DODECAHEDRON {5, 3}

  Simple dodecahedron properties
The name dodecahedron comes from the Greek dodeca + hedron meaning twelve bases (faces). The basic properties of the regular tetrahedron are as listed below.
Name   Symbol      V       E       F       V – E + F   
   Dodecahedron    {5, 3} 20 30 12 2

We notice that the conditions for this to be a regular polyhedron are all satisfied: congruent regular pentagons for faces, and vertices all alike. Each vertex figure is again an equilateral triangle. The face angles are clearly all 108°.

At right is the Schlegel diagram for the dodecahedron. Does it have the right number of vertices, edges, and faces? What does it tell us about this polyhedron?

All the usual measurements can be calculated by the enthusiastic: surface area, volume, inradius, midradius, circumradius, dihedral angle. There seems to be no great merit in doing this; why not just read how someone else did it?!  Here ...


For the record, here is a summary of some of the metric attributes of a regular dodecahedron of side length s. The values indicate some hard calculations!

Volume
Surface area
Dihedral angle
Inradius
Midradius
Circumradius


  Model making

A model of the dodecahedron can be an attractive addition to our set, and is easy to make. Go to this construction page for some face templates.

A net for the regular dodecahedron is given at right. It is known that for the dodecahedron (and for the icosahedron) there are 43380 distinct nets!

  Further properties

The dodecahedron has a number of interesting properties. We shall use our applet to investigate some of these.

•  The projections of the dodecahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a (non-regular) hexagon, an octagon, and a regular decagon {10}.

The cross-sections make us work rather harder. Working from a model (if you have one) is probably easier than working from the applet.

• You can try to find the sequence of cross-sections as a plane moves through the dodecahedron cutting it from vertex to vertex, edge to edge, and face to face.

Alternatively, see if you can find a cross section which is a {3}, a {4}, a {5}, a {6} (there are two distinct types of these) and a {10}.

Note where these shapes occur.




Now check your answers ...




























 The projections of the dodecahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a (non-regular) hexagon, an octagon, and a regular decagon {10}.

• You can try to find the sequence of cross-sections as a plane moves through the dodecahedron cutting it from vertex to vertex, edge to edge, and face to face.

Alternatively, see if you can find a cross section which is a {3}, a {4}, a {5}, a {6} (there are two distinct types of these) and a {10}.

This first diagram shows one of the projections, where we obtain {10} as a shadow figure.

Then below are diagrams showing the regular cross-sections. It is perhaps surprising that we can obtain so many. The blue pentagonal section is of course no surprise, as each face of the dodecahedron is a regular pentagon. We can obtain a ‘proper} section by taking a parallel plane which cuts the solid. The magenta decagonal section is obtained by cutting the solid suitably through its centre.

Is the green section actually square? The edge lengths are all equal (diagonals of congruent faces), so it must be at least a rhombus. Further, since the faces are congruent and regular, the two diagonals of the faces with a common edge are both ‘centred’ with respect to that edge. So the green section is a square.

The factor of 3 in the triangular and hexagonal sections seems a little out of place, except that we have three faces meeting at each vertex. So the equilateral triangle section is just a vertex figure of the solid. Notice that the two regular hexagonal sections are different in nature: the orange section has vertices at edge midpoints, but the purple section has to have its vertices at carefully chosen points on the edges of the dodecahedron. My calculation shows that such points have to be at a ratio of 5 – 1 : 5 + 1 along an edge.

                         

  Inscribed polyhedra

You might remember how we inscribe a tetrahedron inside a cube. We could now ask whether it is possible to inscribe a tetrahedron, a cube, or an octahedron nicely inside a dodecahedron.

• (a) Let us begin with the cube. (If we can show that it is possible for the cube, the tetrahedron result will follow.) Look at the green square above. Will this serve as a face of an inscribed cube? Go back to the applet in outline form to explore this possibility.

(b) Now think about an inscribed octahedron. Here we have six vertices to place. As a hint (take it!), think about placing them at the midpoints of some of the dodecahedral edges.

(c) Finally, can you see how to inscribe a dodecahedron within a cube?

Below are diagrams showing that each of these inscribed polyhedra exists. You might like to ask how many of each type of polyhedra can be inscribed in a given regular dodecahedron. We shall come back to this later.

          


In the case of the dodecahedron inscribed in the cube, we notice that the dodecahedron has three pairs of mutually orthogonal edges. These edges can be placed in the faces of an encompassing cube, as illustrated below.

  Vertex coordinates

To obtain coordinates for the vertices of a dodecahedron, we have to work rather harder.

First of all, we know from above that we can inscribe cubes inside a regular dodecahedron, so this suggests taking the coordinates (1, 1, 1) for eight of them (the green cube in the figure at right). This places the origin at the centre of the cube. The remaining 12 vertices of the dodecahedron occur as the ends of three pairs of congruent and parallel line segments which lie in the coordinate planes. Using the symmetry of the dodecahedron, we may thus assume that the coordinates of these points take the form (0, a, ka), (ka, 0, a), and (a, ka, 0).

We note that the green cube has edge length 2, and that PQ and QR are edges of a congruent cube. Hence

                                 PQ2 = (0 – ka)2 + (–a)2 + (ka – a)2 = 4,
whence
                                                                  a2(k2 – k + 1) = 2.        (*)

Also, P, Q and R form the vertices of a right angled triangle, so by Pythagoras’ theorem,

PR2  =  PQ2 + QR2,    or

                                                             a2+ (k + 1)2a2 + k2a2   =   2[k2a2 + a2 + (k – 1)2a2],
whence      
                                                                               k2 – 3k + 1 = 0. (**)

Recall that 2 = + 1, and = (1 + 5)/2.   

Solving the above quadratic equation (**), we find that k = (3 + 5)/2  =  1 +  = 2  is a solution.

Substituting back in (*) gives a2  = 1/(1 + ) = 1/2. Thus a = 1/. Also ka = .

It follows that we can take as coordinates for the 20 vertices of the dodecahedron:

(1, 1, 1),   (0, –1, ),  (, 0, –1),  and  (–1, , 0).

The relationship between the numbers 20 (vertices) and 3 (coordinates) this time is  20  =  23 + 3 x 22.

  Real life occurrences

•  Can you think of any real life occurrences of the regular tetrahedron? Make a list here:

Here are some ideas.

                

  References

Properties

MathWorld : http://mathworld.wolfram.com/Dodecahedron.html

Wikipedia : http://en.wikipedia.org/wiki/Dodecahedron

Model making

An older classic: Cundy, H. M., Rollett, A. P., Mathematical Models, Oxford (2nd Edition 1961)

The excellent model book: Wenninger, M. J., Polyhedron Models, Cambridge (1971).

Number of nets:

Bouzette, S., Vandamme, F., ‘The regular dodecahedron and icosahedron unfold in 43380 ways’, Unpublished manuscript.

Buekenhout, F., Parker, M., ‘The number of nets of the regular convex polytopes in dimension 4’, Discrete Mathematics 186 (1998) 69 – 94.

Inscribed polyhedra

Animated cube in a dodecahedron : http://www.math.umn.edu/~roberts/java.dir/JGV/cube-in-dodeca.html