[math-fun] Dodecahedron packing conjecture
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet <http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>. Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)? I'd like to see that plane sliding along the major diagonal, like the Menger Sponge movie <https://www.youtube.com/watch?v=fWsmq9E4YC0>. —rwg
Bill writes: Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet
<http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>.
I'm amazed. (And I still don't see it.) Is this explained anywhere? I'd love to see a physical model that I can walk around so I can understand this.
Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)?
I don't understand this question. The airtight sheet is a 3-dimensional structure; a plane section of a 3-d honeycomb is a 2-dimensional structure. Anyway, it'd be interesting to classify polyhedra according to whether translates of the given polyhedron can fill a plane. More generally, for any n-dimensional convex polytope K we can ask whether translates of K can fill a k-dimensional subspace. Has anyone looked at this? Jim Propp
On May 9, 2020, at 4:00 PM, James Propp <jamespropp@gmail.com> wrote:
Bill writes:
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet
<http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>.
I'm amazed. (And I still don't see it.) Is this explained anywhere? I'd love to see a physical model that I can walk around so I can understand this.
A MacGyver episode? He’s stuck in the desert and needs to collect condensation to stay hydrated, but all he has to work with are a bunch of D&D dice ...
Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)?
I don't understand this question. The airtight sheet is a 3-dimensional structure; a plane section of a 3-d honeycomb is a 2-dimensional structure.
Jim, look at a the cut-away dodec. on the left of <http://gosper.org/dodex.gif>. There’s a special plane orthogonal to the 3-fold axis where the cross-sections are regular hexagons. You need to check that when extended above and below this plane the dodec’s don’t interpenetrate, but it seems to me that part of the construction also holds water. -Veit
Anyway, it'd be interesting to classify polyhedra according to whether translates of the given polyhedron can fill a plane. More generally, for any n-dimensional convex polytope K we can ask whether translates of K can fill a k-dimensional subspace. Has anyone looked at this?
Jim Propp
Veit, Thanks for explaining what the cross section was for. Jim, look at a the cut-away dodec. on the left of <
http://gosper.org/dodex.gif>. There’s a special plane orthogonal to the 3-fold axis where the cross-sections are regular hexagons. You need to check that when extended above and below this plane the dodec’s don’t interpenetrate, but it seems to me that part of the construction also holds water.
:-) Jim
On Sat, May 9, 2020 at 11:46 AM Bill Gosper <billgosper@gmail.com> wrote:
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet <http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>. Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)? I'd like to see that plane sliding along the major diagonal, like the Menger Sponge movie <https://www.youtube.com/watch?v=fWsmq9E4YC0>. —rwg
If you have Mathematica, the Manipulate in gosper.org/rhombosis.nb seems to uphold the conjecture. But it is ugly: Bill Gosper <billgosper@gmail.com> [image: Attachments]8:39 AM (2 hours ago) to Wolfram, In the attached Manipulate, the dodecahedra have been shrunk 3% and have visible air gaps, if you tumble them carefully. So how do I exorcise those groups of one, two, or three rhombi painted on the insides of the cutaway dodecs?? This was made by ClipPlanes of GeometricTransformations of PolyhedronData@"Dodecahedron", which I can provide if you're curious. —Bill
Jim — the key idea is that if you balance a dodecahedron on a vertex, then a horizontal plane halfway between the north and south pole intersects the dodecahedron in a regular hexagon. Hexagons of course tile the plane, and if you grow a dodecahedron out of every hexagon by translating the dodecahedron, the adjacent dodecahedra abut at faces that tilt in or out of vertical at the same angle, so they fit smoothly. This is also true of a cube balancing on a vertex, an octahedron balanced on a face, or an icosahedron balanced on a face — all their equators are regular hexagons, and adjacent faces fit together smoothly (because opposite faces are parallel). The analogous construction for a regular tetrahedron is to balance the tetrahedron on an edge — the equator is then a square, which tiles the plane. to form a sheet of tetrahedra, you have to grow tetrahedra out of the tiling squares in alternating directions, twisting by 90°, to get them to abut properly. In all cases the polyhedra form an "airtight" sheet, as Gosper says. Hope that helps. — Scott On Wed, May 13, 2020 at 11:20 AM Bill Gosper <billgosper@gmail.com> wrote:
On Sat, May 9, 2020 at 11:46 AM Bill Gosper <billgosper@gmail.com> wrote:
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet <http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>. Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)? I'd like to see that plane sliding along the major diagonal, like the Menger Sponge movie <https://www.youtube.com/watch?v=fWsmq9E4YC0>. —rwg
If you have Mathematica, the Manipulate in gosper.org/rhombosis.nb seems to uphold the conjecture. But it is ugly: Bill Gosper <billgosper@gmail.com> [image: Attachments]8:39 AM (2 hours ago) to Wolfram, In the attached Manipulate, the dodecahedra have been shrunk 3% and have visible air gaps, if you tumble them carefully. So how do I exorcise those groups of one, two, or three rhombi painted on the insides of the cutaway dodecs?? This was made by ClipPlanes of GeometricTransformations of PolyhedronData@"Dodecahedron", which I can provide if you're curious. —Bill _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
