Crown Jewels

[JMBR]

Hi. I don't know where I should post this, and the CRF main topic is closed. Once more, I was searching for 4d equivalents of last Johnson solid and came the idea of a rotunda, not like the pentagonal, but like bilunabirotunda. The teddy has three types of triangles: top, sides and base. If two of them would be joined by its bases, the lateral can be filled with tetrahedra, a structure that resebles the J91. Besides this, i didn't found how to change it into a J91, but I found a (maybe) valid CRF.
This would have two symmetric halves, each one startd with a pentagonal pyramid. On each triangle, a "luna" appears with 5-fold symmetry. Each teddy is connected to another by its pentagonal faces, and closed by another 5-gon pyramid. I don't know if it closes, but each vertex made it convex. Maybe other CRF have this structure...

[Klitzing]

What you're aiming for can indeed be closed.

As building blocks you'll have teddies pairwise adjoined by those triangles, which are adjacent to the 3 flag ones. And at these flag triangles in turn you can attach tetrahedra. The outer shape aka so far un-adjoined faces then are 3x pairs of triangles of the tetrahedra and the 2x 3 pentagons of the teddies, plus the base triangles of the teddies for sure. The 2 teddies then will make up for a dihedral angle of 120 degrees around their joining triangle. - But it's quite hard to see how these blocks could be adjoined further.

For that purpose it's much easier to recognize that we are dealing within the geometry of ex (600-cell). Infact in that post the required diminishing already was described. Just chop off 2 antipodal vertices, i.e. within this axial direction the vertex layers no. 1 + 9. Further chop off as well all vertices of layers 3 + 7. Then the former section planes introduce polar icosahedra. The icosahedral vertex figures at the other 2 mentioned layers have the right distance to be tangent only, but not intersecting with the formers. Whereas, in contrast, those vertex figures surely will intersect within each layer. This is how the teddies come into the play. In fact this 42-diminishing of ex then introduces 2 ikes and 2x 20 teddies. And only 60 out of the former 600 tets will remain. All then are equatorial.

Here is the remaining lace city, for reference:

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        x3o   o3f f3o   o3x       
                                         with
                                         F=ff=f+x
                                   
                                   
   f3o       o3F   F3o       o3f   
                                   
                                   
o3x   x3f F3o   f3f   o3F f3x   x3o      - id
                                   
                                   
   f3o       o3F   F3o       o3f         - f-ike
                                   
                                   
                                   
                                   
        x3o   o3f f3o   o3x              - ike
                                   
           \
            +---------------------- teddi

Thus, after all, it more resembles to a barrel than to a further rhombochoron (as thawrorh once proved to be one).

--- rk

[Klitzing]

I even considered to start with the leftmost part of the lace city in the previous post and add its reflection. Then the 4 teddies would outline a rhomb indeed. But then it remains the task to fill in the hollow part in between. - I then further continued and managed to come up with this one:

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                   o3x                   
                                         
            f3o           f3o           
                                         
                                         
x3o         o3o           o3o         x3o
                                         
                                         
            f3o           f3o           
                                         
                   o3x                   

You easily can spot the outer teddies. At any corner tetrahedra would be required. Further the "o3." parts of the central vertex positions will complement the teddi pentagrams to peppies. Further you can spot the 6 central ".3o" positions outlining 2 edge-to-edge trips. It also looks like that the remainder can be closed with squippies.

But then, OTOH, the circumradius of "o3o" clearly is smaller than that of any of its surrounding vertex position symbols. Therefore the whole figur ought be most probably somehow concave... - Is it?

None the less I'll post it here. Maybe someone else might benefit therefrom.

--- rk

[quickfur]

Today I got another crown jewel idea.

First, observe some patterns in 3D:
1) Start with an octahedron, pick a vertex, and a pair of opposite incident edges. Imagine cutting the octahedron with scissors along these two edges, then pulling apart the polyhedron, keeping all faces attached except at the cut edges. The polyhedron would deform as the vertex, now divided into two vertices, are pulled apart. The hole is in the shape of a pair of isosceles triangles, which, when the vertices are a unit length apart, become equilateral triangles that can be filled up with new faces. The result is a pentagonal bipyramid.
2) Start with an octahedron again, and this time cut it along a 3 edges lying on a plane, keeping the last edge at the back attached. It now becomes two square pyramids joined at an edge, and as the pyramids are pulled apart the result gap can be filled up with a triangular prism. The result is a biaugmented triangular prism.
3) Start with an icosahedron, and pick an edge. Cut it at this edge, and pull it apart into a square-shaped gap. The two triangles at the far ends of the edge deform into squares. The result is a hebesphenocorona.

