The orthocenter $H$ of a triangle $ABC$ is not on the sides of the triangle and the distance $AH$ equals the circumradius of the triangle. Find the measure of $\angle A$.
Problem
Source: Moldova MO 2000 Grade 7 P4
Tags: geometry, Triangle
thanosaops
23.04.2021 04:44
60 degrees or 120 degrees. Proof: Use a vector system with O as origin. Problem follows extremely easily due to applications of dot product and law of cosines. Corollary: Let N be the nine-point center of ABC. the extension of AN leads to an excenter of triangle ABC if and only if the problem holds. Proof: O and H are isogonal conjugates, so the problem holding implies that O and H are symmetric about either the internal or external angle bisector of A. This means that the midpoint of OH (which is N) lies on either internal or external bisector of A, which leads to an excenter.
Kagebaka
23.04.2021 07:22
Let $M,$ $O,$ and $G$ be the midpoint of $BC,$ the circumcenter, and the centroid, respectively. Since $A,G,M$ are collinear with $\tfrac{AG}{GM}=2$ and $H,G,O$ are collinear with $\tfrac{HG}{GO}=2$ by Euler Line, it follows that $\triangle AGH\sim\triangle MGO$ with scale factor $2.$ Thus, $AH=2OM=2R\cos A.$ $\blacksquare$
) so $\cos \angle A = \tfrac{1}{2},$ giving $\angle A = 60^\circ$ or $\angle A = 120^\circ.$ $\blacksquare$