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Problem 821

AMC 12 late, AIME early

Geometry Difficulty 4.4 Prove it Junior Balkan Mathematical Olympiad · 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.

This one wants a proof. Work it on paper, then read the official solution and mark yourself. Be honest about it: the record is only any use to you if it is.

Official 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 $AE$ .

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