A cool property of the regular icosahedron. Prove it!?

Question

The regular icosahedron is a solid whose 20 faces are congruent, equilateral triangles. It has 12 vertices. You can divide the vertices into three groups of four vertices that each define a rectangle going through the middle of the solid. The three rectangles are perpendicular to each other, and their one common point is the center of the icosahedron.

Prove that the three rectangles are Golden rectangles (their length/width ratio is phi = 1.61803...).

Each rectangle has as its short sides two opposite edges of the icosahedron.

Answer 1

The coordinates of the vertices of the 3 rectangles, by symmetry, would be of the form, (0, +- 1, +-x), (+-1, +- x, 0), and (+-x, 0, +-1), where x is unknown. (0,1,x), (0,1,-x), and (1,x,0) forms one equilateral triangle. Therefore we know that 4x^2 = 1+(x-1)^2 +x^2 . This reduces to x^2 + x -1 = 0. Solving for x, we have x = 1/2(-1 + sqrt(5)). From this, we can arrive at the ratio of rectangle sides at 1/2(1 + sqrt(5)), which is the golden ratio.

Answer 2

Here's the simplest solution I can think of:
See picture: http://mudandmuck.com/str2/ico.JPG

The long side of each rectangle (blue) is a diagonal of a pentagon (red) of side length = edge length of the icosahedron.
Therefore if e = edge length of the icosahedron,
diagonal length = 2e*cos(π/5) = e(√5+1)/2 = eφ