Show that for any tetrahedron the sum of the squares of the lengths of two opposite edges is at most the sum of the squares of the other four.
Solution
Let the vertices be A, B, C, D. We will show that AB2 + CD2 ≤ AC2 + AD2 + BC2 + BD2. Use vectors with origin A. Let the vectors from A to B, C, D be B, C, D respectively. We have to show that B2 + (C - D)2 ≤ C2 + D2 + (B - C)2 + (B - D)2. Rearranging, this is equivalent to (B - C - D)2 ≥ 0.
© John Scholes
jscholes@kalva.demon.co.uk
11 May 2002