IMO 1989 SL 32

Origin: USA

Problem

The vertex A of the acute triangle ABC is equidistant from the circumcenter O and the orthocenter H. Determine all possible values for the measure of angle A.

Solution

Let CC′ be an altitude, and R the circumradius. Then, since AH = R, we have AC′ = |R sin B| and hence (1) CC′ = |R sin B tan A|. On the other hand, CC′ = |BC sin B| = 2|R sin A sin B|, which together with (1) yields 2| sin A| = | tan A| ⇒| cosA| = 1/2. Hence, \angleA is 60◦. (Without the condition that the triangle is acute, \angleA could also be 120◦.) Second Solution. For a point X, let X denote the vector OX. Then |A| = |B| = |C| = R and H = A + B + C, and moreover, R2 = (H −A)2 = (B + C)2 = 2B 2 + 2C 2 −(B −C)2 = 4R2 −BC2. It follows that sin A = BC 2R = \sqrt 3/2, i.e., that \angleA = 60◦. Third Solution. Let A1 be the midpoint of BC. It is well known that AH = 2OA1, and since AH = AO = BO, it means that in the right- angled triangle BOA1 the relation BO = 2OA1 holds. Thus \angleBOA1 = \angleA = 60◦.