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Japan Mathematical Olympiad

Geometry Difficulty 5.8 AIME, harder Prove it Japan

For a triangle $ABC$ the line tangent to its circum-circle at $A$ and the line $BC$ intersects at a point $P$ . Let $Q$ and $R$ be the points which are symmetrically located from $P$ with respect to the lines $AB$ and $AC$ , respectively. Prove that the lines $BC$ and $QR$ intersect perpendicularly.

Solution

Without loss of generality, we may assume that the points $B$ , $C$ , $P$ lie on the same straight line in this order. Let us denote by $Q'$ the point of intersection of the lines $AB$ and $PQ$ and by $R'$ the point of intersection of the lines $AC$ and $PR$ .
If $\angle ACB \neq 90^\circ$ , we see from $PQ' \perp AQ'$ and $PR' \perp AR'$ that the 4 points $A$ , $Q'$ , $P$ , $R'$ lie on the circumference of a circle. Therefore, if we let $S$ be the point of intersection of the lines $BC$ and $Q'R'$ , then we get
$\begin{align*} \angle BSQ' &= \angle BPQ' + \angle PQ'S \\ &= \angle BPQ' + \angle PAR' \\ &= \angle BPQ' + \angle PBQ' \\ &= \angle BQ'Q = 90^\circ, \end{align*}$
which means that $BC \perp Q'R'$ .
If $\angle ACB = 90^\circ$ , we see that the points $Q'$ and $R'$ coincide with the points $A$ and $C$ , respectively, and therefore, $BC \perp Q'R'$ holds in this case as well.
$Q'$ , $R'$ are the mid-points of the line segments $PQ$ and $PR$ , respectively, and therefore, we must have $QR \parallel Q'R'$ , and thus we have shown that $BC \perp QR$ must hold.

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