### IMO 1995

**Problem B2**
Let ABCDEF be convex hexagon with AB = BC = CD and DE = EF = FA, such that ∠BCD = ∠EFA = 60^{o}. Suppose that G and H are points in the interior of the hexagon such that ∠AGB = ∠DHE = 120^{o}. Prove that AG + GB + GH + DH + HE ≥ CF.

**Solution**

BCD is an equilateral triangle and AEF is an equilateral triangle. The presence of equilateral triangles and quadrilaterals suggests using Ptolemy's inequality. [If this is unfamiliar, see ASU 61/6 solution.]. From CBGD, we get CG·BD ≤ BG·CD + GD·CB, so CG ≤ BG + GD. Similarly from HAFE we get HF ≤ HA + HE. Also CF is shorter than the indirect path C to G to H to F, so CF ≤ CG + GH + HF. But we do not get quite what we want.

However, a slight modification of the argument does work. BAED is symmetrical about BE (because BA = BD and EA = ED). So we may take C' the reflection of C in the line BE and F' the reflection of F. Now C'AB and F'ED are still equilateral, so the same argument gives C'G ≥ AG + GB and HF' ≤ DH + HE. So CF = C'F' ≤ C'G + GH + HF' ≤ AG + GB + GH + DH + HE.

36th IMO 1995

© John Scholes

jscholes@kalva.demon.co.uk

25 June 2002

Last corrected/updated 23 Aug 03