are, nicely enough, 2 ArcTan[GoldenRatio⁵] = π/2 - ArcTan[11/2]. The good news: Mathematica 12 has polyhedron vertex solid angle. The bad news: In[473]:= ByteCount[PolyhedronAngle[PolyhedronData["Dodecahedron", "Polyhedron"], 12]] // tim During evaluation of In[473]:= 0.689688,0 (* secs *) Out[473]= 44152 (* bytes *) In[474]:= FullSimplify[PolyhedronAngle[PolyhedronData["Dodecahedron", "Polyhedron"], 12]] // tim During evaluation of In[474]:= 14018.13381,2 Out[474]= 2 ArcTan[11/2 + (5 Sqrt[5])/2] In[475]:= 14018./3600 Out[475]= 3.89388888888889 3.9 hours! (Another 12.x produced 180000 bytes and the FullSimplify never answered.) Also nice: Recall that a dodecahedron face subtends about 1 steradian wrt the center, because 4π/12 ~ 1 <http://gosper.org/1steradian.png>. Conwayish question: Given that alternating Dodecahedra and EndoDodecahedra (EndoBob StarPants <http://gosper.org/endogram.png>) "checker" the 3D grid, what are the two vertex solid angles (convex and nonconvex trihedrals) of the latter? (Kids: Remember that 4π steradians is a "full sphere" in the same sense that 2π radians is a "full circle".) Lastly, some numerology with yesterday's DisdyakisTriacontahedron ("d120" die): Its faces are all congruent triangles with the two shortest edges in ratio very nearly π/2: In[471]:= PolyhedronData["DisdyakisTriacontahedron", "EdgeLengths"] Out[471]= {1, 3/10 (3 + √5), 7/5 + 1/√5} In[472]:= 2 N@% Out[472]= {2., 3.14164078649987, 3.69442719099992} In[1718]:= ContinuedFraction[3 (3/5 + 1/√5)] Out[1718]= {3, {7, 16, 1, 1, 1, 2, 1, 6, 1, 2, 1, 1, 1, 16, 7, 2, 1, 1, 5, 1, 1, 66, 1, 1, 5, 1, 1, 2}} (periodic, vs 3 7 15 1 292 1 1 1 ... fer sure not periodic.) —rwg