I blathered: We can dissect a regular n-gonal pyramid of height h and base circumradius r into n congruent tetrahedra with trihedral components v, v, and 2 asin(sin(v) sin(pi/n)), where v = atan(r/h), i.e., half the apex angle. This gives, after much simplification, two expressions for the solid angle of the apex: 2 %pi 2 %pi 2 %pi 2 cos (---) (2 sin (---) cos(v) + cos(-----)) n n n (d3) n acos(--------------------------------------------- 2 %pi 2 2 %pi sin (---) cos (v) + cos (---) n n 2 %pi - cos(-----)) n n = acos((- 1) n - 1 /===\ 2 %pi 2 | | %pi %pi k 2 2 cot (---) n | | (cos(v) + cot(---) cot(-----)) n | | n n k = 1 (1 - ------------------------------------------------------)) 2 2 %pi n (cos (v) + cot (---)) n Shame on me. The product over a half-period of a polynomial in trigs of angles in arithmetic progression always simplifies, in this case, mondomiraculously: n 2 %pi acos((- 1) (1 - 2 sin (n acot(tan(---) cos(v))))) (!) n The (-1)^n looks very strange. But only because I'm losing it. 1-2 sin^2 x = cos 2x, and simplification continues to an incredible 2 n atan (cos(v) tan(pi/n)), which is wrong, because the rhs(d3) takes the wrong branch when cos v < 1/(1+sec pi/n) = (1-tan^2 2pi/n)/2. It's obviously wrong, since the answer needs to approach 2 pi for small cos v, but acos is normally limited to pi. So, fresh start: the solid angle at the apex of a regular n-gonal pyramid with slant edges 1 and height h is 2 2 (h t - h t + t + 1) (h t + h t - t + 1) n acos(-----------------------------------------) 2 2 2 (t + 1) (h t + 1) n - 1 pi k /===\ cot ---- 2 | | n 2 2 n | | (--------- + h) | | t n k = 1 = acos((- 1) (1 - -----------------------------)) 1 2 n 2 (-- + h ) t 2 t = 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). There must be some obvious insight for the incredibly simple rhs. --rwg <explement deleted>, but there is a brief CO2 remark at gosper.org/co2.txt