On 2015-05-25 11:09, James Propp wrote:
How many of you already know (or can figure out) [w->t?]hat four solid angles associated with a general tetrahedron add up to 4 pi?
I've been told that George Polya, when asked, replied with the correct answer instantly (it was part of his geometric toolkit), but that few of his contemporaries, when polled in an informal survey, were aware of this three-dimensional analogue of the familiar fact about the angles of a triangle summing to pi. I'm guessing that things haven't changed much in the intervening half-century.
Jim Propp I don't get it. Consider the regular case. The four vertex trihedrals obviously sum to far short of 4pi. What other "associated" solid angles are there? Subdividing into four tetrahedra by joining the four vertices to an interior point creates four solid angles which obviously sum to 4pi. But for the sum of the four vertex solid angles I get 4pi/3. I.e., you could squeeze 12 tetrahedra (non-regular, but with a pi/3 solid angle) around a point.
Puzzle: Can a non-regular tetrahedron have equal solid angles? --rwg Now seems a good time to repeat some formulæ for the solid angle at a vertex formed with face angles a, b, c: solidAngle[a_, b_, c_] := {ArcCos[-1 + (1 + Cos[a] + Cos[b] + Cos[c])^2/((1 + Cos[a]) (1 + Cos[b]) (1 + Cos[c]))], 2 ArcCos[1/4 (1 + Cos[a] + Cos[b] + Cos[c]) Sec[a/2] Sec[b/2] Sec[c/2]], 2 ArcSin[1/(4 Sqrt[2]) Sqrt[-1 - Cos[2 a] - Cos[2 b] + 4 Cos[a] Cos[b] Cos[c] - Cos[2 c]] Sec[a/2] Sec[b/2] Sec[c/2]], 4 ArcTan[\[Sqrt](Tan[1/4 (a + b - c)] Tan[1/4 (a - b + c)] Tan[ 1/4 (-a + b + c)] Tan[1/4 (a + b + c)])]} There's some brown belt trig in showing those four equivalent. sphericalPolygonArea[r_, n_, anglesum_] := r^2*(anglesum - (n - 2)*Pi) Largest possible solid angle with vertex angles a,b,c,d: maxSolidAngle[a_, b_, c_, d_] := 2*ArcCos[((Cos[d] + Cos[c] + Cos[b] + Cos[a])/(4*Cos[a/2]*Cos[b/2]*Cos[c/2]* Cos[d/2])) - Tan[a/2]*Tan[b/2]*Tan[c/2]*Tan[d/2]] Dihedral between faces with vertex angles a and b: dihedral[a_, b_, c_] := ArcCos[Csc[a]*Csc[b]*Cos[c] - Cot[a]*Cot[b]] Apex and base solid angles for regular n-gon base, height h pyramidSolidAngle[n_, circumRadius_, h_] := List[2*Pi - 2*n*ArcTan[((h*Tan[Pi/n])/(Sqrt[circumRadius^2 + h^2]))], 2*ArcCos[((h^2*Sin[Pi/n])/(2*(circumRadius^2*(Cos[Pi/n])^2 + h^2))) + Csc[Pi/n]/2]] Apex and base solid angles for regular n-gon base, cos of angle between edge and altitude pyramidSolidAngle[n_, cos_] := {2*Pi - 2*n*ArcTan[cos*Tan[Pi/n]], 2*ArcCos[(Sin[Pi/n]/(2*(1 + Cos[\[Pi]/n]^2*(cos^-2 - 1)))) + Csc[Pi/n]/2]}