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Junior Balkan Mathematical Olympiad

Geometry Difficulty 6.5 National Olympiad Prove it JBMO

Problem:
Given a rectangle $ABCD$ such that $AB = b > 2a = BC$ , let $E$ be the midpoint of $AD$ . On a line parallel to $AB$ through point $E$ , a point $G$ is chosen such that the area of $GCE$ is
$(GCE) = \frac{1}{2}\left(\frac{a^{3}}{b} + ab\right)$
Point $H$ is the foot of the perpendicular from $E$ to $GD$ and a point $I$ is taken on the diagonal $AC$ such that the triangles $ACE$ and $AEI$ are similar. The lines $BH$ and $IE$ intersect at $K$ and the lines $CA$ and $EH$ intersect at $J$ . Prove that $KJ \perp AB$ .

Solution

Solution:
Let $L$ be the foot of the perpendicular from $G$ to $EC$ and let $Q$ be the point of intersection of the lines $EG$ and $BC$ . Then,
$(GCE) = \frac{1}{2} EC \cdot GL = \frac{1}{2} \sqrt{a^{2} + b^{2}} \cdot GL$
So, $GL = \frac{a}{b} \sqrt{a^{2} + b^{2}}$ .

Observing that the triangles $QCE$ and $ELG$ are similar, we have $\frac{a}{b} = \frac{GL}{EL}$ , which implies that $EL = \sqrt{a^{2} + b^{2}}$ , or in other words $L \equiv C$ .

Consider the circumcircle $\omega$ of the triangle $EBC$ . Since
$\angle EBG = \angle ECG = \angle EHG = 90^{\circ}$
the points $H$ and $G$ lie on $\omega$ .

From the given similarity of the triangles $ACE$ and $AEI$ , we have that
$\angle AIE = \angle AEC = 90^{\circ} + \angle GEC = 90^{\circ} + \angle GHC = \angle EHC$
therefore $EHCI$ is cyclic, thus $I$ lies on $\omega$ .

Since $EB = EC$ , we get that $\angle EIC = \angle EHB$ , thus $\angle JIE = \angle EHK$ . We conclude that $JIH K$ is cyclic, therefore
$\angle JKH = \angle HIC = \angle HBC$
It follows that $KJ \parallel BC$ , so $KJ \perp AB$ .

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