Number of Half-Unit Pentachora in a Unit Hexacosichoron

Question

As the title summarizes, I am unable to find out how many 5-cells {3,3,3} (pentachora) with a circumscribed diameter (d) of 1/2 can fit into a 600-cell {3,3,5} (hexacosichoron) with d=1, {3,3,3,6}. The method is analogous to fitting 6 triangles {3} (d=1/2) into a hexagon {6} (d=1), {3,6}, and 20 tetrahedrons {3,3} (d=1/2) into an isocahedron {3,5} (d=1), {3,3,6}. I am new to this and my knowledge of polytopes is still limited, so it would be helpful if you reference a URL in your answer.

Answer 1

Your reasoning is understandable, and maybe even logical, yet it is incorrect. Triangular tiling can be achieved in 2 dimensions without a problem, but tetrahedrons cannot be used for 3-space filling due to the fact that their dihedral angle is arccos(1/3)≈70.5 so the combined angles of five tetrahedra would be about equal to 352.5, just short of 360. Therefore, a regular icosahedron cannot be composed of regular tetrahedra in the way you describe. This principle applies to every simplex after that as well. Don't feel too bad though, Aristotle made pretty much the same mistake...

http://www.ams.org/notices/201211/rtx121101540p.pdf

~CSD~

Answer 2

After further research, I found the solution to be simpler than I had previously thought. All you need to recognize is that for each n-dimensional polytope of {6}, {3,5}, {3,3,5}, ... {3^(n-2),5} type (d=1), the number of (n-1)-simplices (d=1/2) on its hypersurface is equal to the number of n-simplices (d=1/2) that can fit into it. So to give a direct answer to my question, 600 pentachora (d=1/2) compose a hexacosichoron, since it has 600 cells (half-unit tetrahedra) on its hypersurface.

http://en.m.wikipedia.org/wiki/Simplex