Prove that every tetrahedron has a vertex whose three edges have the right lengths to form a triangle.
Solution
The trick is to consider the longest side. That avoids getting into lots of different possible cases for which edge is longer than the sum of the other two.
So assume the result is false and let AB be the longest side. Then we have AB > AC + AD and BA > BC + BD. So 2AB > AC + AD + BC + BC. But by the triangle inequality, AB < AC + CB, AB < AD + DB, so 2AB < AC + CB + AD + DB. Contradiction.
Solutions are also available in: Samuel L Greitzer, International Mathematical Olympiads 1959-1977, MAA 1978, and in István Reiman, International Mathematical Olympiad 1959-1999, ISBN 189-8855-48-X.
© John Scholes
jscholes@kalva.demon.co.uk
3 Oct 1998
Last corrected/updated 3 Oct 1998