IMO 1998 SL 5

Origin: FRA

Problem

Let ABC be a triangle, H its orthocenter, O its circumcenter, and R its circumradius. Let D be the reflection of A across BC, E that of B across CA, and F that of C across AB. Prove that D, E, and F are collinear if and only if OH = 2R.

Solution

Let G be the centroid of \triangleABC and H the homothety with center G and ratio −1 2. It is well-known that H maps H into O. For every other point X, let us denote by X′ its image under H. Also, let A2B2C2 be the triangle in which A, B, C are the midpoints of B2C2, C2A2, and A2B2, respectively. It is clear that A′, B′, C′ are the midpoints of sides BC, CA, AB re- spectively. Furthermore, D′ is the reflection of A′ across B′C′. Thus D′ must lie on B2C2 and A′D′ \perp A B C D E F A′ B′ C′ A2 B2 C2 G H O D′ E′ F ′ B2C2. However, it also holds that OA′ \perpB2C2, so we conclude that O, D′, A′ are collinear and D′ is the projection of O on B2C2. Analogously, E′, F ′ are the projections of O on C2A2 and A2B2. Now we apply Simson’s theorem. It claims that D′, E′, F ′ are collinear (which is equivalent to D, E, F being collinear) if and only if O lies on the circumcircle of A2B2C2. However, this circumcircle is centered at H with radius 2R, so the last condition is equivalent to HO = 2R.