2007
Problem - 2432
Let $\triangle{ABC}$ be an acute triangle. If the distance between the vertex $A$ and the orthocenter $H$ is equal to the radius of its circumcircle, find the measurement of $\angle{A}$.
Let $O$ be the circumcenter. Connect $BO$ and let its extension intersect the circumcircle at point $D$. Connect $AD$, $CD$ and $CH$. $AH \perp BC$ and $CD \perp BC \implies AH \parallel CD$. Similarly, $CH \parallel AD$. Therefore $ADCH$ is a parallelogram. It follows that $CD=AH=CO=DO \implies \angle{BAC} = \angle{BDC} = 60^\circ$.
