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A question on the disnub mentions

  1. golden ($x^2-x-1=0$) gives the dodecahedron + much more.
  2. tribonacci ($x^3-x^2-x-1=0$) gives the snub cube.
  3. plastic ($x^3-x-1=0$) gives the snub icosidodecadodecahedron.

All of these polyhedra can be built by generating 3D points in root $R$'s number with values $a + b R^c$ for integers $a$, $b$, $c$. After that, pick points with the same norm and use those to make a polyhedron.

Is there a nice $R$ for generating the snub dodecahedron?

Is there an interesting polyhedron generated by the number field of the Narayana cow sequence constant (root of $x^3-x^2-1=0$)? I found many pretty ones with octahedral symmetry, but none particularly noteworthy.

Take $R$ as the real root of ($x^3-6 x^2 + 4 x-2=0$). This value deserves a name, since $(R^{24} -24) - e^{\pi \sqrt{163}} <10^{-14}$. That $R$ can generate this figure, which seems more interesting than polyhedra I made with Narayana's cow.

moonshine polyhedron

When generating from $a + b R^c$, the bounds on $a$, $b$, $c$ matter. If all values are between -3 and 3, then we might say a particular polyhedron is easily generated by $R$. The higher the bound, the greater the difficulty. With a bound of 17, $$(0, 5, 17), (1, 12, 13), (3, 4, 17), (3, 7, 16), (5, 8, 15), (7, 11, 12), (8, 9, 13)$$ all have the same norm ($\sqrt{314}$), and can generate this figure:

points at 314

What $R$ values yield a particularly high number of vertices with a relatively low bound? Are there $R$ values that make highly interesting polyhedra?

Ed Pegg
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    For the snub dodecahedron, from this entry, will the number field generated by $R := x \approx 1.7155$ of $-1 + 2 x + 4 x^2 - x^3 - 4 x^4 + x^6 = 0$ do the job? – Tito Piezas III Feb 22 '16 at 22:25
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    You mention by "...these polyhedra can be built by generating 3D points in root $R$'s number with values $a+bR^c$..." I was wondering:: What happens if we use the complex root $R$ of $x^3-x^2-x-1=0$? Can we make polyhedra from complex number fields? – Tito Piezas III Dec 10 '17 at 16:44

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