17th Chinese 2002 Problem B1 Given four distinct points in the plane such that the distance between any two is at least 1, what is the minimum possible value for the sum of the six distances between them?
We strongly suspect that the configuration of two equilateral triangles with a common side is minimal. That gives 6 + (√3 - 1). Hence, in particular, no two points in a minimal configuration can be more than = (1 + √3) 2 - 2(1 + √3)AB + AB 2 with equality only in the same two cases. Taking the square root, AB + √(4 - AB 2 ) >= 1 + √3, and so 4 + AB + √(4 - AB 2 ) >= 5 + √3 with equality iff AB = 1 or √3. So the unique minimal configuration is two equilateral triangles with a common side. 17th Chinese 2002 © John Scholes jscholes@kalva.demon.co.uk 5 June 2002 Last corrected/updated 4 Aug 2003