There's a plane geometry theorem that the sum of the vertex angles of a polygon is pi*(#vertices-2). Is there a 3D analog? Rich -----Original Message----- From: math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com [mailto:math-fun-bounces+rschroe=sandia.gov@mailman.xmission.com] On Behalf Of R. William Gosper Sent: Wednesday, May 25, 2005 11:17 PM To: math-fun@mailman.xmission.com Subject: [math-fun] fun with A := solid_angle(a,b,c) Recall cos(c) + cos(b) + cos(a) + 1 A = 2 acos ----------------------------, a b c 4 cos(-) cos(-) cos(-) 2 2 2 where a,b,c are the (trihedral) angles at the vertex. Problem 1: What is t, the tetrahedral bond angle? Dissect tbe tetrahedron into four congruent squat ones whose apex angles are t. Then each has (c136) solid_angle(t,t,t) = 4*%pi/4) 3 cos(t) + 1 (d136) 2 acos(------------) = %pi 3 t 4 cos (-) 2 taking cos of half of both sides, (c137) cos(%/2) 3 cos(t) + 1 (d137) ------------ = 0 3 t 4 cos (-) 2 I.e., t = asec(-3). Problem 2: What is the solid angle of the apex of a regular polygonal pyramid with all n vertex angles = v? Dissect the pyramid into n congruent tetrahedra, then multiply n by the apical solid angle of one of them: (c138) (asin(sin(v/2)/sin(%pi/n)),n*solid_angle(v,%%,%%))$ (Cleaning up cos^2 of a half angle) (c139) subst(v/2,v,halfangles(subst(2*v,v,%))) 2 v sin (-) 2 cos(v) + 2 sqrt(1 - ---------) + 1 2 %pi sin (---) n (d139) 2 n acos(----------------------------------) 2 v sin (-) v 2 2 cos(-) (sqrt(1 - ---------) + 1) 2 2 %pi sin (---) n Problem 3: What is the circumradius of a regular dodecahedron with unit face circumradius? Use the preceding formula for one of the twelve component pyramids, with n=5, r unknown: (c140) subst([n=5,v=2*asin(print(side_regular_polygon(5,1))/2/r)],%) =4*%pi/12 %pi 2 sin(---) 5 %pi sin(---) 1 5 2 sqrt(1 - --) + cos(2 asin(--------)) + 1 2 r r %pi (d140) 10 acos(------------------------------------------) = --- 2 %pi 3 sin (---) 1 5 2 (sqrt(1 - --) + 1) sqrt(1 - ---------) 2 2 r r Clear the acos and square: (c141) halfangles(cos(trigexpand(%/10)))^2 2 %pi 2 sin (---) 5 1 2 (----------- - 2 sqrt(1 - --) - 2) %pi 2 2 cos(---) + 1 r r 15 (d141) ------------------------------------- = ------------ 2 %pi 2 sin (---) 1 2 5 4 (sqrt(1 - --) + 1) (1 - ---------) 2 2 r r Express (sin pi/5)^2 ito cos pi/15: (c142) (sin(3*x)^2,subst(%pi/15,x,%%=desinify(factor(trigexpand(%%))))) 2 %pi 2 %pi 2 2 %pi (d142) sin (---) = (1 - 4 cos (---)) (1 - cos (---)) 5 15 15 Find an polynomial equation for cos pi/15: (c143) cos(%pi/15) = c %pi (d143) cos(---) = c 15 (c144) block([%piargs:all],factor(multiply_conjugates(?meqhk(''%)))) 4 3 2 16 c + 8 c - 16 c - 8 c + 1 (d144) ------------------------------ 16 (The function ?meqhk is just lhs-rhs.) Find a polynomial equation for r in terms of c: (c145) factor(multiply_conjugates(ratnumer(subst([d142,d143],?meqhk(d141)))),%) 3 2 3 (d145) - (24 c - 8 c - 16 c + 1) (2 r - 8 c + 10 c + 1) 3 2 3 2 (2 r + 8 c - 10 c - 1) (4 r + 24 c + 8 c - 16 c - 7)/64 Which factor makes numeric sense? (c146) sfloat(subst(reverse(d143),args(part(%,1,1)))) 2 (d146) [0.15621, 2.0 r + 3.29456, 2.0 r - 3.29456, 4.0 r + 7.46457] Only the third, so (c147) combine(expand(trigreduce(subst(reverse(d143),solve(part(d145,1,1,3),r)) ))) %pi 1 - sqrt(5) (d147) [r = 2 cos(---) + -----------] 15 4 Check: (c148) subst(%,d140) %pi %pi (d148) --- = --- 3 3 (How did Macsyma do that?-) --rwg PS, DPThurston: (Recall that if you look at a cube from a corner you see a hexagon.) If you tile the plane with those hexagons, their cubes mechanically lock together in an airtight sheet. Similarly for octagons resting on a face, and dodecahedra stood on a vertex (www.ippi.com/rwg/dodex.gif). Tetrahedra stood on edge lock together in a square grid. Icosahedra on a face lock but leak. _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com http://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun