IMO 1998 SL 6

Origin: POL

Problem

Let ABCDEF be a convex hexagon such that \angleB +\angleD +\angleF = 360◦and AB BC \cdot CD DE \cdot EF FA = 1. Prove that BC CA \cdot AE EF \cdot FD DB = 1.

Solution

Let P be the point such that \triangleCDP and \triangleCBA are similar and equally oriented. Since then \angleDCP = \angleBCA and BC CA = DC CP , it follows that \angleACP = \angleBCD and AC CP = BC CD, so \triangleACP ∼\triangleBCD. In particular, BC CA = DB PA . Furthermore, by the conditions of the problem we have \angleEDP = 360◦− \angleB −\angleD = \angleF and PD DE = PD CD \cdot CD DE = AB BC \cdot CD DE = AF F E . Therefore \triangleEDP ∼\triangleEFA as well, so that similarly as above we conclude that \triangleAEP ∼\triangleFED and consequently AE EF = PA F D. Finally, BC CA \cdot AE EF \cdot F D DB = DB PA \cdot PA F D \cdot F D DB = 1. Second solution. Let a, b, c, d, e, f be the complex coordinates of A, B, C, D, E, F, respectively. The condition of the problem implies that a−b b−c \cdot c−d d−e \cdot e−f f−a = −1. On the other hand, since (a −b)(c −d)(e −f) + (b −c)(d −e)(f −a) = (b−c)(a−e)(f −d)+(c−a)(e−f)(d−b) holds identically, we immediately deduce that b−c c−a \cdot a−e e−f \cdot f−d d−b = −1. Taking absolute values gives BC CA \cdot AE EF \cdot F D DB = 1.