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.
Proposition
The following
#(+1/2, +1/2varphi, 0)#
#(0, +1/2, +1/2varphi)#
#(+1/2varphi, 0, +1/2)#
where
Each of the three sets of
Proof
Note that
#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
The distance between
#d = sqrt((x_2x_1)^2+(y_2y_1)^2+(z_2z_1)^2)#
#color(white)(d) = sqrt((01/2)^2+(1/21/2varphi)^2+(1/2varphi0)^2)#
#color(white)(d) = sqrt(1/4+1/4(1varphi)^2+1/4varphi^2)#
#color(white)(d) = sqrt(1/4+1/4(2varphi)+1/4(varphi+1))#
#color(white)(d) = sqrt(1)#
#color(white)(d) = 1#
Outer radius of an icosahedron
The outer radius of an icosahedron with edges of length
#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)#
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
Area of an equilateral triangle with unit sides
An equilateral triangle with sides of length
Surface area of icosahedron
The surface area of our icosahedron with edges of length
#20*sqrt(3)/4 = 5sqrt(3)#
If we had an icosahedron of inner radius
#(5sqrt(3))/(1/2sqrt(2/3+varphi))^2r^2 = (20sqrt(3))/(2/3+varphi)r^2#
If we had an icosahedron of outer radius
#(5sqrt(3))/(1/2sqrt(2+varphi))^2R^2 = (20sqrt(3))/(2+varphi)R^2#
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The surface area (A) of an icosahedron as a function of its radius (r) can be expressed as:
[ A(r) = 5 \sqrt{3} \cdot r^2 ]
Where (r) is the radius of the circumscribed sphere around the icosahedron.
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When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
When evaluating a onesided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.
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