IMO 2003 SL G6

Origin: POL | Category: Geometry

Problem

Each pair of opposite sides of a convex hexagon has the following property: The distance between their midpoints is equal to \sqrt 3/2 times the sum of their lengths. Prove that all the angles of the hexagon are equal.

Solution

Let ABCDEF be the given hexagon. We shall use the following lemma. Lemma. If \angleXZY \geq60◦and if M is the midpoint of XY , then MZ \leq \sqrt 2 XY , with equality if and only if \triangleXY Z is equilateral. Proof. Let Z′ be the point such that \triangleXY Z′ is equilateral. Then Z is inside the circle circumscribed about \triangleXY Z′. Consequently MZ \leq MZ′ = \sqrt 2 XY , with equality if and only if Z = Z′. Set AD \capBE = P, BE \capCF = Q, and CF \capAD = R. Suppose \angleAPB = \angleDPE > 60◦, and let K, L be the midpoints of the segments AB and DE respectively. Then by the lemma, \sqrt 2 (AB + DE) = KL \leqPK + PL < \sqrt 2 (AB + DE), which is impossible. Therefore \angleAPB \leq60◦and similarly \angleBQC \leq60◦, \angleCRD \leq60◦. But the sum of the angles APB, BQC, CRD is 180◦, from which we conclude that these angles are all equal to 60◦, and moreover that the triangles APB, BQC, CRD are equilateral. Thus \angleABC = \angleABP + \angleQBC = 120◦, and in the same way all angles of the hexagon are equal to 120◦.