IMO 1997 SL 18

Origin: GBR

Problem

The altitudes through the vertices A, B, C of an acute-angled triangle ABC meet the opposite sides at D, E, F, respectively. The line through D parallel to EF meets the lines AC and AB at Q and R, re- spectively. The line EF meets BC at P. Prove that the circumcircle of the triangle PQR passes through the midpoint of BC.

Solution

By symmetry, assume that AB > AC. The point D lies between M and P as well as between Q and R, and if we show that DM \cdotDP = DQ\cdotDR, it will imply that M, P, Q, R lie on a circle. Since the triangles ABC, AEF, AQR are similar, the points B, C, Q, R lie on a circle. Hence DB \cdot DC = DQ \cdot DR, and it remains to prove that DB \cdot DC = DM \cdot DP. However, the points B, C, E, F are concyclic, but so are the points E, F, D, M (they lie on the nine-point circle), and we obtain PB \cdot PC = PE \cdot PF = PD \cdot PM. Set PB = x and PC = y. We have PM = x+y and hence PD = 2xy x+y. It follows that DB = PB −PD = x(x−y) x+y , DC = y(x−y) x+y , and DM = (x−y)2 2(x+y), from which we immediately obtain DB \cdot DC = DM \cdot DP = xy(x−y)2 (x+y)2 , as needed.