Marek14 wrote:quickfur wrote:Keiji wrote:Thanks for the hard work quickfur!

I don't suppose we have coordinates/images/.def/.off for CJ4.5.2 anywhere? That's now the only 4D crown jewel without its own page.

I didn't build it because basically it's CJ4.5.1 with the top part of the rectified 120-cell glued on both ends. It's just extra vertices (=slower convex hull times, slower rendering, etc.) away from the exciting action (i.e. where the J92's are). You could still create a page for it, and mention the cell counts, etc., which are all already known. It's just that it's not that much more interesting than CJ4.5.1 itself.

If the top part of rectified 120-cell is glued on "both ends", does it mean you can get anohter CRF that would count as crown jewel by gluing it on only one end?

Yes, and since the caps glued on are themselves made of two separable layers, you can also glue half a cap on one end or half a cap on the other end, and any combination thereof. Also, any diminishing/augmentation that applies to the upper part of the rectified 120-cell also applies here. I consider these variations as uninteresting, since they don't relate to the region of interest where the J92's are. In that sense, CJ4.5.1 is the most "fundamental" form, and CJ4.5.2 and the other unnumbered possibilities are just cut-n-paste variations on it. I assigned CJ4.5.2 a number because it represents the simplest derivation from the rectified 120-cell that contains the CJ4.5.1 core, but other than that, it's not fundamentally interesting.

quickfur wrote:Hmm. Today I was reading up on the snub disphenoid on Wolfram, and it claims that the analytic solution for its coordinates requires solving a cubic equation?? How did Marek get a solution in terms of square roots??

Hm. Let's have a look.

Snub disphenoid has 8 vertices which can be separated into 4 layers. Let's use edge length 2 here:

1st layer: (1,0,0) and (-1,0,0)

2nd layer: (0,x,y) and (0,-x,y)

3rd layer: (x,0,z) and (-x,0,z)

4th layer: (0,1,w) and (0,-1,w)

Each layer has two vertices and the line of these two is always perpendicular between layers. Further, snub disphenoid has dihedral symmetry, meaning that distance between layers 1 and 2 is the same as distance between 3 and 4, so we can replace w with sum y+z.

Now then, snub disphenoid has four kinds of edges:

1. Edge joining two vertices in layers 1 and 4 -- this is 2 by definition.

2. Edge joining layers 1-2 or 3-4 -- this means distance (1,0,0) and (0,x,y) is 2.

3. Edge joining layers 1-3 or 2-4 -- this means distance (1,0,0) and (x,0,z) is 2.

4. Edge joining layers 2-3 -- this means distance (0,x,y) and (x,0,z) is 2.

So all the conditions mean that it's sufficient to ensure the triangle (1,0,0), (0,x,y), (x,0,z) is equilateral with side 2.

We have equations:

x^2+y^2+1 = 4

(x-1)^2+z^2 = 4

2x^2+(z-y)^2 = 4

By solving these three equations in Mathematica, I now get cubic solutions as well (there are quadratic solutions, but these have z = 0). The positive solution has x = 1.28917, y = 1.15674 and z = 1.97898. Not sure what went wrong before...

So the snub disphenoid has CVP 3 after all?