What is the surface area of an icosahedron as a function of its radius?

Warning: long answer...

Synopsis

• First, I will show that the corners of three intersecting golden rectangles with width #1# and height #varphi = 1/2(1+sqrt(5))# lie at the vertices of an icosahedron with edges of length #1#.

• Second, I will determine the outer radius of such an icosahedron, that is the distance from the centre to each vertex.

• Third, I will determine the inner radius of the same icosahedron, that is the distance from the centre to the centre of each face.

• Fourth, I will determine the surface area of an icosahedron with edges of length #1#.

• Fifth, I will use that to write down formulae for the surface area in terms of the inner and outer radii.

#color(white)()#
Proposition

The following #12# points (all combinations of #+-#) in #RR^3# form the vertices of a regular icosahedron with edges of length #1#:

#(+-1/2, +-1/2varphi, 0)#

#(0, +-1/2, +-1/2varphi)#

#(+-1/2varphi, 0, +-1/2)#

where #varphi = 1/2+sqrt(5)/2#

Each of the three sets of #4# points are the corners of a golden rectangle with width #1# and length #varphi#.

Proof

Note that #varphi# is the limit of the ratio between consecutive terms of the Fibonacci sequence, and satisfies:

#varphi^2 = varphi+1#

Due to the symmetrical specification, we just need to check that the distance between a corner of one of the golden rectangles and a selected corner of another is #1#.

The distance between #(x_1, y_1, z_1) = (1/2, 1/2varphi, 0)# and #(x_2, y_2, z_2) = (0, 1/2, 1/2varphi)# is given by the distance formula:

#d = sqrt((x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2)#

#color(white)(d) = sqrt((0-1/2)^2+(1/2-1/2varphi)^2+(1/2varphi-0)^2)#

#color(white)(d) = sqrt(1/4+1/4(1-varphi)^2+1/4varphi^2)#

#color(white)(d) = sqrt(1/4+1/4(2-varphi)+1/4(varphi+1))#

#color(white)(d) = sqrt(1)#

#color(white)(d) = 1#

#color(white)()#
Outer radius of an icosahedron

The outer radius of an icosahedron with edges of length #1# is the distance between #(0, 0, 0)# and #(1/2, 1/2varphi, 0)#, namely:

#sqrt((1/2)^2+(1/2varphi)^2+0^2) = sqrt(1/4+1/4varphi^2)#

#color(white)(sqrt((1/2)^2+(1/2varphi)^2+0^2)) = sqrt(1/4(2+varphi))#

#color(white)(sqrt((1/2)^2+(1/2varphi)^2+0^2)) = 1/2sqrt(2+varphi)#

#color(white)()#
Inner radius of an icosahedron

One of the faces of the icoshedron described above has corners:

#(1/2, 1/2varphi, 0)#, #(0, 1/2, 1/2varphi)#, #(1/2varphi, 0, 1/2)#

The centre of this face is therefore located at:

#(1/6(1+varphi), 1/6(1+varphi), 1/6(1+varphi))#

The distance of this point from the origin is:

#sqrt(3(1/6(1+varphi))^2) = sqrt(1/12(1+varphi)^2) = 1/2sqrt(2/3+varphi)#

This is the inner radius of an icosahedron with edges of length #1#.

#color(white)()#
Area of an equilateral triangle with unit sides

An equilateral triangle with sides of length #1# has height #sqrt(3)/2# and therefore area #sqrt(3)/4#

#color(white)()#
Surface area of icosahedron

The surface area of our icosahedron with edges of length #1# is:

#20*sqrt(3)/4 = 5sqrt(3)#

If we had an icosahedron of inner radius #r#, then its surface area would be:

#(5sqrt(3))/(1/2sqrt(2/3+varphi))^2r^2 = (20sqrt(3))/(2/3+varphi)r^2#

If we had an icosahedron of outer radius #R#, then its surface area would be:

#(5sqrt(3))/(1/2sqrt(2+varphi))^2R^2 = (20sqrt(3))/(2+varphi)R^2#