Above is a picture of central cross section of a regular 120-cells
polytope. See a page of 'Literka' Cross section of 120-cells regular polytope
for details. We'll use this cross section as a help to find volume formula for this polytope.
For simplicity let us assume that a regular 120-cells polytope W built of 120 regular dodecahedrons has
edges of length 1. Assume that the picture above shows a central cross section of W and that it is a
42-faces polyhedron Q. Hexagonal faces of Q are described on the page of 'Literka'
How to compute volume of a regular dodecahedron.
As it was mentioned on the page of 'Literka' Cross section of 120-cells regular polytope,
polyhedron Q is a truncated polyhedron of a polyhedron K built of 30 congruent rhombuses. Let us present picture of a face of K
(which is a rhombus). This picture shows the way that faces of K are cut:
Rhombus ANDM is a face of a polyhedron K, hexagon AFEDCB is a face of a polyhedron V, blue triangles are the parts
cut by truncation. Faces of a polyhedron with 30 congruent rhombuses are described on the page of 'Literka'
Remarks about polyhedrons built of rhombuses.
It follows from the remarks of this page (See also a page of 'Literka'
Volume of a polyhedron built of 30 congruent rhombuses) that
Moreover, since BC and FE is parallel to AD
However BF=AD (see page of 'Literka' How to compute volume of a regular dodecahedron)
and BC=1, since we assume that edges of W have length 1 and BC is an edge of W.
Using above two equalities we find that
A page of 'Literka' Volume of a polyhedron built of 30 congruent rhombuses
gives us a formula for the radius of inscribed sphere into a polyhedron built of 30 congruent rhombuses with lengths of sides equal 1.
Transforming this formula for the polyhedron K, we have (r is radius of the inscribed sphere of K):
Finally, W is a union of 120 (4-dimensional) pyramids with vertex in the center of W
and bases - cells of W. Each pyramid has height equal r. Let S be (3-dimensional) volume of each
cell and V - volume of W.
If we don't assume that lengths of edges of W are 1, but an orbitrary number L>0, then we conclude a
general formula.
Final volume formula for 120-cells regular polytope with lengths of edges equal to L:
