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The golden ratio,

$$\phi =\frac{1+\sqrt{5}}{2}$$

appears (among other polyhedra) in the Platonic solids icosahedron and dodecahedron, and it's quite easy to see the significance of the discriminant $d_\phi = 5$ in those two solids. Both $\phi$ and the plastic constant,

$$P=\frac{(9+\sqrt{3d})^{1/3}+(9-\sqrt{3d})^{1/3}}{2^{1/3}\cdot9^{1/3}}=1.324717\dots$$

with $d = 23$, can be found in the vertices of the snub icosidodecadodecahedron,

$\hskip2.8in$Snub icosidodecadodecahedron

$\hskip1.3in$(Image courtesy of wikipedia and Robert Webb's Stella software)

This may be a silly question, but why does $P$ appear here? And does $d = 23$ have any geometric significance to that solid? Note that $P$ is also,

$$P = \frac{e^{\pi i/24}}{\sqrt{2}}\frac{\eta(\tau)}{\eta(2\tau)}=1.324717\dots$$

$$\tau = \frac{1+\sqrt{-23}}{2}$$

where $\eta(\tau)$ is the Dedekind eta function.

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    $\begingroup$ Have you tried going through a computation of the coordinates yourself? $\endgroup$ Sep 8, 2013 at 18:29
  • $\begingroup$ No, but someone here might be able to derive those vertex coordinates from first principles, and explain why $P$ pops up. $\endgroup$ Sep 8, 2013 at 18:34
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    $\begingroup$ Please note that $P$ is the unique real zero of $x^3= x+1$. I would expect this equation rather than $d=23$ to appear in the calculation of the vertex coordinates. The ocurrence of $d=23$ would then be more something like a secondary effect. $\endgroup$ Sep 8, 2013 at 19:12
  • $\begingroup$ Yes, I am familiar with that equation, and I agree it will appear when deriving the coordinates. But regarding the aspects of that polyhedron with integer values, I'm embarrassed to admit I actually tried to find small integer relationships between its edges, faces, and vertices to see if $d=23$ is there somewhere. But all I've come up with seem to be arbitrary ones. $\endgroup$ Sep 8, 2013 at 20:01
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    $\begingroup$ I assume that it is not $23$ that appears, but $24 - 1$ — the number $24$ appears throughout mathematics. Note that $\tau$ solves $\tau^2 - \tau + 6=0$, if I am not mistaken, and so that appearance of $-23$ is probably just an appearance of $1 - 4\times 6$. $\endgroup$ Sep 12, 2013 at 3:07

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