Competition problem — Geometry — Maths Olympiad Prep

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Geometry Difficulty 4.4 AIME Prove it Romania

Let $ABCDE$ be a convex pentagon with $AB + CD = BC + DE$ , and $k$ a circle centered on side $AE$ , tangent to sides $AB$ , $BC$ , $CD$ and $DE$ at points $P$ , $Q$ , $R$ and $S$ respectively. Prove that lines $PS$ and $AE$ are parallel.

Solution

$AB + CD = (AP + PB) + (CR + RD) = AP + (BP + CR + DR)$
$BC + DE = (BQ + QC) + (DS + SE) = ES + (BQ + CQ + DS).$
Using the fact that tangents from a point to a circle are of equal length, one gets $AP = ES$ . Denoting by $O$ and $r$ the center, respectively radius of $k$ , one gets that right-angled triangles $OPA$ , $OSE$ are congruent, since $AP = ES$ , $OP = OS = r$ , and $OA = \sqrt{OP^2 + AP^2} = \sqrt{OS^2 + ES^2} = OE$ . Therefore the corresponding altitudes of these triangles, from $P$ , respectively $S$ , are of equal length, hence $PS$ is parallel to $AE$ .

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