How to find the maximum diagonal length inside a dodecahedron? I am trying to find the maximum length of a diagonal inside a dodecahedron with a side length of $2.319914107\times10^{89}$ meters.
I am not sure if any other information than that is needed, if it is I probably have it and will give it, but if you could help that would be great.
I am really just trying to find the distance between opposite vertices.
 A: This is an old question but here is an answer that is easy to understand, without using trigonometry.  This construction comes from Euclid's Elements (XIII.17).

Consider a cube of edge length $e = 2\varphi$ where $\varphi = (1+\sqrt{5})/2$ is the golden ratio.  Note that $\varphi^2 = \varphi+1$.  Now, consider one face of this cube, say $ABCD$.  Let the center of this face be called $F$, and let the midpoint of $AB$ be $M$.  On $FM$ locate $P$ such that $PF = 1$, hence $MP = \varphi - 1$.  Now consider an outward normal to the face $ABCD$ from $P$, and locate the point $K$ on this normal such that $KP = 1$.
Similarly, if we locate midpoint $M'$ on $CD$, $P'$ on $FM'$, and $K'$ similarly to the above procedure, we now have $KK' = 2$.  We now endeavor to show that $KA = KB = K'C = K'D = 2$.  This is easy to see since we have by the Pythagorean relationships $$\begin{align*} (MK)^2 &= (MP)^2 + (KP)^2 = (\varphi-1)^2 + 1^2 = 3 - \varphi, \\ (KA)^2 &= (MK)^2 + (MA)^2 = 3-\varphi + \varphi^2 = 4. \end{align*}$$  From here, it is easy to show that $A, B, C, D, K, K'$ are vertices of a regular dodecahedron of edge length $2$ in which a cube of edge length $2\varphi$ is inscribed.  (For example, we can show that $AB = AK'$ using more Pythagorean relationships, and consequently isosceles $\triangle ABK \cong \triangle AK'K$.)  Consequently, the distance between diametrically opposite vertices of both polyhedra is simply $2 \varphi \sqrt{3}$, and the ratio of this diameter to the dodecahedral edge length is simply $$(2 \varphi \sqrt{3})/2 = \varphi \sqrt{3}.$$
A: 
I haven't derived the diagonal here, but this is a strategy that may work. Sometime ago I tried it with calculating the diagonal of an icosahedron.
In the 1st image that I have attached, there are 2 identical 3-d shapes, with their major bases put together and then turned $\frac{1}{10}$ of a revolution. Each shape consists of a smaller pentagon on the top with a side length of $s$, a larger pentagon on the bottom with a side length of $\phi s$, where $\phi$ is the golden ratio. The sides of the shape consist of 5 identical trapezoid whose lengths sides are $s,s,s,$ and $\phi s$.

I'm going to build a docdecahedron by attaching red elastics to the vertices of the 2 larger bases, one end of each elastic attached to a vertex on the top figure, and the other end one attached to a vertex on the bottom figure. Once all $10$ elastics have been attached, I pull the 2 objects apart (keeping the 2 bases parallel, without rotation) until the length of each elastic is $s$. Then I calculate how far apart the larger bases of the 2 objects are (vertically), and add that to twice the height of each object, giving me how far apart the 2 smaller pentagons are from each other. Once I have that information, it's just a simple matter of finding the distance between opposing vertices of the docdecahedron.
