hypervolume of truncated tesseract?

Jonathan Post jvospost3 at gmail.com
Wed Mar 28 00:13:00 CEST 2007

In geometry, a truncated tesseract is a uniform polychoron
(4-dimensional uniform polytope) which is bounded by 24 cells: 8
truncated cubes, and 16 tetrahedra.  It has 64 vertices, 128 edges, 88
faces (64 triangles and 24 octagons).

We know that: "The truncated tesseract may be constructed by
truncating the vertices of the tesseract at 1 / (2 + sqrt(2)) of the
edge length."

1 / (2 + sqrt(2)) = 0.292893219


Does that mean that the hypervolume of a truncated tesseract is
[1 / (2 + sqrt(2))]^4 of the hypervolume of a tesseract
= 0.00735931288 times L^4 where L is the edge length?

That just seems absurdly small to me.


1 - (1 / (2 + sqrt(2))) = 0.707106781

(1 - (1 / (2 + sqrt(2))))^4 = 0.707106781^4 = 0.992640687?

That seems too large.

This comes up as a case of the integer sequence problem that I've been
working on, as a 4-D analogue of the 3-dimensional problem which was
posed and solved by Wouter Meeussen in 2001, as given in OEIS.

What is the content (hypervolume) of the convex hull
of the vertices of P(n) (i.e. of the polytope), as a
function of n? P(n), for a nonnegative integer n, is
the convex hull of all vertices which have integer
coordinates in the Euclidean space Z^4, defined by all
permutations of: (+-h, +-i, +-j, +-k) such that h^2 +
i^2 + j^2 + k^2 = n.

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