I said, Recall the odd (mod 3) squarefree sequence d(k) = 0 1 2 1 0 1 2 0 2 1 0 2 0 1 2 1 0 1 2 0 1 0 2 1 2 0 2 1 0 ... defined as the left-to-right alternating sum of the nonzero trits of k. In *balanced* ternary. (Thanks, Neil!) Theorem: if any trapezohedron be scaled to inscribe in a sphere, its faces are "right kites", i.e., they have two square vertices. Mathworld's trapezohedron article just lists areas and volumes of a few special cases (scaled inconsistently). For an n-trapezohedron inscribed in a spheroid with polar radius a, equatorial radius b, 2 2 2 2 2 2 8 c (b c - b + 2 a ) 2 sqrt(c (a c + b )) s = sqrt(4 a - ----------------------), S = ---------------------, 2 c + 1 (c + 1) 2 2 2 %pi 8 b sqrt(b c + a c) sin(---) n a b c n inradius = ------------------, area = ---------------------------------, 2 2 2 2 sqrt(b c + a c) (c + 1) 2 %pi 8 a b c sin(---) n n 2 b sqrt(c) volume = -------------------, tradius = -----------, 2 c + 1 3 (c + 1) 2 2 2 2 2 2 (c + 1) ((2 - 2 c) S - s ) (c + 1) ((c - 1) S - c s ) sqrt(---------------------------) sqrt(-----------------------------) (1 - c) (2 c - 1) (c - 1) c (2 c - 1) a = ---------------------------------, b = -----------------------------------, 2 2 2 2 %pi 2 b cos(-----) + a c n angles = [acos(--------------------), 2 2 a c + b 2 2 (b - a ) sqrt(1 - c) sqrt(c) acos(-----------------------------------------), 2 2 2 2 sqrt(a c + b ) sqrt(2 b c + a (1 - c)) 2 2 2 2 b c - a (1 - c) %pi - acos(--------------------)], 2 2 2 b c + a (1 - c) 2 3/2 b c --------------------------- + a 2 2 sqrt(a c + b ) + a sqrt(c) apex_solid_angle = 2 acos(-------------------------------) n, 2 2 sqrt(b c + a ) 2 2 %pi 2 girdpt_solid_angle = acos(((a cos(-----) + b c) n 4 2 %pi 4 2 2 2 (a cos(-----) + b c - 2 a b c (2 c + 1)) n 2 2 2 3/2 2 2 + 4 a (b - a ) sqrt(1 - c) c (c + 1) sqrt(a c + b ) 2 2 2 2 3 sqrt(2 b c + a (1 - c)))/(b c + a ) ), c + 1 natural_a = --------------------------------------------, %pi 2 sqrt(2) sqrt(1 - c) sqrt(2 c + 1) sin(---) n c + 1 natural_b = ------------------, %pi 4 sqrt(c) sin(---) n where s:=short side, S:=long side, c:=cos(%pi/n), tradius := the "tubular" or girth radius, and natural_a and natural_b are the spheroid radii of the trapezohedron which is the dual of the antiprism with unit edges. The face angle order is [apex, upper girdle, lower girdle]. Note vanishing cosine of upper girdle when a=b. Maybe somebody can simplify the girdle vertex solid angle. When a=b, it boils all the way down to acos(1 - 2 c). (This all started when I added prisms, pyramids, etc. to Macsyma's rather meager PLOT_3D_FIGURE for a geomety lesson.) --rwg PS, while the cube is the only regularhedron whose vertex solid angle is a rational fraction (1/8) of the sphere, the tetrahedral and octahedral vertex solid angles are 2 atan(sqrt(2)/5) and 4 atan(sqrt(2)/4). Since 3 atan(sqrt(2)/4) + 2 atan(sqrt(2)/5) = pi/2, six octahedra and eight tetrahdra come together at each vertex in the octahedral-tetrahedral tessellation of 3-space. This is a (rather lame) example of a pi arctan formula a la Machin's, but with sqrt(rational) series multipliers, which are considered kosher in the game of maximizing digits/term of hypergeometric pi series. So, with sqrts, (but not surds), can we beat Machin?