And as you move up or down from the equator, the hexagons go irregular with short/long edges alternating around, and triangular gaps appear between three hexagons where they used to join at a corner. Thus, while it might be water tight in a sense, it's not pressure tight, as the water tight plane has triangular pyramidal voids opposite each other meeting at the same point. -tom On Wed, May 13, 2020 at 2:14 PM Scott Kim <scott@scottkim.com> wrote:
Jim — the key idea is that if you balance a dodecahedron on a vertex, then a horizontal plane halfway between the north and south pole intersects the dodecahedron in a regular hexagon. Hexagons of course tile the plane, and if you grow a dodecahedron out of every hexagon by translating the dodecahedron, the adjacent dodecahedra abut at faces that tilt in or out of vertical at the same angle, so they fit smoothly. This is also true of a cube balancing on a vertex, an octahedron balanced on a face, or an icosahedron balanced on a face — all their equators are regular hexagons, and adjacent faces fit together smoothly (because opposite faces are parallel). The analogous construction for a regular tetrahedron is to balance the tetrahedron on an edge — the equator is then a square, which tiles the plane. to form a sheet of tetrahedra, you have to grow tetrahedra out of the tiling squares in alternating directions, twisting by 90°, to get them to abut properly. In all cases the polyhedra form an "airtight" sheet, as Gosper says. Hope that helps. — Scott
On Wed, May 13, 2020 at 11:20 AM Bill Gosper <billgosper@gmail.com> wrote:
On Sat, May 9, 2020 at 11:46 AM Bill Gosper <billgosper@gmail.com> wrote:
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet <http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>. Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)? I'd like to see that plane sliding along the major diagonal, like the Menger Sponge movie <https://www.youtube.com/watch?v=fWsmq9E4YC0>. —rwg
If you have Mathematica, the Manipulate in gosper.org/rhombosis.nb seems to uphold the conjecture. But it is ugly: Bill Gosper <billgosper@gmail.com> [image: Attachments]8:39 AM (2 hours ago) to Wolfram, In the attached Manipulate, the dodecahedra have been shrunk 3% and have visible air gaps, if you tumble them carefully. So how do I exorcise those groups of one, two, or three rhombi painted on the insides of the cutaway dodecs?? This was made by ClipPlanes of GeometricTransformations of PolyhedronData@"Dodecahedron", which I can provide if you're curious. —Bill _______________________________________________ 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://cube20.org/ -- http://golly.sf.net/ --
Yes, that helps a lot. Why isn’t this better known? Has anyone built models? (Did Gardner ever write about it? Aside from the cubes case?) Jim On Wed, May 13, 2020 at 5:14 PM Scott Kim <scott@scottkim.com> wrote:
Jim — the key idea is that if you balance a dodecahedron on a vertex, then a horizontal plane halfway between the north and south pole intersects the dodecahedron in a regular hexagon. Hexagons of course tile the plane, and if you grow a dodecahedron out of every hexagon by translating the dodecahedron, the adjacent dodecahedra abut at faces that tilt in or out of vertical at the same angle, so they fit smoothly. This is also true of a cube balancing on a vertex, an octahedron balanced on a face, or an icosahedron balanced on a face — all their equators are regular hexagons, and adjacent faces fit together smoothly (because opposite faces are parallel). The analogous construction for a regular tetrahedron is to balance the tetrahedron on an edge — the equator is then a square, which tiles the plane. to form a sheet of tetrahedra, you have to grow tetrahedra out of the tiling squares in alternating directions, twisting by 90°, to get them to abut properly. In all cases the polyhedra form an "airtight" sheet, as Gosper says. Hope that helps. — Scott
On Wed, May 13, 2020 at 11:20 AM Bill Gosper <billgosper@gmail.com> wrote:
On Sat, May 9, 2020 at 11:46 AM Bill Gosper <billgosper@gmail.com> wrote:
Ages ago I mentioned that regular dodex can interlock in an "airtight" sheet <http://gosper.org/dodex.gif>, analogous to a sheet of cubes ("Martin's Marbles") <http://gosper.org/martinsmarbles.png>. Isn't this just a plane section perpendicular to (1,1,1) through the 3D endo-dodec checkerboard <http://gosper.org/Endo-dodecahedron_honeycomb_1.png> (absent the endos)? I'd like to see that plane sliding along the major diagonal, like the Menger Sponge movie <https://www.youtube.com/watch?v=fWsmq9E4YC0>. —rwg
If you have Mathematica, the Manipulate in gosper.org/rhombosis.nb seems to uphold the conjecture. But it is ugly: Bill Gosper <billgosper@gmail.com> [image: Attachments]8:39 AM (2 hours ago) to Wolfram, In the attached Manipulate, the dodecahedra have been shrunk 3% and have visible air gaps, if you tumble them carefully. So how do I exorcise those groups of one, two, or three rhombi painted on the insides of the cutaway dodecs?? This was made by ClipPlanes of GeometricTransformations of PolyhedronData@"Dodecahedron", which I can provide if you're curious. —Bill _______________________________________________ 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
JP: "Why isn’t this better known? Has anyone built models?" There's a 2009 paper in the Moscow Mathematical Journal (Interlocking of convex polyhedra: towards a geometric theory of fragmented solids) by Kanel-Belov, Dyskin, Estrin, Pasternak, & Ivanov-Pogodaev that appears to be what we are discussing here. https://www.researchgate.net/publication/23709852_Interlocking_of_Convex_Pol...
