Show that if the angle between each pair of faces of a tetrahedron is equal, then the tetrahedron is regular. Does a tetrahedron have to be regular if five of the angles are equal?
Solution
Answer: no.
Let the tetrahedron be ABCD. Let the insphere have center O and touch the sides at W, X, Y, Z. The OW, OX, OY, OZ are the normals to the faces. But the angle between each pair of normals is equal. So OW, OX, OY and OZ are vectors of equal length at equal angles. Hence each side of WXYZ is equal (eg WX = 2 OW sin(WOX/2) ). So WXYZ is a regular tetrahedron. But the faces of ABCD are just the tangent planes at W, X, Y, Z. So if we rotate through an angle 120o about the line OW, then X goes to Y, Y to Z and Z to X. Hence AB = AC, AC = AD, AD = AB and BC = CD = DB (assuming appropriate labeling). Similarly, for rotation about the other axes. So ABCD has equal edges and hence is regular.
Consider the four normals OW, OX, OY, OZ. We can move X, Y, Z slightly closer together so that XYZ remains an equilateral triangle. Then move W so that WX = WY = XY. So five of the distances are equal, but the sixth is unequal. The reason for slightly is that all the angles between pairs of normals must remain less than 180o. The corresponding tetrahedron will now have the angles between only five pairs of faces equal.
© John Scholes
jscholes@kalva.demon.co.uk
20 Aug 2002