I'll respond to George Hart's highly embarrassing hexagon question as soon as I unwedge my head on the subject. (In case anyone besides Shel is interested.) Given a tetrahedral apex with face angles (trihedral components) a, b, c, the solid angle is just the area on the unit sphere of the projection of the tetrahedron's base (opposite face) when the apex is at the center of the sphere. By the Spherical Triangle entry in Eric's Encyclopedia, this area is just the spherical excess, Omega = C + B + A - pi, where A, B, C are the vertex angles of the spherical triangle, i.e., the dihedral angles between the other three faces. Erics' Spherical Trigonometry entry derives the dihedrals from the face angles: cos(a) A = acos(------------- - cot(b) cot(c)), sin(b) sin(c) symmetrically for B and C. So Omega = acos + acos + acos - pi, which is maybe what most people use. To get 2 (cos(c) + cos(b) + cos(a) + 1) Omega = acos(-------------------------------------- - 1) (cos(a) + 1) (cos(b) + 1) (cos(c) + 1) I TRIGEXPANDed cos(Omega), then tediously massaged the resulting mess with TRIGSIMP, FACTOR, and my own trivial DESINIFY and DECOSIFY, and eventually got lucky and tried subtracting and adding (1+cos+cos+cos)^2 from/to the numerator. --rwg