A tetrahedron of edge length e has volume e^3/sqrt(72). So the 600-cell has surface volume 600e^3/sqrt(72). Let r be the distance from the centre of the 600-cell to the centre of one of its boundary tetrahedra. If we increase r to r+epsilon, the content increases by a shell around the 600-cell of width epsilon. This shell therefore has content epsilon*600e^3/sqrt(72) (plus terms of order epsilon^2 and smaller). Therefore we can find the content of the whole 600-cell by integrating 600e^3/sqrt(72) from r=0 to whatever r is when e=1. It remains to calculate r in terms of e.
Suppose we have a vertex of a polyhedron where n angles of size theta meet symmetrically at a point. Then the angle between adjacent faces (the so called “dihedral angle”) is given by 2arcsin(cos(pi/n)/cos(theta/2)). This isn’t too hard to see if you draw a little diagram. Thus the angle between two faces in a tetrahedron is 2arcsin(cos(pi/3)/cos(pi/6)). Since five tetrahedra meet at each edge of the 600-cell we can calculate the angle between two tetrahedra to be 2arcsin(cos(pi/5)/cos(theta/2)) where theta is the angle we already calculated between two triangles in a tetrahedron. Call this new angle phi. In a tetrahedron of edge length e the distance from the centre to the centre of a face can be seen to be e/sqrt(24). By considering the triangle formed by the centre of a 600-cell, the centre of one of its boundary tetrahedra, and the centre of one of the tetrahedron’s faces, we can therefore see that r = (e/sqrt(24))*tan(phi/2).
After much calculation, all of the trigonometry cancels and we have r=e(sqrt(5/8)+sqrt(3/2)). Then doing the integration we get the content to be (25/4)(sqrt(5)+2)e^4.
When I saw this site I have linked, I thought what a shame that the bulk of the cell-600 is “unknown”. This would be a perfect problem for those clever guys who solved some problems already!
I don’t know how Wolfram Alpha knows; I’m fearfully ignorant of this sort of thing myself. Perhaps there’s a well-known decomposition of the 600-cell into nice simple bits whose 4-volume is easy to calculate?
The method I use below (integrate its surface area) works for every regular polytope, so it could be the method Wolfram Alpha uses. The only difficult part is simplification of a complicated trigonometric expression, but Wolfram Alpha eats those for breakfast.
On the other hand there are only three families of regular polytope above dimension 4, so maybe it just knows a general formula for those three families and then just has the five exceptional regular polytopes programmed in as special cases.
What do you mean by “general formula for the bulk”? I mean, (1) are you using “bulk” as a synonym for hypervolume or does it mean something else, and (2) along what axes are you wanting answerers to generalize? Obviously allowing for different edge length is trivial. Are you looking for a formula for arbitrary convex polytopes, or what?
You may try it!
https://protokol2020.wordpress.com/2017/02/13/cell-600-bulk-hypervolume/
A tetrahedron of edge length e has volume e^3/sqrt(72). So the 600-cell has surface volume 600e^3/sqrt(72). Let r be the distance from the centre of the 600-cell to the centre of one of its boundary tetrahedra. If we increase r to r+epsilon, the content increases by a shell around the 600-cell of width epsilon. This shell therefore has content epsilon*600e^3/sqrt(72) (plus terms of order epsilon^2 and smaller). Therefore we can find the content of the whole 600-cell by integrating 600e^3/sqrt(72) from r=0 to whatever r is when e=1. It remains to calculate r in terms of e.
Suppose we have a vertex of a polyhedron where n angles of size theta meet symmetrically at a point. Then the angle between adjacent faces (the so called “dihedral angle”) is given by 2arcsin(cos(pi/n)/cos(theta/2)). This isn’t too hard to see if you draw a little diagram. Thus the angle between two faces in a tetrahedron is 2arcsin(cos(pi/3)/cos(pi/6)). Since five tetrahedra meet at each edge of the 600-cell we can calculate the angle between two tetrahedra to be 2arcsin(cos(pi/5)/cos(theta/2)) where theta is the angle we already calculated between two triangles in a tetrahedron. Call this new angle phi. In a tetrahedron of edge length e the distance from the centre to the centre of a face can be seen to be e/sqrt(24). By considering the triangle formed by the centre of a 600-cell, the centre of one of its boundary tetrahedra, and the centre of one of the tetrahedron’s faces, we can therefore see that r = (e/sqrt(24))*tan(phi/2).
After much calculation, all of the trigonometry cancels and we have r=e(sqrt(5/8)+sqrt(3/2)). Then doing the integration we get the content to be (25/4)(sqrt(5)+2)e^4.
Well done, I think.
When I saw this site I have linked, I thought what a shame that the bulk of the cell-600 is “unknown”. This would be a perfect problem for those clever guys who solved some problems already!
In a sense, I wasn’t wrong.
Yeah, it is strange that they don’t have it.
It was still a fun problem even if WolframAlpha already knew the answer!
Vs lbh glcr “ibyhzr bs 600-pryy” vagb Jbysenz Nycun, gur nafjre V nffhzr lbh’er ybbxvat sbe pbzrf fgenvtug bhg.
This was quick, if true.
How does Wolfram Alpha knows that? And since when?
http://hi.gher.space/wiki/Hydrochoron
Those guys should update, and also should Wikipedia.
I don’t know how Wolfram Alpha knows; I’m fearfully ignorant of this sort of thing myself. Perhaps there’s a well-known decomposition of the 600-cell into nice simple bits whose 4-volume is easy to calculate?
The method I use below (integrate its surface area) works for every regular polytope, so it could be the method Wolfram Alpha uses. The only difficult part is simplification of a complicated trigonometric expression, but Wolfram Alpha eats those for breakfast.
On the other hand there are only three families of regular polytope above dimension 4, so maybe it just knows a general formula for those three families and then just has the five exceptional regular polytopes programmed in as special cases.
The second of those seems extremely likely. (But, I repeat, I don’t really know anything about this stuff.)
What do you mean by “general formula for the bulk”? I mean, (1) are you using “bulk” as a synonym for hypervolume or does it mean something else, and (2) along what axes are you wanting answerers to generalize? Obviously allowing for different edge length is trivial. Are you looking for a formula for arbitrary convex polytopes, or what?