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:
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
This can be simplified as follows
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
From the previous discussion of the regular pentagon, Equation 5.7), we know that
From Figure 9.4 it is clear that
A standard trigonometric identity for half-angles is:
And so we have
Substituting for r’ in this equation we have
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
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
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
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
So finally we have a in terms of R as
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
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.