Prove that a regular tetrahedron has five distinct spheres each tangent to its six extended edges. Conversely, prove that if a tetrahedron has five such spheres then it is regular.
Solution
First part is obvious. The wrong way to do the second part is to start looking for the locus of the center of a sphere which touches three edges. The key is to notice that the tangents to a sphere from a given point have the same length.
Let the tetrahedron be A1A2A3A4. Let S be the sphere inside the tetrahedron, S1 the tetrahedron opposite A1, and so on. Let the tangents to S from Ai have length ai. Then the side AiAj has length ai+aj. Now consider the tangents to S1 from A1. Their lengths are a1 + 2a2, a1 + 2a3, and a1 + 2a4. Hence a2 = a3 = a4. Similarly, considering S2, we have that a1 = a3 = a4.
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
19 Sep 1998
Last corrected/updated 24 Sep 2003