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It's well-known/documented that you can construct the vertices of a regular dodecahedron from the following Cartesian coordinates (with appropriate rotation/scaling):

$(±1, ±1, ±1)\\ (0, ±\varphi, ±\frac{1}{\varphi})\\ (±\frac{1}{\varphi}, 0, ±\varphi)\\ (±\varphi, ±\frac{1}{\varphi}, 0)$

It's also well-known that you can construct the regular icosahedron in spherical coordinates with one point at each of the latitudes $±\pi/2$, and 5 points at each of the latitudes $±\arctan(1/2)$, rotationally symmetric except with the top and bottom hemispheres offset by $\pi/5$ radians.

Since the icosahedron and dodecahedron are duals of each other, this approach clearly should extend to the dodecahedron. For two different latitudes $±x$ and $±y$, each will have 5 vertices evenly spaced around them, for a total of the 20 vertices. The upper hemisphere will be offset in longitude $\pi/5$ from the lower hemisphere. The question is, what are $x$ and $y$?

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The orientation of the dodecahedron in Cartesian coordinates you cite is of course not the same as the orientation that would be produced by taking the dual of the icosahedron in spherical coordinates.

The desired latitude angles are, $$\pm \tan^{-1} \frac{3-\sqrt{5}}{4}, \quad \pm \tan^{-1} \frac{3+\sqrt{5}}{4}.$$ Equivalently, these are $$\pm \cot^{-1} (3 + \sqrt{5}), \quad \pm \cot^{-1} (3 - \sqrt{5}).$$ The computation is rather tedious to derive from the spherical coordinates of the icosahedron, but the way to do it is to observe that the centroid of an icosahedral face may be computed as the average of its three vertices in Cartesian coordinates, and after converting this to spherical coordinates, ignore the radius and longitudinal angle.

So instead of doing such a calculation, it may be better to use the Cartesian coordinates for the dodecahedron that you cited. Specifically, we want to calculate the angle at the origin between the centroid of a dodecahedral face to any of its vertices. If we call this angle $\alpha$, then $\pi/2 - \alpha$ will be the more extreme latitude of the two pairs of five vertices. Then if we calculate the angle at the origin between any two adjacent vertices, and call this $\beta$, then $\pi/2 - (\alpha + \beta)$ will be the less extreme latitude of the two pairs. Since for your given coordinates the dodecahedral circumradius is $R = \sqrt{3}$, in my opinion these calculations are much more tractable than the first approach.

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  • $\begingroup$ Great! I did in fact take the "tedious" approach of calculating the centroid of an icosahedronal face from the spherical coordinates, and got arctan((3+sqrt(5))/4), but then I tried to double-check it by computing the dot-product of the midpoint of the centroid of a dodecahedron face with one of its vertices, and got something totally different. I must have made a mistake somewhere, but it made me doubt my approach. $\endgroup$
    – D0SBoots
    Sep 16 '20 at 0:49
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The dodecahedron's vertices are partitioned into 4 groups of 5, each of which forms a regular pentagon. We have two pentagonal faces, whose edges are the edges of the dodecahedron, and two "diagonal pentagons" whose edges are the edges of an inscribed cube.

If we take Euclid's construction, the inscribed cube has edge length 2, the pentagonal faces have side length $\sqrt5-1$, and the circumradius is $\sqrt3$.

Then, the circumradii of the two pentagons are $\frac{\sqrt{5}-1}{2\sin(\pi/5)}$ and $\frac{2}{2\sin(\pi/5)}$, corresponding to latitudes: $$x = \arccos(\frac{\sqrt{5}-1}{2\sqrt{3}\sin(\pi/5)})$$ $$y = \arccos(\frac{2}{2\sqrt{3}\sin(\pi/5)})$$

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  • $\begingroup$ Validated that these do line up with the arctan-based formulas, although simplifying them down is a pain... $\endgroup$
    – D0SBoots
    Sep 16 '20 at 1:04
  • $\begingroup$ If it helps, these are essentially the same 4 planes as for the dual icosahedron (rotated a bit). See the gif on wikipedia: en.wikipedia.org/wiki/… $\endgroup$ Sep 16 '20 at 1:17
  • $\begingroup$ The existence of the four planes is relatively obvious, it's their exact location that is tricky. :) $\endgroup$
    – D0SBoots
    Sep 16 '20 at 1:30
  • $\begingroup$ Right, what I meant was that another approach is just to take the latitudes you have for the icosahedron and rotate them all to get the latitudes for the dodecahedron. $\endgroup$ Sep 16 '20 at 1:57
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    $\begingroup$ Ah, I get what you're saying now. You're not rotating the latitudes themselves (because the latitudes define planes or circles, depending on which object you're talking about), but rather rotating a set of model points chosen from those latitudes, which is basically equivalent to saying "take the vertices of the icosahedron, line them up along a line of longitude, and rotate by a certain amount." This works, again, because of the dual relationship between the dodecahedron and the icosahedron, in particular that the face-vertex central angle is the same. $\endgroup$
    – D0SBoots
    Sep 16 '20 at 2:46

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