IMO 1995 SL G5

Origin: NZL | Category: Geometry

Problem

Let ABCDEF be a convex hexagon with AB = BC = CD, DE = EF = FA, and ∡BCD = ∡EFA = \pi/3 (that is, 60◦). Let G and H be two points interior to the hexagon such that angles AGB and DHE are both 2\pi/3 (that is, 120◦). Prove that AG + GB + GH + DH + HE \geqCF.

Solution

Triangles BCD and EFA are equilateral, and hence BE is an axis of symmetry of ABDE. Let C′, F ′ respectively be the points symmetric to C, F with respect to BE. The points G and H lie on the circumcircles of ABC′ and DEF ′ respectively (because, for instance, \angleAGB = 120◦= 180◦−\angleAC′B); hence from Ptolemy’s theorem we have AG+GB = C′G and DH + HE = HF ′. Therefore AG + GB + GH + DH + HE = C′G + GH + HF ′ \geqC′F ′ = CF, with equality if and only if G and H both lie on C′F ′. Remark. Since by Ptolemy’s inequality AG+GB \geqC′G and DH+HE \geq HF ′, the result holds without the condition \angleAGB = \angleDHE = 120◦.