What is the edge length of the regular decagon that is the external boundary of the face-centered orthogonal projection of a regular dodecahedron of edge length 1?
In other words, in this diagram of a regular dodecahedron:
If the edge length of the pentagons in the middle are 1, then what is the edge length of the decagon on the outside?
On
The Cartesian coordinates of the vertices of one regular dodecahedron with edge length $\sqrt 5-1$ are $(\pm 1, \pm 1, \pm 1), (0, \pm \phi, \pm \phi^{-1}), (\pm \phi^{-1}, 0, \pm \phi),(\pm \phi, \pm \phi^{-1}, 0),$ with $\phi$ being the Golden Ratio: $\phi = \frac {1+\sqrt 5}{2}$. So scale the edge length back to 1, drop the last coordinate and get distances between outer vertices.
You can see that the outer vertices belong to both a pentagon of edge $\phi$, and a decagon. Since the pentagon edge is the shortchord of the decagon, we can derive the edge direct as
$\frac{\phi}{\sqrt{\phi\sqrt{5}}}$, or $\sqrt{\frac{\phi}{\sqrt 5}}$.
Putting in values, we get 1.61803398875/1.90211303259.