Let's say we have ourselves a regular Icosahedron, constructed entirely out of equilateral triangles:
If you were to "flatten" out one vertex star-point of that Icosahedron, you'd have a star with five lines coming off of it, like so:
To successfully flatten this Icosahedron piece, two concessions must be made, the angle between the lines increase from 60° to 72°, and the outermost line of each equilateral triangle must "grow" from 100 to ~117.557.
I can calculate all of the pieces of that triangle fine, and here:
What I need to figure out is at what angle do I need to raise each line so that the inner angle returns to 60°. To understand what I mean, let's look at a unit-sphere in reference to two of the lines from our Icosahedron: [![enter image description here][5]][5] The drawing on the left shows that at 90° (completely horizontal) we're at the 72° angle, but as we raise them up (equally, together) they will reach 60°. On the left we are at the "maximum" of the orange slices width. As we move up (or down) that will decrease, and at some point it will return to 60°, which is the angle we need for the Icosahedron to return to its proper shape.
As I understand it, as the two lines reach 90° offset from the horizontal slice, they near (and then reach) 0. This is some function of SIN(), since that goes from 1 to 0 in this direction. I have a pretty good understanding of how to break all of these problems down when they are separated into their 2-dimensional components, but I don't know how to derive an equation that relates the various perspectives together. In what way does the upward angle of each line affect the inner-angle between the lines?
Update: Okay, I found a way to find the answer, but it's via "unconventional" means. I used Fusion 360 to produce the orange slice with the parameters that I have been using. I then sketched a 100 unit rectangle angled and above the slice such that when I used it to cut away the slice, it would reveal two "fang" points at which were the incident of intersection where the distance between the two points was the exact length of the rectangle. I then measured the angle between the point and the horizontal:
So, the direct answer to the question is 31.717 deg... but I still don't know how to solve this without using Fusion 360 and hacking away at geometry.
Suppose $r$ is the outer radius of a regular pentagon with edge length 1. (It is easiest to keep the keep the outer edge lengths constant to match the edges of the icosahedron.) Then it is not too hard to see that $\sin(36) = \frac{1/2}{r}$, so $r = \frac1{2\sin36}$.
Now looking at the the pentagonal pyramid, you have a vertical right triangle with $r$ as the base, and hypotenuse $1$ as this is another edge of the icosahedron. The angle you are looking for has $\cos\alpha= \frac r1$. So $\alpha = \cos^{-1}(\frac1{2\sin36}) \approx 31.7174$.