UkraineMO

Geometry Difficulty 7.1 National Olympiad, round 2 Prove it Ukraine

Let $AM$ be a median in an acute triangle $ABC$ . Its extension intersect the circumcircle $w$ of $ABC$ at $P$ . Let $AH_1$ be an altitude of $\triangle ABC$ , $H$ - its orthocenter. The rays $MH$ and $PH_1$ intersect $w$ at $K$ and $T$ respectively. Prove that the circumcircle of $\triangle AKTH_1$ is tangent to $BC$ .

(Khilko Danylo)

Solution

It suffices to show that $\angle TKH_1 = \angle TH_1B$ (Fig. 14). Let us extend $KH_1$ and intersect if with $w$ at $S$ . Then
$\angle TKS = \angle TAB + \angle BAS, \quad \angle TH_1B = \angle TAB + \angle PAC,$
So it is sufficient to show that $\angle PAC = \angle BAS$ .
Denote by $A_1$ the point such that $AA_1$ is the diameter of $w$ . Then $\angle A_1CA = \angle ABA_1 = 90^\circ$ , so $BH \parallel A_1C$ and $CH \parallel A_1B$ . Then $BHCA_1$ is a parallelogram, so $HA_1$ passes through the point $M$ .

Then $K$ lies on $HA_1$ , hence $\angle A_1KA = 90^\circ$ . Then the quadrilateral $AKH_1M$ is inscribed. So $\angle KH_1A = \angle KMA$ . Suppose $AH_1$ intersect $w$ secondly at $F$ . Then
$\angle KH_1A = \angle KCA + \angle FAS, \quad \angle KMA = \angle KCA + \angle PAA_1.$
Hence, $\angle FAS = \angle PAA_1$ , also $\angle ABC = \angle AA_1C$ , so we derive $\angle BAF = 90^\circ - \angle ABC = \angle A_1AC$ . Then
$\angle BAS = \angle BAF + \angle FAS = \angle PAA_1 + \angle A_1AC = \angle PAC.$

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