On 11/14/08, rwg@sdf.lonestar.org <rwg@sdf.lonestar.org> wrote:
Tetrahedral minifact: Resting on a flat surface, the (dihedral) angle between the surface and a (sloping) face equals the bond angle (between rays from the center to two vertices), namely pi - asec 3. Is this equality geometrically obvious?
Consider a "stella octangula" comprising a pair of interpenetrating regular tetrahedra with common centre, and edges bisecting one another perpendicularly in pairs. The joins of the centre to any two vertices of one tetrahedron, together with the perpendiculars from the corresponding edge mid-point along the faces of the other tetrahedron, form a plane cyclic quadrilateral --- the other two angles being right-angles, by symmetry. Hence the sum of the (interior) dihedral and bond angles equals \pi; and of course sum of the former and exterior dihedral --- which presumably you had in mind above --- also equals \pi. WFL