## What is this solid called?

Discussion of tapertopes, uniform polytopes, and other shapes with flat hypercells.

### What is this solid called?

Hi!

Having recently gotten a missing piece of insight into 4D visualization I'm revisiting the regular polytopes trying to observe them mentally. I'm fine with the 5-cell, 8-cell and 16-cell, working on the 24-cell.

Anyway, after the 16-cell I was considering the following figure:

- Start with a cube in 3 space. (Say at ±1 on x, y, z and w=0).
- Now add two vertices placed on each side of the w axis, at the center of the cube in xyz.
- Connect the vertices of the cube to form pyramids towards each side of the w axis. So we have square pyramids coming in and out the w axis, meeting in these two "apices".

Now if the vertices on each side of the cube are distant from each other of the same length than the diagonals of the cube faces, these back to back pyramids actually make up regular octahedra (right?).

I'm not sure that description was the best way to convey the figure. Hopefully someone else can see it.

When I observe it mentally I count the following:
- 10 vertices.
- 28 edges.
- 6 cells (all regular octahedra).

I feel this figure to be somewhat special because it's made entirely of regular polyhedra while still being rather simple.

The vertex figure of the vertices at each end are cubes.
The vertex figure of the other vertices are the figure formed by two tetrahedra back to back.

I've reviewed the cubic pyramid and the octahedral pyramid but they only ever go into one direction.
Did I mixup something while imagining this solid?
Is there a family for polytopes made of regular polyhedra but that don't have interchangeable vertices?

Thanks
joan
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### Re: What is this solid called?

Hi Joan.

It's a cubic tegum, but the square pyramids do not line up to form octahedra.

Wendy
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the dream we dream together is reality.

wendy
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### Re: What is this solid called?

Thanks!

in the meantime I looked at analogies in 3D and found the bipyramids.

But trying to think about more figures like this, for example by replacing all the faces of a dodecahedron by pentagonal bipyramids (J13), yes, I think there is a problem.
It seems I'm still having trouble visualizing the angles between adjacent faces.
joan
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### Re: What is this solid called?

The trouble with the name 'bi-pyramid' is that this word could mean a pyramid whose base is a pyramid. For example, if you have a pentagon, you can in 3d, erect a pyramid on it. In 4d, you erect a pyrimid on that, but there are two pentagonal-pyramid faces that can function as a base.

Tegum is a kind of product, in the same manner as prisms. The figure that you describe in your first post amounts to a covering of an orthogonal line and cube. But in 4d, you can do something like a covering of a pentagon in the wx plane, and the pentagone in the yz plane. The faces become 25 tetrahedra (disphenoid tetrahedra), such as you can see by squeezing a tube at one point to a line, and further up at right angles to it.
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the dream we dream together is reality.

wendy
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### Re: What is this solid called?

Wrt. bipyramids first consider the better accessible 3D case:
Here we have the trigonal bipyramid (= 2 "attached" tetrahedra), the tetragonal bipyramid (= octahedron), and the pentagonal bipyramid. (The hexagonal bipyramid already would be flat.) In all these cases you can see already that the pairs of lacing triangles, joining each the (then becoming internal) base to the tips, would have some dihedral angle (= angle between faces), which definitely is not 180 degrees.

The same then holds in 4D. You well can have a bipyramid based on a tetrahedron (= 2 "attached" pentachora), based on an octahedron (= hexadecachoron), and based on an icosahedron. All these will have tetrahedral cells only, cause we always have pyramids based on trigons. Again these pairs of tetrahedra are mutually inclined, bending on either side from the rim towards the tips. And, as you already observed, you can have a similar bipyramid based on the cube. That one then has square pyramids for lacing cells.

But there is no such dodecahedral bipyramid, at least none with all unit edges. This is a consequence of the theorem of Pythagoras: The circumradius of the dodecahedron already is larger than unity (i.e. larger than edge size). Therefore any height of either tip atop the hyperplane of the base polyhedron would just increase that asked for lacing edge size from that circumradius: (lacing edge size)2 = (circumradius of base)2 + height2.

--- rk
Klitzing
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### Re: What is this solid called?

What the OP has constructed is indeed the cubic bipyramid. The square pyramids do not merge into regular octahedra, because they are bent in different directions at the square faces of the cube, since otherwise one set of them could not possibly converge on the opposite apex. This construction is analogous to the 3D case of constructing the octahedron as a square bipyramid. The triangular faces connecting the square to either apex remain as triangles, not rhombuses, because they must be bent where they meet the edges of the square, otherwise they could not meet at both apices.

With respect to the impossibility of a unit-edged dodecahedral bipyramid, a similar thing happens in going from 4D to 5D: the tesseract has the peculiar property that its circumradius is equal to its edge length. That means that the tesseract pyramid in 5D, if we constrain all edges to be unit length, must have zero height! So the n-cube (bi)pyramid construction only produces unit edges up to the 4D case of the OP's cube bipyramid. In 5D this construction becomes degenerate (height zero, so it is subdimensional), and in 6D the 5-cube (bi)pyramid cannot have unit edges, as the 5D cube's circumradius is greater than its edge length. However, the n-cross bipyramid always exists, and is in fact none other than the (n+1)-cross.

Among the pentagonal polytopes, the icosahedral bipyramid exists because the icosahedron's circumradius is (slightly!) less than its edge length. However, the 600-cell clearly has a far larger circumradius than edge length, so the 5D 600-cell pyramid cannot have unit edge length. The 120-cell also has a far larger circumradius than edge length, so the same thing applies to the 120-cell (bi)pyramid in 5D. Past that point, there are no more regular pentagonal polytopes (unless you get into hyperbolic geometry), so the trend stops here.

The other family is the simplex family, which always have bipyramids with unit edges -- the n-simplex pyramid is simply the (n+1)-simplex, and it is always possible to glue two (n+1)-simplices together at a common facet to form a bipyramid of the n-simplex. Interestingly enough, as n increases without bound, the difacetal angle between the facets of the n-simplex approaches 90°, so the n-simplex bipyramid approaches having (n-1)-simplex bipyramid cells, but never quite gets to that point. Supposedly you would get coplanar facets if you're willing to take the leap to n=infinity, but at n=infinity some very strange things happen that causes such generalizations from finite dimensions to be questionable at best, or outright wrong at worst.
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