Saudi Booklet

Geometry Difficulty 5.4 AIME, harder Prove it Saudi Arabia

Let $ABC$ be a triangle inscribed in circle $(O)$ with $\angle A = 45^\circ$ . Two rays $BO$ , $CO$ intersect $AC$ , $AB$ at $E$ , $F$ respectively. The circumcircles of triangles $BOC$ and $EOF$ intersect at $K$ . Let $J$ be circumcenter of triangle $AEF$ . Prove that $JK$ passes through the orthocenter of triangle $ABC$ .

Solution

We have
$\angle BOC + \angle EOF = 90^\circ + 90^\circ = 180^\circ$
so according to the familiar property of isogonal conjugates in quadrilaterals, we see that there exists a point $O'$ which is the isogonal conjugate of $O$ in $BFEC$ . On the other hand, $BH$ , $BO$ and $CH$ , $CO$ are isogonal pairs in the angles $\angle B$ , $\angle C$ so clearly $O' \equiv H$ . It follows that $\angle BFH = \angle OFE$ and $\angle CEH = \angle OEF$ .
Suppose $JH$ intersects the circle $(EJF)$ at point $K$ .

We will prove that $K$ belongs to the circle ( $BOC$ ). Indeed, we have $\angle HBF = 45^\circ = \angle JEF = \angle JKF$ so $F$ , $K$ , $H$ , $B$ are congruent. Similarly, $E$ , $K$ , $H$ , $C$ are congruent.
It follows that
$\angle BKC = \angle BKH + \angle CKH = \angle BFH + \angle CEH = \angle OFE + \angle OEF = 90^\circ.$
From here it follows that $K$ belongs to $BOC$ .
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