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Singapore Mathematical Olympiad (SMO)

Geometry Difficulty 4.8 AIME Prove it Singapore

Let the diagonals of the square $ABCD$ intersect at $S$ and let $P$ be the midpoint of $AB$ . Let $M$ be the intersection of $AC$ and $PD$ and $N$ the intersection of $BD$ and $PC$ . A circle is inscribed in the quadrilateral $PMSN$ . Prove that the radius of the circle is $MP - MS$ .

Solution

Let $O$ be the centre and $r$ the radius of the circle. Let $X, Y$ be its points of contact with the sides $PM, MS$ , respectively.

Since $OY \perp MS$ and $\angle YSO = \angle ASP = 45^\circ$ , $SY = YO = r$ . Also $\angle OPX = \angle PDA$
(since $OP \parallel DA$ ) and $\angle OXP = \angle PAD = 90^\circ$ . Therefore $\triangle OXP \sim \triangle PAD$ . Hence
$OX/XP = PA/AD = 1/2$ . Hence $PX = 2r$ . Therefore $PM - MS = 2r + MX - MY - r = r$ .

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