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.

A math problem

https://protokol2020.wordpress.com/2017/02/20/landaus-problem/

This one is a real one, but somewhat transformed and potentially solvable.