Question of 3D geometry and some polyhedra

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Question of 3D geometry and some polyhedra

Postby hy.dodec » Thu Dec 28, 2017 3:52 pm

Hello, I ashamed to ask this question releated with 3D in this forum,
but I think that there're many members who're smart of 3D geometry, So I ask this question.

OK, Let's go the point.

1. Marked vertices on a icosahedron are coplanar?
2. Marked vertices on a snub dodecahedron are coplanar?
3. Marked vertices on a tetrakis hexahedron are coplanar?

Could you answer these questions only using "logical" geometry? (Easily speaking, not via coordinates, but using axioms or 3D Euclidean Vector)

Because I study these polyhedrons for first time...
and don't know how to set coordinates of each vertice simply (for example, express only in terms of golden ratio and its reciprocal)
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Re: Question of 3D geometry and some polyhedra

Postby Mercurial, the Spectre » Fri Dec 29, 2017 6:48 am

1 and 2: they are isogonal. The marked points are their vertex figures.
#1 is always coplanar because it is, by definition, regular, and there can be only one form. The shape being formed is a regular pentagon.
#2 is coplanar in specific cases, and is true for the uniform snub dodecahedron. But generally under symmetry, these are not generally coplanar, because the edge lengths may not be equal.

3: Think of a tetrakis hexahedron as a cube augmented with 6 square pyramids. Four of its highlighted vertices are also the vertices of the cube, in this case, a 1:sqrt(2) rectangle which is easily coplanar. Now look at the cube standing on its edge. The rectangle is exactly centered on the origin of the xy plane, so edge-first, it represents the 2D projection of the cube. The two remaining highlighted vertices are opposite and represent two square pyramids. Since there are two faces in the edge-first projection of the cube that are perpendicular to the xy plane while their centers are located at the y-axis, the 2 vertices must lie on the opposite sides of the y-axis for symmetry, so by definition, they must also lie on the xy plane. In general, it is always coplanar.
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Re: Question of 3D geometry and some polyhedra

Postby wendy » Fri Dec 29, 2017 1:33 pm

The answer to this is that you are looking at two intersecting spheres. These intersect in a circle, which lies in a plane.

All of the polytopes given are uniform, and thus the circumsphere passes through all vertices.

The edges that approach a given sphere can be read as the radii of the second sphere.

If either of these conditions fail, then you are left with the possibility that the marked verteces (the vertex figure) do not lie on the same plane.

This is not a silly question. It is essentially the gist of much of my geometry.

In the case of the tetrakis hexahedron, or apiculated cube, the margins or walls between faces are necessarily parts of the mirror-plane of the figure, and here because it looks like a line, and there are equal number of faces on either side, the vertices must lie on the same mirror, and hence the same plane.
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Re: Question of 3D geometry and some polyhedra

Postby Mercurial, the Spectre » Fri Dec 29, 2017 4:33 pm

When finding the coordinates of polyhedra with high symmetry, it is best to use the octahedral model (all permutations of x,y,z with all sign changes) since it's relatively easy. For every vertex (x,y,z), there is an octahedral symmetry polyhedron (generally isogonal) that has this coordinate.
Say, you have the coordinate set (1,2,3). You have these 6 permutations:
and, accounting for 8 sign changes (+++, ++-, +-+, -++, +--, -+-, --+, ---), you have 6*8 = 48 possible coordinates, representing a truncated cuboctahedron.
In fact that is the order of octahedral symmetry.

Now, when building icosahedral symmetry figures, the coordinates generally involve the golden ratio (phi) because the regular icosahedron has 3 golden rectangles (1:phi) that are identical in its pyritohedral subsymmetry plus a regular pentagon embedded inside the icosahedron; this is marked by edges but not considered a face.

In fact, you can look at the icosahedron standing on its edge. Two opposite edges form a golden rectangle of coordinates (±1,±phi,0). Then there is another of coordinates (0,±1,±phi), and another of coordinates (±phi,0,±1). These rectangles lie on the xy, xz, and yz planes, so they are mutually perpendicular. This represents the pyritohedral subsymmetry of the icosahedron which has half the order of octahedral symmetry.

One thing is, the icosahedron does not have full octahedral subsymmetry. It doesn't have any 4-fold rotations the cube and octahedron has. This means that the coordinates of the icosahedron need to be in cyclical order, like abc -> cab -> bca where the symbols are interpreted as vertices of a triangle and labeled by moving in only one direction like a circle. So you have to base on pyritohedral symmetry to construct an icosahedral-symmetric object. In fact there are two variants, one from either abc or cba. However, you still have the 8 possible sign permutations.

For tetrahedral-symmetric polyhedra, you have the 6 permutations but only four sign changes (+++, +--, --+, -+-, or ++-, +-+, -++, ---), distinguished by whether the number of minus signs in a given coordinate is even or odd. In fact, the tetrahedron can be inscribed in a cube, which is indeed the cube's own alternation.

Both symmetries have half the order of full octahedral symmetry, but they are not the same. Pyritohedral symmetry has no 4-fold projections but central inversion, while tetrahedral symmetry has 4-fold projections but no central inversion. Central inversion is the property that for a coordinate (+a,+b,+c), you have (-a,-b,-c).

That is all I can explain to you regarding the construction of polyhedra through coordinates. Hope you can learn more about these funky solids!
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