IMO 1990 SL 5

Origin: FRA

Problem

Given \triangleABC with no side equal to another side, let G, K, and H be its centroid, incenter, and orthocenter, respectively. Prove that \angleGKH > 90◦.

Solution

Let O be the circumcenter of ABC, E the midpoint of OH, and R and r the radii of the circumcircle and incircle respectively. We use the following facts from elementary geometry: −−\to OH = 3−−\to OG, OK2 = R2 −2Rr, and KE = R 2 −r. Hence −−\to KH = 2−−\to KE −−−\to KO and −−\to KG = 2−−\to KE+−−\to KO . We then obtain −−\to KH \cdot −−\to KG = 1 3(4KE2 −KO2) = −2 3r(R −2r) < 0 . Hence cos \angleGKH < 0 ⇒\angleGKH > 90◦.