IMO 1982 SL 5

Problem

A5 (NET 2)IMO5 Let A1A2A3A4A5A6 be a regular hexagon. Each of its diagonals Ai−1Ai+1 is divided into the same ratio \lambda 1−\lambda, where 0 < \lambda < 1, by a point Bi in such a way that Ai, Bi, and Bi+2 are collinear (i \equiv 1, . . . , 6 (mod 6)). Compute \lambda.

Solution

We first observe that \triangleA5B4A4 ∼= \triangleA3B2A2. Since \angleA5A3A2 = 90◦, we have \angleA2B4A4 = \angleA2B4A3 + \angleA3B4A4 = (90◦−\angleB2A2A3) + (\angleB4A5A4 + \angleA5A4B4) = 90◦+ \angleB4A5A4 = 120◦. Hence B4 be- longs to the circle with center A3 and radius A3A4, so A3A4 = A3B4. O A1 A2 A6 A3 A4 A5 B4 B2 Thus \lambda = A3B4/A3A5 = A3A4/A3A5 = 1/ \sqrt 3.