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Annual Harvard-MIT Mathematics Tournament

Geometry Difficulty 5.3 AIME, harder Prove it United States

Problem:
Let $ABCD$ be a cyclic quadrilateral, and let $P$ be the intersection of its two diagonals. Points $R$ , $S$ , $T$ , and $U$ are feet of the perpendiculars from $P$ to sides $AB$ , $BC$ , $CD$ , and $AD$ , respectively. Show that quadrilateral $RSTU$ is bicentric if and only if $AC \perp BD$ . (Note that a quadrilateral is called inscriptible if it has an incircle; a quadrilateral is called bicentric if it is both cyclic and inscriptible.)

Solution

Solution:
First we show that $RSTU$ is always inscriptible. Note that in addition to $ABCD$ , we have cyclic quadrilaterals $ARPU$ and $BSPR$ . Thus,
$\angle PRU = \angle PAU = \angle CAD = \angle CBD = \angle SBP = \angle SRP$
and it follows that $P$ lies on the bisector of $\angle SRU$ . Analogously, $P$ lies on the bisectors of $\angle TSR$ and $\angle UTS$ , so is equidistant from lines $UR$ , $RS$ , $ST$ , and $TU$ , and $RSTU$ is inscriptible having incenter $P$ .

Now we show that $RSTU$ is cyclic if and only if the diagonals of $ABCD$ are orthogonal. We have
$\begin{aligned} & \angle APB = \pi - \angle BAP - \angle PBA = \pi - \angle RAP - \angle PBR = \pi - \angle RUP - \angle PSR \\ & \quad = \pi - \frac{1}{2}(\angle RUT + \angle TSR) \end{aligned}$
It follows that $\angle APB = \frac{\pi}{2}$ if and only if $\angle RUT + \angle TSR = \pi$ , as desired.

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