Here are some comments on these constructions: in case (1), the chosen vertex expands into an edge, whereas the two incident edges expand into triangles. In case (2), the middle edge expands into a square, while the two side edges expand into triangles. In case (3), the chosen edge expands into a square, whereas two existing triangles deform into other squares.

So here's the idea for an analogous 4D construction: start with a 16-cell, and choose a triangular face. Cut the polychoron along this face, as well as along 3 adjacent triangles. Pull this triangle apart, producing a hole in the shape of a triangular prism. The 3 adjacent triangles should distort into square pyramids, and the original vertices should stretch into new edges. These new edges are where 3 new tetrahedral cells would appear, sitting between the new square pyramids (there were triangular faces originally where these new tetrahedra lie). Assuming the distortion can be done in a CRF way, the resulting polychoron should have 1 triangular prism, 3 square pyramids, 3 + 16 = 19 tetrahedra.

Does this construction work? Can the result be CRF? As far as I can tell, this works topologically, and the result ought to be convex, since 19 tetrahedra plus a few odd cells seem far short of the 600-cell, which is also convex, so the curvature of the result ought to be well within convex bounds. The only thing I'm unsure about is if all the edges can be made unit length: we may assume a rigid triangular prism, so the existence of the square faces is assured, so the question of whether it's possible to close up the polytope with tetrahedra in a CRF way, given the initial configuration of a triangular prism + 3 square pyramids, is reduced to whether all edges can be made unit length.

[quickfur]

P.S. On further consideration, it would appear that this prospective CRF should be an augmentation of 3prism || gyro-triangle, a member of the infinite family of n-prism || gyro-n-gon. This should make computation of the coordinates relatively easy.

[quickfur]

Unfortunately, after trying to calculate the coordinates for the biaugmented 3prism || gyro-triangle, which is what this construction would have produced, I found that the resulting shape is concave at the antipode of the triangular prism.

But perhaps an analogous construction applied to the icosahedral dipyramid might yield a CRF? Since the dipyramid would be shallower in that case, and there might be enough room to fit in a trigonal prism and square pyramid cells without making the antipode concave. The result should have 1 triangular prism, 3 square pyramids, 6 + 40 = 46 tetrahedra. But there's another danger here, that inserting the triangular prism would make the polytope hyperbolic around its top/bottom edges. Fortunately, a preliminary calculation indicates that the dichoral angles (3 tetrahedra + triangular prism at 90-degree edge + square pyramid) add up to 356.36 degrees, a very small angle defect just barely short of being flat/hyperbolic. Phew! So perhaps this might yield a CRF? But computing the coordinates will be a lot harder.

[quickfur]

Some further thoughts about this construction.

An icosahedral bipyramid has the following lace city, when analysed along a 3-fold axis of symmetry:

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   o3x
   f3o
o3o   o3o
   o3f
   x3o


The modified form, therefore, ought to have a lace city that looks something like this:

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    o3x   o3x
       A3o
   o3o     o3o
       o3B
       x3o


for some unknown values of A and B. There are, of course, additional unknowns in this system, as the relative heights of the vertex layers would not be the same as in the original icosahedral bipyramid.

Can this lace city be made CRF?

[quickfur]

Bad news: I realized yesterday that this lace city can't be CRF, because of two things: the o3o ... o3o apices of the original icosahedral bipyramid basically force A3o || o3B || x3o to be orbiform, since otherwise it's impossible to have unit edge length to all the vertices. However, since A3o || o3B || x3o by construction must lie on the same hyperplane, the only possible solution is A=F, B=F, i.e., they have to be identical to (a subset of) the original icosahedral vertices. This is the only solution because otherwise some of the lateral edges will be non-unit length. So that leaves only the o3x ... o3x || A3o part. If we replace the original o3x of the icosahedral bipyramid with o3x || o3x without any lateral displacement, then the edges between F3o and o3x || o3x will be longer than unit length, which implies that o3x || o3x needs to be moved closer to F3o in order to retain unit edge length. However, the distance between o3o and o3x || o3x will be shorter than unit length without lateral displacement, which implies that o3x || o3x needs to move away from F3o in order to retain unit edge length with o3o. This is a contradiction. Therefore, the proposed lace city cannot be made CRF.

[quickfur]

Wonder if the following modified lace city might work?

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   x3o   x3o
      o3A
  o3o     o3o
      A3o
   o3x   o3x