Balkan Shortlist, 2012

Geometry Difficulty 7.9 National Olympiad, round 2 Prove it Balkan Mathematical Olympiad

Let $ABC$ be a triangle, $O$ its circumcenter and $AD$ the bisector of the angle $A$ where $D \in BC$ . Let $\ell$ be the line passing through $O$ and parallel to the bisector $AD$ . Prove that $\ell$ passes through the orthocenter $H$ of the triangle $ABC$ if and only if $ABC$ is isosceles or $\angle BAC = 120^{\circ}$ .

Solution

Let $G$ be the center of mass of the triangle $ABC$ .
Assume that $OH \parallel AD$ . Since $G, O, H$ are collinear, then $OG \parallel AD$ . Suppose that $AD$ meets the circumcircle of $ABC$ again at point $M$ and consider the midpoints $K, N$ of $MC, AC$ , respectively. Let the lines $OG$ and $BK$ meet at $S$ and let the lines $MO$ and $BK$ meet at $T$ . Then $NK \parallel AD$ and $T$ is the center of mass of the isosceles triangle $MBC$ . Therefore we have
$OG \parallel AD \Rightarrow OG \parallel NK \Rightarrow \frac{BS}{SK} = 2 \Rightarrow \frac{BS}{SK} = \frac{BT}{TK} \Rightarrow T = S.$
If $S \neq O$ , then the lines $OS, OT$ and $AD$ coincide and $ABC$ is isosceles. If $T = S = O$ , then the triangle $MBC$ is equilateral, $\angle BMC = 60^{\circ}$ and $\angle BAC = 120^{\circ}$ .

To prove the converse assume that $\angle BAC = 120^{\circ}$ . Then $\angle BMC = 60^{\circ}$ and the triangle $BMC$ is equilateral. Hence $O$ coincide with the center of mass $T$ of the triangle $BMC$ .
Since $\frac{BG}{GN} = 2 = \frac{BO}{OK}$ , it follows that $OH \parallel OG \parallel KN \parallel AD$ .

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