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

We notice that the conditions for this to be a regular polyhedron are all satisfied: congruent equilateral triangles for faces, and vertices all alike. Each vertex figure is a regular pentagon. The face angles are clearly all 60°.

At left is the Schlegel diagram for the icosahedron. Try counting the numbers of vertices, edges and faces. Are you satisfied this represents the icosahedron? Is it a planar graph? What is the importance of this?

All the usual measurements can be calculated by the enthusiastic: surface area, volume, inradius, midradius, circumradius, dihedral angle. Again you can look up the calculations for these here ...

One way of looking at the icosahedron is to see a pentagonal pyramid of triangular faces on top.
Beneath this lies a pentagonal antiprism. Then beneath this again lies an inverted pentagonal pyramid.
We can check that this gives 5 + 10 + 5 = 20 triangular faces, as expected.

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

Surface area
Dihedral angle

  Model making

Making a model of the icosahedron is a little fiddly with 20 faces, but it is not difficult to make. Go to this construction page for some face templates.

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

  Projections and sections

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

 As with the dodecahedron, the projections of the icosahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a {6}, a non-regular octagon, and a {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 {5}, a {10}.

Note where these shapes occur.

Now check your answers ...

  As with the dodecahedron, the projections of the icosahedron are less interesting than those of some of the other Platonic solids but you should be able to obtain a {6}, an octagon, and a {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 {5}, a {10}.

This left diagram below shows one of the projections, where we obtain {6} as a shadow figure. The view looks directly down through one of the triangular faces. The outline of the regular hexagon arises from a set of six linked edges of the icosahedron, three of which are shown in bold. The fact that these lie evenly between the top and bottom hexagonal faces shows that the resulting hexagonal projection is in fact regular. The figure at right shows the decagonal projection. We obtain this by looking down through one of the long diagonals of the icosahedron from one vertex to the opposite vertex. The shadow vertices are cast by the successive vertices of the antiprism forming the main body of the icosahedron.


The triangular section {3} is given by any of the faces of the icosahedron. We will have to be satisfied with an ‘improper’ section. The figure below left shows a regular pentagonal section, occurring as the base of one of the pentagonal pyramid caps forming the icosahedron. At right is shown the regular decagonal section, with vertices at midpoints of edges of the equilateral triangles bounding the pentagonal antiprism.

How do you think this decagon relates to the decagonal projection above? Are they the same?

Check your answer.


  Inscribed polyhedra

As we did with the dodecahedron, we now ask whether it is possible to inscribe a tetrahedron, a cube, or an octahedron nicely inside a dodecahedron.

(a) The simplest is the octahedron. How many vertices does the octahedron have? Can you find six nicely related edges of the icosahedron? Check your ideas from one of the above figures, or using the applet. Briefly describe your solution below.

(b) Next investigate the cube. Use the applet to find four points which will serve as vertices of a face. Possible suitable points might include icosahedral vertices, midpoints of edges, and centres of faces. Briefly describe your solution below. Are you certain it is a solution?!

(c) It should now be trivial to find a regular tetrahedron inscribed in the icosahedron. Why is this? Describe your solution below.

The diagrams below show that each of these inscribed polyhedra is possible. For the octahedron on the right, we observe that we can choose three parallel pairs of edges of the icosahedron which are mutually perpendicular. The midpoints of these six edges form the vertices of a regular inscribed octahedron.

[Are the three pairs of edges parallel and mutually perpendicular? Consider one of the pentagonal pyramids forming the icosahedron, and let e denote one of the edges from the vertex. The plane of symmetry of the pyramid containing e passes through the altitude of the opposite triangular face, cutting the base f of that triangle perpendicularly. Thus edges e and f are perpendicular. Our result follows by applying this argument to each of the three pairs of edges in turn.]

In the case of the cube, we choose as vertices eight triangular face centres of the icosahedron. But is the resulting figure a cube? Consider a triad of three adjoining triangular faces, with the centres of the outer two triangles located at the endpoints of a cube edge. By symmetry of the icosahedron, such triads will be congruent wherever they are placed on the icosahedron. It follows that the edge lengths of the ‘cube’ are all equal. Symmetry arguments also tell us that each ‘cube’ edge is parallel to the base of the central triangle of its triad. So the 12 ‘cube’ edges are parallel, in fours, to a set of three mutually perpendicular (and equal) edges of the icosahedron. Hence the ‘cube’ edges are not only equal but meet at right angles as required.

To inscribe a regular tetrahedron, we simply use our known result about inscribing a tetrahedron in a cube.


You might like to think about the number of different ways each of these solids can be inscribed in an icosahedron.

We defer consideration of inscribing a dodecahedron inside an icosahedron until later when we look at duality.

However, there is an easy way to place the icosahedron inside a cube, using those three pairs of equal, parallel and mutually orthogonal edges of the icosahedron:

  Golden rectangles

With the dodecahedron and the icosahedron having so many regular pentagons associated with them, it shouldn’t surprise us to find some golden rectangles occurring. We can show that the rectangles having as opposite edges those three pairs of equal, parallel and mutually orthogonal edges of the icosahedron, are in fact golden rectangles. Furthermore, the rectangles themselves are mutually perpendicular.

Can we easily show that these rectangles are golden rectangles? It looks as though we might have to tackle the dreaded dihedral angles! Another idea is to work backwards, starting with three such golden rectangles of dimensions 2 x 2. Choosing our axes in the obvious way, this would mean that the vertices of our enclosing icosahedron (the vertices of the golden rectangles) would lie at (, 1, 0), (0, , 1), ( 1, 0, ). Suitably joining up the rectangle vertices gives us an icosahedron, but is it regular? We need to check that the faces are equilateral triangles; if we can prove this, the symmetry of the configuration will establish our result. The short edges of the rectangles have length 2. Let us check the distance s between (, 1, 0) and (1, 0, ). We have

s2 = ( – 1)2 + (1 – 0)2 + (0 – )2 = 22 – 2 + 2 = 4   (since 2 – 1 = 0).

Thus s = 2. The other distances are established similarly. Hence the vertices of the rectangle coincide with the vertices of a regular icosahedron.

Finally, since each of these golden rectangles can be easily inscribed in a (diagonally placed) square, we find a nice way of placing an icosahedron inside an octahedron. Pick out the rectangles in the figure at right; notice how they are inscribed in the three determining squares of the octahedron. The vertices of the golden rectangles divide the edges of the octahedron in the golden ratio : 1.

  Vertex coordinates

From our above work with the three mutually orthogonal golden rectangles, we have obtained a nice symmetric set of vertex coordinates for the icosahedron:

(, 1, 0), (0, , 1), ( 1, 0, ).

The 12 vertices fall naturally into 3 sets of 4, and so relate to the 3 axes by 12 = 3 x 4.

  Real life occurrences

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

Here are some ideas.




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

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

Model making

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

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.

Projections and sections

Various: http://whistleralley.com/polyhedra/icosahedron.htm

Inscribed polyhedra

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

Various: http://whistleralley.com/polyhedra/icosahedron.htm

Golden rectangles

The golden section: Coxeter, H.S.M., Introduction to geometry, Wiley (2nd Edition 1969), page 160.

In the projection, the edges of the projection arise from the zig-zag edges of the antiprism. Hence the midpoints of the zig-zag edges map to the midpoints of the edges of the regular decagon projection. Thus the decagon section is congruent to the regular decagon with vertices at the midpoints of the decagon projection.