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