IMO 1982 SL 17

Problem

C5 (USS 5) The right triangles ABC and AB1C1 are similar and have opposite orientation. The right angles are at C and C1, and we also have ∡CAB = ∡C1AB1. Let M be the point of intersection of the lines BC1 and B1C. Prove that if the lines AM and CC1 exist, they are perpendic- ular.

Solution

Let A be the origin of the Cartesian plane. Suppose that BC : AC = k and that (a, b) and (a1, b1) are coordinates of the points C and C1, respectively. Then the coordinates of the point B are (a, b)+k(−b, a) = (a−kb, b+ka),

while the coordinates of B1 are (a1, b1) + k(b1, −a1) = (a + kb1, b1 −ka1). Thus the lines BC1 and CB1 are given by the equations x−a1 y−b1 = x−(a−kb) y−(b+ka) and x−a y−b = x−(a1+kb1) y−(b1−ka1) respectively. After multiplying, these equations transform into the forms BC1 : kax + kby = kaa1 + kbb1 + ba1 −ab1 −(b −b1)x + (a −a1)y CB1 : ka1x + kb1y = kaa1 + kbb1 + ba1 −ab1 −(b −b1)x + (a −a1)y. The coordinates (x0, y0) of the point M satisfy these equations, from which we deduce that kax0 + kby0 = ka1x0 + kb1y0. This yields x0 y0 = −b1−b a1−a, implying that the lines CC1 and AM are perpendicular.