concurrent wanted, starting with a cyclic ABCD and two circumcircles

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concurrent wanted, starting with a cyclic ABCD and two circumcircles

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Source: 2020 Austrian Federal Competition For Advanced Students, Part 2, p1

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#1 h Nov 22, 2020, 10:25 PM

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Let $ABCD$ be a convex cyclic quadrilateral with the diagonal intersection $S$ . Let further be $P$ the circumcenter of the triangle $ABS$ and $Q$ the circumcenter of the triangle $BCS$ . The parallel to $AD$ through $P$ and the parallel to $CD$ through $Q$ intersect at point $R$ . Prove that $R$ is on $BD$ .

(Karl Czakler)

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#3 h Nov 25, 2020, 11:45 AM • 1 Y

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Austrian MO 2020/Advanced/p1 wrote:

Let $ABCD$ be a convex cyclic quadrilateral with the diagonal intersection $S$ . Let further be $P$ the circumcenter of the triangle $ABS$ and $Q$ the circumcenter of the triangle $BCS$ . The parallel to $AD$ through $P$ and the parallel to $CD$ through $Q$ intersect at point $R$ . Prove that $R$ is on $BD$ .

Karl Czakler

It's clear that $CD\perp PS$ and so $PS\perp QR$ and analogously $QS\perp PR$ . Due to fact that $BS$ is the radical axis of $(ABS)$ , $(BSC)$ , we have $\overline{BSD}\perp PQ$ , and since $\{P,S,Q,R\}$ form an orthocentric system , $RS\perp PQ$ . Hence $R\in BD$ . $\textbf{Q.E.D.}$

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