Let A_{1}, A_{2}, A_{3}, A_{4}, A_{5}, A_{6} be distinct points in the plane. Let D be the longest distance between any pair, and d the shortest distance. Show that D/d ≥ √3.

**Solution**

Let ABC be a triangle with sides a = BC ≥ b = CA ≥ c = AB. Suppose that angle A ≥ 2π/3. It follows immediately from the cosine formula we have a^{2} = b^{2} + c^{2} - 2bc cos A ≥ b^{2} + c^{2} + bc ≥ 3c^{2}, and hence that a/c ≥ √3. So it is sufficient to find a triangle with an angle of at least 2π/3.

If the 6 points form a convex hexagon, then the six angles of the hexagon sum to 4π so at least one is at least 2π/3. If not then one point is inside the convex hull of the others. Draw diagonals to triangulate the hull. Then the inside point must lie inside (or on) one of the triangles. But if P lies inside (or on) the triangle ABC, then at least one of the angles APB, BPC, CPA is at least 2π/3.

© John Scholes

jscholes@kalva.demon.co.uk

5 Feb 2002