The icosahedron is a Platonic solid having 12 vertices,† 20 faces and 30 edges.† Each face is an equilateral triangle.† So each face has equal angles of 60 degrees and sides of equal length.† The icosahedron is the dual of the dodecahedron which has 20 vertices and 12 pentagonal faces.
Consider the five faces that meet at a single vertex. This forms a pyramid with a regular pentagonal base. At any vertex on the base, three surfaces meet. Two are equilateral triangles and the third is a regular pentagon. The dihedral angle between the two triangles is the same as that between any other two faces of the icosahedron.
The formula for the dihedral angle when there are three planes involved is, as we saw in the discussion of the dodecahedron,
This formula can be applied to the pyramid described above with a = b = 60 degrees and g = 108 degrees.
cos(60) = 1/2†† ,† sin(60) = †††††††† and† cos(108) =
So the formula becomes
Hence q †= 138.19 degrees.
The circle inscribed within one of the equilateral triangle faces has the radius, from Equation 3.1)
††††††††††† rí =
To calculate the radius, r, of the inscribed sphere, consider Figure 10.1
In this figure, two adjoining faces are shaded and the point O is at the center of both the inscribed and circumscribed spheres.† Point T is on the surface of one of the triangular faces and is at the centroid of that face.† The distance OT is then the desired radius r.† Point M is at the midpoint of the common edge of the two adjoining faces and the distance MT is the radius, rí, of the circle inscribed in either of the two faces.† Figure 11.2 shows this triangle in a more convenient view.† It is clear that the angle OTM is a right angle.† It is also clear that the angle OMT bisects the dihedral angle q . †So from Figure 10.2, we have that
††††††††††††† or† r† =† rí
Proceeding as in the discussion for the dodecahedron, we use the half-angle formulas to arrive at the equation
Further simplifications leads to the following:
So finally we have
where a is the length of an edge and r is the radius of the inscribed sphere.
To find the value of the radius, R, of the circumscribed sphere we proceed exactly as we did in the case of the dodecahedron.† The distance from the center point, O, to the point at the vertex, N, is the distance required.† In Figure 11.1, the distance OM is given by the Pythagorean theorem to be
To get from point M to point N, the distance MN is half of one side or .† So the distance ON, which is R, is given by the Pythagorean theorem to be
So we have this formula for R:
From here, we just substitute known values for r and rí and then simplify the result.† We know from Equation 10.1) that
We know from Equation 3.1)† that
So the equation for R becomes
And so finally we have
To find the length of an edge, a, and the radius of the inscribed sphere, r, in terms of the radius of the circumscribed sphere, R, we just do some rearranging in the Equations 10.1 and 10.2.
From Equation 10.2 we have that
And so we have it!
To find r in terms of R we start with Equation 10.1.
From Equation 10.1† we have that
From Equation 10.3† we have that
So we have, by substituting the value of a:
Icosahedron with two outside adjacent faces shaded.
Triangle through the center of the icosahedron with its base in the plane that bisects the dihedral angle between the two shaded sides in Figure 11.1