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Problem 1842

IMO Shortlist mid-range; USAMO P2/P5

Geometry Difficulty 8.6 Prove it Baltic Way 2011 Problem Shortlist · Baltic Way · 2011

Let $E$ be an interior point in the convex quadrilateral $ABCD$ . Let $F$ , $G$ , $H$ , and $I$ be points opposite the quadrilateral with respect to the lines $AB$ , $BC$ , $CD$ , and $DA$ , respectively, such that $\triangle ABF \sim \triangle DCE$ , $\triangle BCG \sim \triangle ADE$ , $\triangle CDH \sim \triangle BAE$ , and $\triangle DAI \sim \triangle CBE$ . Let $P$ , $Q$ , $R$ , and $S$ be the projections of $E$ on the lines $AB$ , $BC$ , $CD$ , and $DA$ , respectively. Prove that if the quadrilateral $PQRS$ is cyclic, then
$EF \cdot CD = EG \cdot DA = EH \cdot AB = EI \cdot BC.$

This one wants a proof. Work it on paper, read the official solution, then mark yourself honestly — the ladder only means something if the record is true.

Official solution

We consider oriented angles modulo $180^\circ$ . From the cyclic quadrilaterals $APES$ , $BQEP$ , $PQRS$ , $CREQ$ , $DSER$ and $\triangle DCE \sim \triangle ABF$ we get
$\begin{align*} \angle AEB &= \angle EAB + \angle ABE = \angle ESP + \angle PQE \\ &= \angle ESR + \angle RSP + \angle PQR + \angle RQE \\ &= \angle ESR + \angle RQE = \angle EDC + \angle DCE \\ &= \angle DEC = \angle AFB, \end{align*}$
so the quadrilateral $AEBF$ is cyclic. By Ptolemy we then have
$EF \cdot AB = AE \cdot BF + BE \cdot AF.$
This transforms by $AB : BF : AF = DC : CE : DE$ into
$EF \cdot CD = AE \cdot CE + BE \cdot DE.$

Since the expression on the right of this equation is invariant under cyclic permutation of the vertices of the quadrilateral $ABCD$ , the asserted equation follows immediately.

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