Above is a picture of central cross section of a regular 600-cells
polytope (see also a page of 'Literka' Cross section of 600-cells regular polytope).
We'll use this cross section as a help to find volume formula for this polytope.
We need more information about the cross section Q of a 600-cells regular polytope W. This cross section Q is shown at the top of this page. Let us repaint
polyhedron Q. We'll use the same color for congruent faces:
There are 20 blue faces and 60 yellow ones. Blue faces - these are equilateral triangles. They are just faces of
tetrahedrons, which are cells of a polytope W. Yellow faces - these are cross sections of tetrahedrons, which are
cells of a polytope W. Planes of these cross sections pass through one edge of tetrahedrons and through
the center of opposite edge. Let us assume that edges of a polytope W have lengths equal 1. Then also lenfths of sides
of blue equilateral triangles have lengths 1.
Yellow faces of Q are isosceles triangles
such that AB=1 and
Let us take a cross section of Q by a plane passing through the center of Q and one of sides of blue equilateral
triangles (one of 20 faces of Q). As a result we receive a regular polygon with 10 sides:
Point O is the center of Q, A and B - vertices of cross section are points of edges of blue faces of Q. Angle AOB has 36 degrees.
But AB=1. Formulas for cos36 from the page of 'Literka' Volume of a polyhedron built of 30 congruent rhombuses
give us
Let us remind: A is a vertex of W, O is the center of W. Let us take a cell J of W such that one of vertices
of J is A. Let P be the center of J. Then triangle APO is a right triangle. A simple consideration shows
that AP=3*h/4, where h is the height of J. Number h is also the height of yellow faces of Q and was computed above.
Hence
and
But OP=r, where r is the radius of the inscribed sphere into W. We have:
Polytope W is the union of 600 4-dimensional pyramids, each of them has a vertex O and a base - one
of cell of W. Let V be a volume of W. Then V=600*r*S/4=150*r*S, where S is a volume of a cell of W.
We know that S=h*p/3, where p is area of a face of a cell of W.
Summing the previous equalities we obtain
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 600-cells regular polytope with lengths of edges equal to L:
