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Problem 1049

AMC 12 late, AIME early

Geometry Difficulty 4.6 Prove it Harvard-MIT Mathematics Tournament · United States

Let the incircle of $ABCD$ be tangent to sides $AB$ , $BC$ , $CD$ , and $AD$ at points $P$ , $Q$ , $R$ , and $S$ , respectively. Show that $ABCD$ is cyclic if and only if $PR \perp QS$ .

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official solution

Solution:

Let the diagonals of $PQRS$ intersect at $T$ . Because $\overline{AP}$ and $\overline{AS}$ are tangent to $\omega$ at $P$ and $S$ , we may write $\alpha = \angle ASP = \angle SPA = \angle SQP$ and $\beta = \angle CQR = \angle QRC = \angle QPR$ . Then $\angle PTQ = \pi - \alpha - \beta$ . On the other hand, $\angle PAS = \pi - 2\alpha$ and $\angle RCQ = \pi - 2\beta$ , so that $ABCD$ is cyclic if and only if
$\pi = \angle BAD + \angle DCB = 2\pi - 2\alpha - 2\beta$
or simply
$\pi/2 = \pi - \alpha - \beta = \angle PTQ$
as desired.

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