IMO 1995 · Problem 5

Let \(ABCDEF\) be a convex hexagon with \(AB = BC = CD\) and \(DE = EF = FA\), such that \(\angle BCD = \angle EFA = \pi/3\). Suppose \(G\) and \(H\) are points in the interior of the hexagon such that \(\angle AGB = \angle DHE = 2\pi/3\). Prove that \(AG + GB + GH + DH + HE \ge CF\).

Official text of the problem. Official problem paper (PDF)

Solution

Since \(BC=CD\) and \(\angle BCD=\pi/3\), triangle \(BCD\) is equilateral. Similarly, triangle \(EFA\) is equilateral. Consequently, \[ AB=BD,\qquad EA=ED. \] Thus \(B\) and \(E\) both lie on the perpendicular bisector of \(AD\), so reflection in \(BE\) interchanges \(A\) and \(D\).

Let \(P\) and \(Q\) be the reflections of \(C\) and \(F\), respectively, in \(BE\). Reflecting the equilateral triangles \(BCD\) and \(EFA\), we see that \(ABP\) and \(DEQ\) are equilateral. Also, \[ PQ=CF. \]

Rotate about \(A\) through an angle of magnitude \(\pi/3\), in the direction that takes \(B\) to \(P\), and let \(G_1\) be the image of \(G\). The rotation preserves distances, so \[ PG_1=BG. \] Moreover, \(AG=AG_1\) and \(\angle GAG_1=\pi/3\); hence triangle \(AGG_1\) is equilateral, giving \[ G_1G=AG. \]

Similarly, rotate about \(D\) through an angle of magnitude \(\pi/3\) taking \(E\) to \(Q\), and let \(H_1\) be the image of \(H\). Then \[ H_1Q=HE,\qquad HH_1=DH. \]

The triangle inequality along the polygonal path \(P,G_1,G,H,H_1,Q\) now gives \[ \begin{aligned} CF=PQ &\le PG_1+G_1G+GH+HH_1+H_1Q\\ &=GB+AG+GH+DH+HE, \end{aligned} \] as required. This argument remains valid if intermediate points coincide, including \(G=H\).

The solution was checked against John Scholes' IMO solution archive (Kalva).