Is this old news?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

Is this old news?

Postby Secret » Tue Feb 14, 2012 11:45 am

While painstakingly trying to build the 600 cell in googlesketch up, using the 24 cell made by rectifying the 16 cell

I check quickfur's site for construction pathways
I also checked the old thread I made here
Wendy and quickfur mentioned of coherent indexing
When I read the method will give you icosahedra which later form the snub 24 cell, I then tried using the same method on an ordinary octehedra
And then bingo! A nice icosahedron ... T4NST5.jpg

I then browse google, and notice most method of making a icosahedron is very troublesome
No I can finally make icosahedron in google sketchup and in my drawings

This also make me wonder what happen if I apply the same method to a 16 cell...

->Back to working on the 600 cell
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Re: Is this old news?

Postby quickfur » Wed Apr 25, 2012 8:23 pm

Yes I'm replying 2 months late, but I'm bored today and this forum has been a tad quiet these days. So here goes.

Deriving an icosahedron by partitioning the edges of an octahedron is quite well known. In fact, it's the basis for deriving "nice" coordinates for the icosahedron in terms of the golden ratio. Coxeter, in his book "Regular Polytopes", uses this fact as the basis for describing the construction of the 600-cell from the 24-cell. I doubt he is the original discoverer of this construction; I believe he was merely incorporating someone else's conclusion into the flow of thought in his book (and in a quite beautiful way, I might add). The 24-cell partitioned in this way gives rise to the snub 24-cell, which can then be turned into a 600-cell by inserting the vertices of the dual 24-cell.

Now unfortunately, this construction cannot be applied to the 16-cell. The requirement appears to be that it works on the rectified alternated n-cube: in 3D, the alternated cube is the tetrahedron, and the rectified tetrahedron just happens to be the octahedron. In 4D, the alternated tesseract is the 16-cell, and the rectified 16-cell just happens to be the 24-cell. In 5D, however, the pattern breaks down: the alternated 5-cube is the 5-demicube, but the rectified 5-demicube is not regular (I don't think it's even uniform), so partitioning it in the golden ratio seems unlikely to yield anything interesting.

The problem with the 16-cell is that its edges cannot be coherently indexed. You can try it... but there will always be a triangular face that isn't cyclical in its edges. So there's no consistent way to do a golden ratio partitioning on it.
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