Let $O$ be the centre of the square $ABCD$. Let $P,Q,R$ be respectively on the segments $OA,OB,OC$ such that $OP=3,OQ=5,OR=4$. Suppose $S$ is on $OD$ such that $X=AB\cap PQ,Y=BC\cap QR$ and $Z=CD\cap RS$ are collinear. Find $OS$.
Problem
Source: India Postal Coaching 2014 Set 3 Problem 2
Tags: geometry, rectangle, angle bisector, india
Luis González
30.11.2014 00:11
From problem condition, $ABCD$ and $PQRS$ are perspective. Thus $O,$ $M \equiv PQ \cap RS$ and $AB \cap CD$ are collinear and $O,$ $N \equiv QR \cap PS$ and $BC \cap DA$ are collinear $\Longrightarrow$ $OM \parallel AB$ and $ON \parallel BC,$ i.e. $OM,ON$ bisect $\angle POQ.$ Hence, by angle bisector theorem and Van Aubel's theorem for the cevian triangle $\triangle OMN$ of $S$ WRT $\triangle PQR,$ we obtain $\frac{OQ}{OP}+\frac{OQ}{OR}=\frac{MQ}{MP}+\frac{NQ}{NR}= \frac{QS}{OS}=\frac{OQ+OS}{OS} \Longrightarrow \frac{1}{OP}+\frac{1}{OR}=\frac{1}{OQ}+\frac{1}{OS}.$ $OS=\frac{1}{ \frac{1}{OP}+\frac{1}{OR}-\frac{1}{OQ}}=\frac{1}{\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}=\frac{60}{23}.$
sunken rock
30.11.2014 15:25
Remark: $ABCD$ could be a rectangle, and we get the same result: Let's lift $O$ above the plan $ABCD$ to $V$ so that $VO\bot ABCD$. $ABCD$ and $PQRS$ being perspective, $P,Q,R,S$ remain coplanar points, i.e. $PR, QS$ intersect the common angle bisector of $\angle PVR,\angle QVS$ at the same point $O'$. Now let's rotate the triangle $\triangle VAC$ over triangle $VBD$, such as $A\equiv B, C\equiv D$ and note $P',Q',R',S'$ the new positions of $P,Q,R,S$ respectively. $P'R'$ intersects $Q'S'$ at $O'$ onto the bisector of angle $\angle AVC$ and let's note $O"=VO'\cap Q'R'$. The segments $VP'$, a.s.o. are proportional to $OP$, a.s.o. and we may apply Ceva: $\frac{VP'}{P'Q'}\cdot\frac{Q'O"}{O"R'}\cdot\frac{R'S'}{S'V"}=1$, with $\frac{Q'O"}{O"R'}=\frac{5}4$ and $\frac{VS'}{S'R'}=\frac{x}{4-x}$. We shall get $60=23x$. Best regards, sunken rock