olymmath

JBMO-SL 2018 G5

Given a rectangle ABCDA B C D such that AB=b>2a=BCA B=b>2 a=B C, let EE be the midpoint of ADA D. On a line parallel to ABA B through point EE, a point GG is chosen such that the area of GCEG C E is

(GCE)=12(a3b+ab)(G C E)=\frac{1}{2}\left(\frac{a^{3}}{b}+a b\right)

Point HH is the foot of the perpendicular from EE to GDG D and a point II is taken on the diagonal ACA C such that the triangles ACEA C E and AEIA E I are similar. The lines BHB H and IEI E intersect at KK and the lines CAC A and EHE H intersect at JJ. Prove that KJ⊥ABK J \perp A B.

Dataset text NuminaMath olympiads. Quoted with source.

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