User talk:Robert Stanforth

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[edit] Cartesian coordinates of snub icosidodecadodecahedron

Hi Robert, If you see this message, there's some question over some edits you added for Cartesian coordinates of snub icosidodecadodecahedron. Talk:Snub_icosidodecadodecahedron Can you look this over? I appreciate it. Tom Ruen 21:03, 8 June 2007 (UTC)

Thanks, Tom! I have checked over the edits, and they did indeed uncover an error that I made in transcribing my results to the Wikipedia page. Robert Stanforth 16:08, 14 October 2007 (UTC)

Hello

Are sure of the coordinates of : http://en.wikipedia.org/wiki/Rhombidodecadodecahedron  ??? and how do you calculate all theses coordinates ?

Thanks !

Robert FERREOL ferreol@mathcurve.com --81.66.134.192 (talk) 20:06, 24 May 2008 (UTC)

I have double-checked these Cartesian coordinates, and I confirm that they are correct. Note that, in my choice of coordinate axes, an axis of 5-fold symmetry passes through the point (1,0,τ); regrettably this is the opposite convention to that used in the Wikipedia article on the icosahedron, in which a vertex is placed at (0,1,τ). Of course, both are valid mathematically.

As to how these are derived, I'll give a brief outline. Imagine you are holding a physical model of this polyhedron. (It's possible to follow this argument with a picture, but it's easier if you have a model. I don't!) Orient it so that six of the squares are aligned with coordinate planes. Now look at the model along the axis of 3-fold rotational symmetry passing through (1,1,1). You'll see that the six vertices nearest to you form an irregular hexagonal arrangement, with opposite corners connected by edges of this polyhedron. Assume a polyhedral edge length of 2. The edges of this hexagon come in two lengths, the longer length equalling 2/τ as can be seen by the adjacent pentagrams. Simple geometry gives the shorter lengths as 2/τ², and simple algebra then gives the distance d between any pair of adjacent-but-one points in the hexagon as 2(√2)/τ.

This distance d is useful as follows. Inscribe an equilateral triangle inside this hexagon, using its three vertices that are incident on the coordinate-plane-aligned squares. Doing so on the other seven similar hexagons yields a non-uniform rhombicuboctahedron incorporating the six coordinate-plane-aligned squares (of side 2) and these eight equilateral triangles (of side d) (and also twelve rectangles). Simple algebra then gives the perpendicular distance between opposite squares as 4τ−2, and hence one vertex of the rhombidodecadodecahedron is (1,1,2τ−1).

To obtain the other vertices, apply the 3-by-3 matrix with the following rows:

[ 1/(2τ) , −τ/2 , 1/2 ]

[ τ/2 , 1/2 , 1/(2τ) ]

[ −1/2 , 1/(2τ) , τ/2 ]

This matrix effects a one-fifth turn about the axis that passes through the icosahedral vertex (1,0,τ), and is derived by considering the effect of such a rotation upon those edges of the icosahedron that are aligned with the coordinate axes.

Hope this helps! Robert Stanforth (talk) 21:53, 24 May 2008 (UTC)