Proof of golden rectangle inside an Icosahedron

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