From: dseal@armltd.co.uk (David Seal)
Newsgroups: sci.math
Subject: Re: Location of polyhedron vertices
Date: 25 Jan 1995 21:18:13 GMT

kewe@bskewe.atr.bso.nl (Cornelis Wessels) writes:

[quoting someone else about the co-ordinates of the vertices of an
 icosahedron:]
>  > (1, 0, TAU)
>  > (TAU, 1, 0)
>  > (0, TAU, 1)
>  >  
>  > and the points resulting from arbitrary sign changes to the coordinates of
>  > these.
>
>An isocahedron has 12 vertices. If I take all permutations of TAU and
>-TAU in your above matrix I come to 2^3 = 8 possibilities. If I also
>vary the signs of the term with value 1 I come to 2^6 = 64 permutations.
>How can I map this on 12 vertices?

That isn't a matrix - it's a list of 3 points, each with three
co-ordinates. If you switch TAU to -TAU in each of them, you get three
more points, giving 6 in total. If you switch 1 to -1 in each of those
6, you get 6 more points.

There are now 12 points, which can reasonably easily be verified to be
the 12 vertices of an icosahedron with edge length 2.

For a dodecahedron, use the same value TAU = (1+SQR(5))/2, then
generate 8 points from the single point:

  (TAU, TAU, TAU)

by every combination of sign changes, and 12 points from the list of
three points:

  (1, 0, TAU^2)
  (TAU^2, 1, 0)
  (0, TAU^2, 1)

by every combination of sign changes. The resulting 20 points are the
20 vertices of a dodecahedron of edge length 2.

David Seal
dseal@armltd.co.uk