Balkan 2012 shortlist

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 $S\ne O$, 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 $OH\parallel OG\parallel KN\parallel AD$.