** **

The regular dodecahedron is a Platonic solid having of **20
vertices**, 30 edges, and 12 faces.
Each face is a regular pentagon.

The dodecahedron is the dual of the icosahedron which has 12 vertices, 30 edges and 20 faces. Each face of an icosahedron is an equilateral triangle.

The dihedral angle is the angle of intersection of two planes. It is the angle whose vertex is on the intersecting edge and one side in each of the two planes. The sides of the angle are at right angles to the intersecting edge. For polyhedra, the dihedral angle is the angle of intersection between two adjacent faces. For all of the Platonic solids, there is only one dihedral angle. This is due to the fact that all of the adjacent faces intersect at the same angle.

A vertex of a dodecahedron is at the intersection of three faces. In this case there is a convenient formula for calculating the dihedral angle.

Let q be the dihedral angle of a dodecahedron and let a, b and g be the angles of each of the pentagonal faces that intersect to for the dihedral angle. The formula for the dihedral angle is:

**9.1)** _{}

** **

In a dodecahedron, all of the angles, a, b, g, are equal and are 108 degrees. This is due to the fact that each of the
faces is a regular pentagon and in a regular pentagon each angle is _{}, where n is 5.

Substituting this into the equation for the dihedral angle, q, gives the following:

_{} _{}

sin( 108 ) = 0.9511 and cos(108) = -.309

Substituting these values in the relation for the dihedral angle gives the following

_{} _{}

and so q =
116.57 degrees.

**The dihedral angle
for a regular dodecahedron is about 116.57 degrees.**

** **

To find a
closed-form value for cos(q) we start from **Equation** **9.1.**

** **

Let the Greek letter
a represent 108 degrees. Then **Equation 9.1** becomes the following:

_{}

This can be simplified by canceling out the factor _{}in the numerator and denominator to give the equation

_{}

But we know that cos(a) = _{}

Substituting this value in the equation above gives

_{}

so _{}

This can be simplified as follows

_{}

so

**9.2)** _{}

And so the dihedral angle is

q = _{} degrees.

The next task is to calculate the radius, **r**, of the inscribed sphere. This is a sphere that touches the solid at
the centroid of each face and at no other point. This is shown in **Figure 9.2**. (Reference 11 ). (All of
the solids discussed here are Platonic Solids and all have both inscribed and
circumscribed spheres.)

In **Figure 9.3**, point **O** is the center of both
the inscribed and circumscribed spheres.
Point **T** is the centroid of the ‘base’ pentagonal face of the
dodecahedron. Point **M** is the
midpoint of an edge of the ‘base’ pentagonal face.

The line **OM** bisects the angle between two faces,
which is the dihedral angle, and is hence equal to _{}. The angle **OTM**
is a right angle (90 degrees). The line
**OT** is the radius, **r**, of the inscribed sphere. Line **MT** is the radius of the **inscribed**
**circle** in the pentagonal face of the ‘base’ and is shown as **r’** in

**Figure 9.4**.

From the previous discussion of the regular pentagon, **Equation
5.7),** we know that

_{} where **a** is the
length of an edge of the pentagon.

From **Figure 9.4** it is clear that

_{}

and
so

_{}

A standard trigonometric identity for half-angles is:

_{}

_{} and so

_{}

And so we have

_{}

Substituting for r’ in this equation we have

**9.3)** _{}

Since we know that

_{}

we then have

_{}

The term _{} can be simplified as
follows:

And the term _{} can be simplified as
follows:

_{}

Substituting these values into the equation for r above leads to the following:

_{}

_{}

Multiplying out the factors inside the radical gives the following simplified result:

_{}

This simplifies to the final result

**9.4)** _{}

where **r** is the radius of the inscribed sphere and **a**
is the length of an edge of one of the pentagonal faces.

To find the radius, **R**, of the circumscribing sphere, consider **Figure 10.3** once
more. In that figure, the distance from
point **O** to point **N** is the radius of the circumscribing
sphere. From the figure we have the
following relation

_{}

And _{}

From **Figure 9.4** we can see that **OT** is **r**
and **TM** = **r'** . And from **Figure
9.3** we can see that **MN** is **a/2** or half the length of a
side. Putting all of this together, we
have

_{}

Substituting the values for **r** and **r’** in this
equation gives the following:

_{}

Some messy algebra will simplify this expression:

_{}

_{}

_{}

_{}

_{}

_{}_{}

_{}_{}

The quantity _{} can be written as _{} since

_{}

Hence we have that

_{}_{}

So finally we have that

**9.5)** _{}

where **R** is the radius of the circumscribed sphere and
**a** is the length of an edge of one of the pentagonal faces.

It is also useful to calculate **a** and **r** in terms
of **R**. To do this we proceed as
follows:

_{} or a = _{}

But since we know that

_{}

we have

_{}

_{}

So finally we have **a** in terms of **R **as

** **

**9.6) **_{}

** **

where **R** is the
radius of the circumscribed sphere and **a** is the length of any side of
any pentagonal face.

** **

To find **r**, the radius of the inscribed sphere, in
terms of **R**, the radius of the circumscribed sphere, we start with **Equation**
**9.4)**.

_{}

We know from **Equation 9.6)** that

_{}

Substituting this value for **a** into the above equation
for **r** we have

_{}

Continuing with the simplification we have:

_{}

_{}

_{}

_{}

_{}

Now move the _{} into the radical on
the right as 3 which finally gives

**9.7)** _{}

where **R** is the radius
of the circumscribed sphere and **r** is the radius of the inscribed sphere.

** **

A regular pentagon with all vertices and angles marked. All edges are of length 1 unit.

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