http://math.ucr.edu/home/baez/golden.html On Mon, Apr 27, 2020 at 2:59 PM James Buddenhagen <jbuddenh@gmail.com> wrote:
I found a file of coordinates: Here are the 20 integer point vertices of an almost regular dodecahedron: 2550409 2550409 2550409 4126648 0 1576240 1576240 4126648 0 2550409 2550409 -2550409 4126648 0 -1576240 0 1576240 4126648 0 -1576240 4126648 2550409 -2550409 2550409 -1576240 4126648 0 -2550409 2550409 2550409 0 1576240 -4126648 0 -1576240 -4126648 2550409 -2550409 -2550409 -2550409 2550409 -2550409 -2550409 -2550409 2550409 -4126648 0 1576240 -2550409 -2550409 -2550409 -1576240 -4126648 0 1576240 -4126648 0 -4126648 0 -1576240
James
On Mon, Apr 27, 2020 at 2:44 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
What intrigues me about that apparently straightforward solution is the number of previous authors who fail to mention it, despite the question plainly having occurred to them.
There is a slightly elusive trap involved: solving for the exact (irrational) vertices, before rationally approximating that solution, initially looks more promising, but instead leads back to the same dead end.
From RA dodecahedra follow easily RA icosidodecahedra (even if one is uncertain how to spell them), via mid-points of edges. However for the dual RA rhombic triacontahedra, straight "dualisation" doesn't work any more: five approximating planes might fail to meet at a common vertex.
Is there another way ... ?
WFL
On 4/27/20, Andy Latto <andy.latto@pobox.com> wrote:
Just taking rational points near the 20 vertices doesn't work, because the result won't be a dodecahedron at all. Five of the vertices that are near 5 vertices of the original dodecahedron will be near to lying in a plane, but they won't be exactly in a plane.
So instead of approximating the vertices, approximate the faces.
The regular dodecahedron's 12 faces lie on 12 planes, and each of the vertices is the intersection of three of those planes. Each of the planes has an equation of the form Ax + By + Cz = D. Taking close rational approximations to A, B, C, and D will yield a combinatorial dodecahedron with all its vertices close to those of the regular dodecahedron.
Andy
On Mon, Apr 27, 2020 at 2:05 PM Fred Lunnon <fred.lunnon@gmail.com> wrote:
To construct a family of pentagonal dodecahedra P_k , all their vertices with rational Cartesian components, which successively approximate a regular specimen P .
More pedantically, for each real error bound e > 0 , there should exist some h such that dist(P, P_k) < e for all k >= h ; where dist(P, Q) is defined as (say) square root of sum of squared distances between corresponding vertices of (combinatorially isomorphic) polyhedra P, Q .
See
Adam Goucher, James Buddenhagen
https://cp4space.wordpress.com/2013/09/22/rational-approximations-to-platoni...
(Have I come across those fellows somewhere before?)
Günter M. Ziegler (2007) "Non-rational configurations, polytopes, and surfaces" https://arxiv.org/abs/0710.4453 (Undefined terms in the latter may not mean quite what one might expect: like "simple", "polyhedron", "polytope"!)
Fred Lunnon
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