IMO Hk TST

Geometry Difficulty 7.7 National Olympiad, round 2 Prove it Hong Kong

From a point $P$ outside a circle centred at $O$ , draw the two tangents to the circle touching it at $A, B$ . Let $M$ be a point on the segment $AB$ and let $C, D$ be points on the circle with midpoint $M$ . Let the tangents to the circle at $C, D$ intersect at $Q$ . Show that $OQ \perp PQ$ .

Solution

This is a simple corollary of Brokard's theorem. Alternatively, we provide an elementary proof as follows.
Note that $O$ , $M$ , $Q$ are collinear since all of them lie on the perpendicular bisector of $CD$ . By the property of tangents, we know that $Q$ , $C$ , $O$ , $D$ are concyclic. This yields
$MQ \times MO = MC \times MD = MA \times MB.$
Thus, $Q$ , $A$ , $O$ , $B$ are concyclic, and hence $P$ , $Q$ , $A$ , $O$ , $B$ are concyclic. Therefore, we obtain
$\angle OQP = \angle OAP = 90^\{\circ\}.$
This means $OQ \perp PQ$ .

Want a route through all this instead of an archive? The track puts 2,604 problems in a working order, from Junior Challenge level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.