A tetrahedron has opposite sides equal. Show that all faces are acute-angled.
Solution
Let the tetrahedron be ABCD. Let M be the midpoint of BC. We have AM + MD > AD (*). Now the triangles ABC and DCB are congruent because AB = DC, BC = CB and AC = DB. Hence AM = DM. Also AD = BC = 2MC. So (*) implies that AM > MC. But that implies that angle BAC is acute. Similarly for all the other angles.
© John Scholes
jscholes@kalva.demon.co.uk
17 Jul 2002