What is the maximum number of planes of symmetry a tetrahedron can have?

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Show that each plane of symmetry must pass through an edge of the tetrahedron.

## Solution

In a regular tetrahedron, there are 6 planes of symmetry: each of them passes through one of the edges and bisects the edge that intersects it. We will show that there cannot be more than six planes of symmetry. Immediately note that a plane of symmetry passes through an even number of vertices of the tetrahedron, as the remaining vertices must be divided into pairs that are symmetric with respect to this plane of symmetry. Suppose a plane of symmetry does not pass through any of the vertices of the tetrahedron. Then the 4 vertices of the tetrahedron must be divided into two pairs of vertices that are symmetric with respect to the plane of symmetry. Consequently, the lines passing through these pairs of vertices must be perpendicular to the plane of symmetry, i.e., parallel to each other. Thus, the four vertices of the tetrahedron must lie in the same plane, which is impossible. It is clear that all four vertices of the tetrahedron cannot lie in the plane of symmetry. Therefore, the plane of symmetry must pass through exactly two of the vertices (i.e., through an edge) of the tetrahedron. Suppose a plane of symmetry passes through one of the edges of the tetrahedron. Then the remaining two vertices are symmetric with respect to this plane, so this plane must pass through the midpoint of the edge connecting the remaining two vertices. The previous
arguments show that there is no more than one plane of symmetry passing through a given edge of the tetrahedron. Thus, the total number of planes of symmetry is no greater than the number of edges of the tetrahedron, i.e., no more than six.

## Answer

6.00