Given four distinct parallel planes, prove that there exists a regular tetrahedron with a vertex on each plane.
Solution
Intuitively, we can place A and B on the two outer planes with AB perpendicular to the planes. Then tilt AB in one direction until we bring C onto one of the middle planes (keeping A and B on the outer planes), then tilt AB the other way (keeping A, B, C on their respective planes) until D gets onto the last plane.
Take A as the origin. Let the vectors AB, AC, AD be b, c, d. Take p as one of the outer planes. Let the distances to the other planes be e, f, g. Now we find a vector n satisfying: n.b = e, n.c = f, n.d = g. This is a system of three equations in three unknowns with non-zero determinant (because b.c x d is non-zero), so it has a solution n. Scale the tetrahedron by |n|, orient p perpendicular to n/|n|, then B, C, D will be on the other planes as required.
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
10 Oct 1998
Last updated/corrected 14 Mar 03