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.
Before learning this term, our Esteemed Moderator used the logically incongruous but delightfully intuitive "tetrahedron of David". By now everyone must know that iterating the Snowflake recursion on this figure produces (something whose closure is) merely a cube instead of a fractal. But the boundary of the construction using open tetrahedra is an interesting Cantor set, (strongly?) resembling eight corners cut off a Menger sponge. (The reason this limit, unlike a "spacefilling" polygon sequence, is defined is that it is a convergent, cumulative union.)
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
Very nice.