Most symmetric coloring of icosahedron/dodecahedron

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Most symmetric coloring of icosahedron/dodecahedron

Postby quickfur » Thu Jan 30, 2014 3:54 am

This is probably well-known, but recently I noticed that it's possible to create a highly-symmetrical 4-coloring of the icosahedron's vertices (resp. dodecahedron's faces). This coloring partitions the icosahedron's vertices into 4 sets of 3, that sit at the vertices of 4 equilateral triangles. These triangles have edges that lie at the chords of pentagonal cross-sections of the icosahedron -- I'm guessing they are subsets of some Kepler-Poinsot solid? No adjacent vertices have the same color (being a 4-coloring). It would appear that these 4 equilateral triangles have tetrahedral symmetry (though they do not themselves form a tetrahedron since they are disjoint).

Is there a similar highly-symmetrical coloring of the dodecahedron's vertices (resp. icosahedron's faces)? In particular, is there a 3-coloring that exhibits a high degree of symmetry?
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Re: Most symmetric coloring of icosahedron/dodecahedron

Postby wendy » Thu Jan 30, 2014 8:04 am

The dodecahedron's vertices are usually mapped onto a 5*5 grid, with the middle diagonal (p,p) empty. All vertices with the same number in one position, eg (3,p) or (p,4), represent a tetrahedron from one of the chiral pair of 5 tetrahedra in a dodecahedron. The full set of points (p,3) (3,p) represent one of the five inscribed cubes.

For the icosahedron, the main group is a set of six opposite vertices, which reflects into four dimesnions, as the six inscribed x10o2o10x in the {5,3,3}. This bi-decagonal prism then contains the vertices of 20 pentachora.

But apart from these two coulourings, i do not know of any partition into 2, 3 or 4, of either of these figures.
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Re: Most symmetric coloring of icosahedron/dodecahedron

Postby quickfur » Fri Jan 31, 2014 3:45 am

wendy wrote:The dodecahedron's vertices are usually mapped onto a 5*5 grid, with the middle diagonal (p,p) empty. All vertices with the same number in one position, eg (3,p) or (p,4), represent a tetrahedron from one of the chiral pair of 5 tetrahedra in a dodecahedron. The full set of points (p,3) (3,p) represent one of the five inscribed cubes.

Aha, so the compound of 5 tetrahedra can be inscribed in a dodecahedron? If so, that would represent a highly-symmetrical 5-coloring -- choose a color for each tetrahedron. Or conversely, perhaps we can convert it into a symmetrical 4-coloring by somehow assigning, in some "coherent" way, one of 4 colors to each tetrahedron's vertices, while maintaining the same relationships between vertices of like color?

My original question was a bit silly, though -- a symmetrical 3-coloring is not possible because 3 doesn't divide 20. :roll:

For the icosahedron, the main group is a set of six opposite vertices, which reflects into four dimesnions, as the six inscribed x10o2o10x in the {5,3,3}. This bi-decagonal prism then contains the vertices of 20 pentachora.

But apart from these two coulourings, i do not know of any partition into 2, 3 or 4, of either of these figures.

What of the 4-coloring that I found, where vertices of each color forms an equilateral triangle?
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Re: Most symmetric coloring of icosahedron/dodecahedron

Postby wendy » Fri Jan 31, 2014 7:10 am

The colouring that you are hinting at (four colours of 3) belong to the vertices of four triangles, these form a tetrahedron inscribed in a great icosahedron. A pair of these triangles form the vertices of Weimholt's hexahedron or bid ike in RK's parlance,
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Re: Most symmetric coloring of icosahedron/dodecahedron

Postby quickfur » Fri Jan 31, 2014 4:08 pm

wendy wrote:The colouring that you are hinting at (four colours of 3) belong to the vertices of four triangles, these form a tetrahedron inscribed in a great icosahedron. A pair of these triangles form the vertices of Weimholt's hexahedron or bid ike in RK's parlance,

Ah you're right, they are indeed 4 of the great icosahedron's faces that span its vertices. Thanks!
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