WFL> normalised so that \sum_j (a_ij)^2 = 0 for i = 1,...,n, Now that was a normalization liberals can only dream of. My turn:

So, fresh start: the solid angle at the apex of a regular n-gonal pyramid with slant edges 1 and height h ... = 2 pi - 2 n atan(h t),

where t = tan pi/n, but explement the middle term (2 pi - acos) when x< 1/(1+sec pi/n).

Should say h < 1/(1+sec pi/n). rwg> There must be some obvious insight for the incredibly simple rhs. WFL>This looks suspiciously like the area formula for the dual spherical polygon: this has angles and sides corresponding to the sides and angles of the of the original polygon (whose angles corrspond to dihedral angles, and sides to apical angles). See Todhunter. Pending a jaunt to the Stanford Math Library, the formula says the following: from the unit southern hemisphere slice off enough of the polar cap to leave a segment of thickness h. Inscribe in-phase regular n-gons on the two circular faces. (The lower n-gon and the sphere center form the pyramid in question.) Then one of the isosceles trapezoidal pyramids with apex at the center and base defined by one side of an n-gon and the corresponding side of the other has apical solid angle 2 atan(h tan pi/n). Lemma: The apical solid angle of an isosceles trapezoidal pyramid with vertex angles a,b,a,c is 2 c b c 2 b - sin (-) + (cos(a) - 1) sin(-) sin(-) - sin (-) + cos(a) + 1 2 2 2 2 (d79) 2 acos(-------------------------------------------------------------) b c (cos(a) + 1) cos(-) cos(-) 2 2 I derived this by splitting the general a,b,a,c hedral vertex into trihedrals a,b,d and d,a,c, then maximizing over d, getting b c (d38) cos(d) = cos(a) - 2 sin(-) sin(-) 2 2 Plugging into d79 the pyramid's particulars: (c80) subst([a = asin(h),b = 2*%pi/n,c = 2*asin(sqrt(1-h^2)*sin(%pi/n))],%) (d80) <big mess> (c81) trigsimp(factor(trigsimp(factor(rootscontract(tan(d80/2)))))) | %pi | abs(h) |sin(---)| | n | (d81) ----------------- %pi cos(---) n I.e., h tan pi/n. QED. Is someone out there who can say if Todhunter has the solid angle from vertex angles formula cos(c) + cos(b) + cos(a) + 1 (d1) 2 acos ---------------------------- ? a b c 4 cos(-) cos(-) cos(-) 2 2 2 Odditum: the height of a square pyramid with unit slant edges and apex angles v is just sqrt(cos(v)). Apropos the question of a 3D analog of the (n-2) pi formula, there's Descartes' total angular defect formula in which a sum over all the 2D angles of a convex polyhedron gives 4 pi. Not a solid result. (But could it test for convexity?) --rwg