Question

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

Answer 1

`(20sqrt(3))/(2/3+varphi)r^2" "` where `r` is the inner radius

`(20sqrt(3))/(2+varphi)R^2" "` where `R` is the outer radius

Explanation:

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`