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

_{}

_{}

And so _{}

Hence q = 138.19 degrees.

The circle inscribed within one of the equilateral triangle
faces has the radius, from **Equation** **3.1)**

r’ = _{}

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, **r’**, 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
= r’_{}

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

**10.1)** _{}

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:

_{} or _{}

From
here, we just substitute known values for **r** and **r’**
and then simplify the result. We know
from **Equation 10.1)** that

_{}

We know from **Equation 3.1)** that

r’ = _{}

So
the equation for **R** becomes

_{}

_{}

_{}

And so finally we
have

**10.2)** _{}

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!

**10.3)** _{}

_{}

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:

_{}

_{}

And so

**10.4)** _{}

** **

** **

**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

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