I'm going to have fun working out equators of regular 4d polytopes. Probably some of you already know the answer. I know from Banchoff's hypercube movie that the balancing-on-a-vertex equator of a hypercube is an octahedron, and I can see that the balancing-on-a-facet equator of the 16-cell (cross-polytope) is a cuboctahedron. It looks to me like the equator of a a balancing-on-a-facet equator of the 24-cell is also a cuboctahedron, but a very special one — the edges of the equatorial cuboctahedron are edges of the 24-cell itself. So...what's the most symmetrical way to balance a 5-cell? It's certainly not on a vertex! On Wed, May 13, 2020 at 7:38 PM Hans Havermann <gladhobo@bell.net> wrote:
JP: "Why isn’t this better known? Has anyone built models?"
There's a 2009 paper in the Moscow Mathematical Journal (Interlocking of convex polyhedra: towards a geometric theory of fragmented solids) by Kanel-Belov, Dyskin, Estrin, Pasternak, & Ivanov-Pogodaev that appears to be what we are discussing here.
https://www.researchgate.net/publication/23709852_Interlocking_of_Convex_Pol... _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
My extremely fuzzy 4D visualization skills suggest that there are no good answers for the 5-cell. The best I've been able to do is set it on a triangular face, with a horizontal edge forming the peak. The horizontal cross-sections have order-12 symmetry; they are triangular prisms with various aspect ratios, and the one halfway up is a semiregular triangular prism (with square sides). Notice that the equator has no more symmetry than any of its parallel cross-sections. On Thu, May 14, 2020 at 1:24 AM Scott Kim <scott@scottkim.com> wrote:
I'm going to have fun working out equators of regular 4d polytopes. Probably some of you already know the answer. I know from Banchoff's hypercube movie that the balancing-on-a-vertex equator of a hypercube is an octahedron, and I can see that the balancing-on-a-facet equator of the 16-cell (cross-polytope) is a cuboctahedron. It looks to me like the equator of a a balancing-on-a-facet equator of the 24-cell is also a cuboctahedron, but a very special one — the edges of the equatorial cuboctahedron are edges of the 24-cell itself. So...what's the most symmetrical way to balance a 5-cell? It's certainly not on a vertex!
On Wed, May 13, 2020 at 7:38 PM Hans Havermann <gladhobo@bell.net> wrote:
JP: "Why isn’t this better known? Has anyone built models?"
There's a 2009 paper in the Moscow Mathematical Journal (Interlocking of convex polyhedra: towards a geometric theory of fragmented solids) by Kanel-Belov, Dyskin, Estrin, Pasternak, & Ivanov-Pogodaev that appears to be what we are discussing here.
https://www.researchgate.net/publication/23709852_Interlocking_of_Convex_Pol...
_______________________________________________ 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
A very nice program for generating and visualizing these cross sections of 4D polytopes is Stella4D. It has a library of the regular 4D polytopes (including the nonconvex ones) and a large number of uniform 4D polytopes. Along with many other generative opertions, it has a built-in cross sectioning operation in the vertex-first, edge-first, face-first, and cell-first directions. It's not free, but well worth the cost if you want to visualize or make models (paper or 3D printed) of polyhedra and polytopes. See: https://www.software3d.com/Stella.php George http://georgehart.com On 5/14/2020 1:23 AM, Scott Kim wrote:
I'm going to have fun working out equators of regular 4d polytopes...
participants (8)
-
Allan Wechsler -
Bill Gosper -
George Hart -
Hans Havermann -
James Propp -
Scott Kim -
Tomas Rokicki -
Veit Elser