# Solution to problem 32 submitted by David Bernstein

## Problem 32

The faces of a given tetrahedron all have the same area. Prove that they are mutually congruent.

## Solution

For each face F of the tetrahedron, define the vector vF to be perpendicular to the face (directed out) and have length equal to the area of the face.

Lemma 1. The sum of these vectors is equal to zero.

Proof. If we put some gas in the tetrahedron, then these vectors would represent the force of the pressure this gas would exert on the faces of the tetrahedron. As the presence of the gas would not cause the tetrahedron to move, the vector sum of the pressures must be zero.

Lemma 2. The tetrahedron can be recovered from the set of such vectors and its volume.

Proof. Every edge e of the tetrahedron is perpendicular to the two vectors that correspond to the faces e is in. So we know the direction e has if considered as a vector. Further, the projection of e onto another one of these vectors (one that corresponds to a face e is not in) is the height of the tetrahedron, so its length can be computed from the area of that side and the volume. Knowing the length of the projection and the direction of e, we can reconstruct the length of e. Thus we can find all the edges of the tetrahedron as vectors, and rebuild it.

If we connect these four vectors vF in space one after another we would get a figure which looks like a rhombus bent on a diagonal (the figure is closed because the vectors sum to zero). It is easy to see that, since its sides have equal lengths, this rhombus is symmetric with respect to some two planes. This means that there is a rigid motion taking this four edged figure to itself, and taking any vector in it to any other. It follows that there exists a rigid motion taking the tetrahedron to itself, and taking any face to any other. Thus, the faces are mutually congruent.

Last revised August 2003