Start with an octahedron, and attach 4 thawros to 4 non-adjacent faces on it (by their "top" triangles, the ones with all 3.5.3.5 vertices). We'll also connect the other "icosidodecahedral" triangles of the thawros (at six places total), bringing their squares closer together. This sets up all 54 vertices of the polytope (6 from the octahedron, 12 (2x6) shared by 2 thawros each, and 36 ([3+6]*4) that are part of only 1 thawro each).

Attach 4 teddis to the remaining exposed faces of the octahedron (by the singular triangles, of course). Let their pentagonal faces connect with those of the thawros.

Fill in the six "trenches" between the thawros with a Trip between the thawros' squares and with two Squippys on each side of the trip. For each squippy, one of its triangles connected to the trip's, the two adjacent triangles connected to the two neighboring thawros, and the opposite triangle connected to the neighboring teddi.

Bring in a truncated octahedron. Connect its 6 square faces to the Trips' remaining faces, and 4 of its hexagonal faces to the "backs" of the thawros. Then on each of the other 4 hexagonal faces, place a tricu. Each tricu's "top" triangle connects to the last exposed triangle of a teddi, the squares connect to the "bases" of the three nearby squippys, and the 3 other triangles of the tricu finish the polytope off by connecting to the last remaining triangles on the thawros (the ones adjacent to the thawro's hexagon).

It has 54 vertices, 150 edges, 124 faces, and 32 cells. This makes its Euler characteristic -4 (https://people.math.osu.edu/fiedorowicz.1/math655/HyperEuler.html), although it's supposed to be 0. In my very limited experience, even a polychoron with angle sums greater than 360° has an Euler characteristic of 0. This indicates to me that it may be "self-intersecting," or whatever the 4D analog of that property is.

Also checked the Excel file with all the CRFs in it. Nothing matches its cell counts.

Not sure how relevant this is, but there are 10 different kinds of edges, all of which have angle sums of less than 360° (although some of them come very close).

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`Many thanks to Marek14 for the complete listing of cell dihedral angles!`

1. 4-6 tricu (54.7356°), 4-6 toe (125.264°), 3-4 trip (90°), 3-4 squippy (54.7356°) (total 324.7352°) (x12)

2. 4-6 toe (54.7356°), 4-6 thawro (110.905°), 4-4 trip (60°) (total 225.6406°) (x12)

3. 6-6 toe (109.471°), 3-6 tricu (70.5288°), 3-6 thawro (138.19°) (total 318.1898°) (x12)

4. 3-4 tricu (125.264°), 3-4 squippy (54.7356°), 3-3 thawro (138.19°) (total 318.1896°) (x24)

5. 3-4 tricu (125.264°), 3-4 squippy (54.7356°), 3-3 teddi (138.19°) (total 318.1896°) (x12)

6. 4-4 trip (60°), 3-4 thawro (110.905°) X2 (total 281.81°) (x6)

7. 3-4 trip (90°), 3-4 thawro (159.095°), 3-3 squippy (109.471°) (total 358.566°) (x24)

8. 3-5 teddi (100.812°), 3-3 squippy (109.471°), 3-5 thawro (100.812°) (total 311.095°) (x24)

9. 5-5 teddi (63.4349°), 3-5 thawro (142.623°) X2 (total 348.6809°) (x12)

10. 3-3 oct (109.471°), 3-5 thawro (142.623°), 3-5 teddi (100.812°) (total 352.906°) (x12)

I'm not good enough with lace-tower notation to get these edge lengths right, but I know the vertices lie on o3x3o, x3x3o, x3o3x, x3x3x. One of the x's in x3x3o should be f, and the x3x3x term should have a u and another one I don't know the exact length of (but one of the x's is correct). o3x3o is the octahedron (with 6 vertices, of course), "x3x3o" has the 12 vertices shared by 2 thawros, "x3o3x" has 12 of the vertices exclusive to certain thawros, and "x3x3x" has the 24 vertices of the thawros' hexagons.

Anyway, I'm almost completely sure this polytope has some kind of self-intersection. Could one of you please use your advanced methods to determine that this thing isn't worth our time?

I don't know if this explanation is sufficient, so I can post some pictures of it if necessary. My methods are laughably primitive, though, so don't expect renders of Quickfur quality.

Thanks in advance, and sorry for wasting your time if this turns out to be an insignificant result.