Small rhombicosidodecahedron spotted at Wal-Mart. I couldn't resist firing up the pyramid solid angle formula, 1 - cos(v) 1 - --------------------------------- 2 %pi sqrt(cos(v) - cos(-----)) n 2 (------------------------- + 1) %pi sqrt(2) sin(---) n 2 n acos(-------------------------------------) v cos(-) 2 (all n vertex angles = v) to confirm Weisstein's radius/edge ratio: r sqrt(4 sqrt(5) + 11) - = -------------------- ~ 2.23295050941569. e 2 Much smaller than I expected, given 62 faces. Confession: although very easy to set up, the formula was too big to clear of radicals, and I resorted to Rich's numeric lattice reducer to get the quartic. The solid angle of a base corner is surprisingly nicer: v sin(-) 2 %pi -------- + sin(---) %pi n sin(---) n 2 acos(-------------------). v sin(-) + 1 2 Adding two squares, a triangle, and a pentagon gives the solid angle of the rhombi...dron vertex: 4 sqrt(5) - 5 acos(-------------) + %pi ~ 4.44630893315853 15 Which (finite) semiregular polyhedron have vertex solid angles a rational multiple of pi? So far, I have found only the regular polygonal prisms, since solid_angle(pi/2,pi/2,a) = a. There seems to be no 3D analog of the polygon sum-of-vertex-angles formula. Except for the prisms and the five regular solids, even the central angles subtended by the faces are rarely nice. The solid angle of the vertex of an ngonal antiprism is %pi %pi sqrt(3) tan(---) (4 cos(---) + 5) 2 n n 2 acos(---------------------------------), 9 a derivation which took inexcusably long. E.g., n=3 sez the vertex solid angle of an octahedron is 2 acos 7/9. --rwg