IMO 1988 SL 15

Origin: ICE

Problem

Let ABC be an acute-angled triangle. Three lines LA, LB, and LC are constructed through the vertices A, B, and C respectively according to the following prescription: Let H be the foot of the altitude drawn from the vertex A to the side BC; let SA be the circle with diameter AH; let SA meet the sides AB and AC at M and N respectively, where M and N are distinct from A; then LA is the line through A perpendicular to MN. The lines LB and LC are constructed similarly. Prove that LA, LB, and LC are concurrent.

Solution

Referring to the description of LA, we have \angleAMN = \angleAHN = 90◦− \angleHAC = \angleC, and similarly \angleANM = \angleB. Since the triangle ABC is acute-angled, the line LA lies inside the angle A. Hence if P = LA\capBC and Q = LB \capAC, we get \angleBAP = 90◦−\angleC; hence AP passes through the circumcenter O of ∆ABC. Similarly we prove that LB and LC contains the circumcenter O also. It follows that LA, LB and LC intersect at the point O. Remark. Without identifying the point of intersection, one can prove the concurrence of the three lines using Ceva’s theorem, in usual or trigono- metric form.