If we inscribed all the 6 regular convex four-dimensional polytopes in a sphere, which one would have the highest volume?

Question

When a dodecahedron is inscribed in a sphere, it occupies more of the sphere's volume (66.49%) than an icosahedron inscribed in the same sphere (60.54%).

But what about for the 6 regular convex four-dimensional polytopes? The 5-cell, the 8-cell, the 16-cell, the 24-cell, the 120-cell and the 600-cell. If we inscribed all of them in a sphere... classify/sort them from the one that would have the highest volume to the one that would have the less volume. And please also provide the % of the volume of the sphere that each one of them would occupy.

Answer 1

Have a look here, this provides a great overview, and 'calculate-yourself'-solutions for your questions: http://davidf.faricy.net/polyhedra/Polytopes.html

For all regular convex polytopes inscribed to spheres the following should hold: The share of the volume covered by the polytope is directly related to the number of it's vertices: --> the more vertices, the more of the spheres volume is covered.

I don't have a formal proof for this, but consider the following: The regular convex polytopes are completely symmetrical, every face is regular and congruent to each other face. That maximizes the distance of each point lying on the surface to it's neighbouring vertices, which is the characteristic edge length (which is a constant for all edges and is directly correlated to the radius of the circumscribed sphere).

For polychora (= 4d-polytopes) I have the following results:

• 5-cell (5 vertices): 2.95%
• 16-cell (8 vertices): 13.51%
• 8-cell (16 vertices): 20.26%
• 24-cell (24 vertices): 40.53%
• 600-cell (120 vertices): 78.27
• 120-cell (600 vertices): 84.96%