\angle BAC = \angle AC'C + \angle ACC' = 2 \cdot \angle BC'C$
The chord $BC$ of circle $k \cup l$ has a fixed length, hence the corresponding inscribed angle $BC'C$ has fixed size and we may conclude.

Fig. 1

Solution 2

Let $O$ be the midpoint of $PQ$ . We will show that $O$ lies on the circumcircle of triangle $ABC$ (Fig. 2). This will imply that $\angle BAC = \angle BOC$ which is clearly fixed.
Observe that $O$ lies on the perpendicular bisector of $BC$ . Moreover, if $O \neq A$ then $AO$ is the external $A$ -angle bisector with respect to triangle $ABC$ . Therefore $O$ is the midpoint of arc $BAC$ .

Fig. 2

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