# Proof of golden rectangle inside an Icosahedron

This is an regular Icosahedron with three golden rectangles inscribed in it. I realize that they are clearly rectangles, I was wondering if someone could prove that they are a golden rectangles, assuming all edges are length $s$.

(I know that it's true from a result and from google but I'm not sure how to show it.)

I am completely fine with a proof from its dual, the dodecahedron.

I believe the longer sides of the rectangles coincide with the diagonals of the regular planar pentagons made up of the edges of the icosahedron, thus the ratio requested here is the ratio of the diagonal in a regular pentagon to its side; this is well known to be the golden ratio.

Assume $s=1$.

The generalized diamete is the max distance between vertices. Consider two of them, opposite wrt the origin $A\left(0,0,\dfrac{5}{\sqrt{50-10 \sqrt{5}}}\right)$ and $B\left(0,0,-\dfrac{5}{\sqrt{50-10 \sqrt{5}}}\right)$

$AB=\sqrt{\dfrac{1}{2} \left(5+\sqrt{5}\right)}$

The side of the golden rectangle is then

$\sqrt{AB^2-1}=\sqrt{\dfrac{3}{2}+\dfrac{\sqrt{5}}{2}}=\dfrac{1}{2} \left(1+\sqrt{5}\right)$

QED

Hope this helps