[math-fun] Rational approximate Platonic dodecahedra
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
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
I did this many hard-disks ago, so the original is long lost, although I may have posted it to this list. I think I did this: take a regular dodecahedron (in R^3) and approximate the normal to one face with a rational normal, i.e. all 3 components rational numbers. I may have used continued fractions for this. Then apply (some? but which?) of the symmetries of the dodecahedron to that to get 12 planes all with rational normals. The intersections of each 3 that correspond to 3 incident to the same dodecahedral vertex then give vertices of a rational approximation of the dodecahedron. Probably I would have to do it all again to get details right. 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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
I'm trying very hard at the moment to _avoid_ getting involved in another current thread ... WFL On 4/27/20, Mike Stay <metaweta@gmail.com> wrote:
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
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Andy.Latto@pobox.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
-- Mike Stay - metaweta@gmail.com http://math.ucr.edu/~mike https://reperiendi.wordpress.com
_______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
participants (4)
-
Andy Latto -
Fred Lunnon -
James Buddenhagen -
Mike Stay