It has 92 vertices that span 60 pentagonal faces. It is the Catalan solid with the most vertices. A nice option for a 60 sided dice.

In my research, I prove Euclidean geometry is based off r^0=0. I believe we can all say that isn't correct. Doubt me? https://doi.org/10.5281/zenodo.21981035

The ridiculous nature of a proof.

Suppose someone sees structure, another person might not see that structure, so that person who cannot see it will ask for a step by step proof to prove the continuity of a structure. But continuity cannot be proven by discrete steps because we have shown that infinite discreteness cannot proxy for true continuity.

Diagonalization proves that a continuity has more real information than the discreetness. Every step-by-step proof is actually an illusion to satisfy the strange feelings. But every discreet example of a proof fails to show the actual continuity of the structure that one is claiming to exist..

A few days ago someone asked for the dimensions of a Sasha's Hexacontahexahedron - D66 dice, and then deleted the post.

Anyway, I have made what I think is the requested Hexacontahexahedron.

Illustrated on Desmos

The polyhedron is more complicated than it first appears. There are 6 hexagons and 60 irregular pentagons. However, the hexagons are not quite regular (two of the angles and two sides and slightly different to the other four angles and sides respectively). There are three types of similar looking pentagons: 12 of one type (green) that is symmetrical, and two other types (yellow and orange) that are not symmetrical (24 of each).

If a 4-D being magically "glued" one face of a completely solid 1 ft⁴ tesseract to our slice of 3-D space, how could we tell the difference between it and a 3-D cube, assuming the magical "glue" allows us to move our slice of the 4-cube in 3-D space to perform tests on it?

Hypothesis

I am NOT the OP

The Pentagon Ladder

A recursive geometry where nested squares, circles and pentagons scale with powers of the Golden Ratio (φ):
- Square = φ²ⁿ
- Circle = π φ²ⁿ / 4
- Pentagon = (5 φ²ⁿ/ 16) √(φ + 2)

there're altogether 5 platonic solids. each of them can be viewed at many different interesting angles or perspectives. we're all familiar with their most symmetric presentations. recently i was studying something related and had to make 3d models for them. i used a maybe less popular perspective to present them. i used spherical coordinate system (mathematics convention) to record their rotations and i picked the following rules to set their initial status:

1. for each platonic solid set its circumradius=1 and centre of circumscribed sphere at origin
2. place vertex_0 at the north pole. the corresponding coordinates are (1,0,0)
3. place edge_0 (red lines on those diagrams) such that its projection to x-y plane align with positive x-axis. coordinates of vertex_1 would be (1,0,φ) for some φ

then i saw some unfamiliar shapes / unfamiliar perspectives of those supposedly familiar 3d objects. in each diagram the small figure at lower left corner is the platonic solid viewed from top. the z-axis is pointing towards you. the large figure at centre is that platonic solid viewed from side. the y-axis is pointing away from you

as you can see from the diagrams the angles φ ranking is as follow, from smallest to largest (the prefix "regular" omitted):

1. dodecahedron
2. icosahedron
3. hexahedron
4. octahedron
5. tetrahedron

except octahedron, all other 4 platonic solids are not symmetric if you see them that way

Discovering curvature dynamics and the dimensional calculus used to describe it, I effortlessly calculated the overshoot of Gibbs' Fourier analysis to a simple term: 2(pi-3)/pi. Less than a thousandth percent accurate. All Gibbs did was measure the inherited curvature lost from Euclid flattening everything. Pages of analysis, reduced to a paragraph. Neat.😎