5.6: Regular and semi-regular polyhedra

We have seen how regular polygons sometimes fit together edge-to-edge in the plane to create tilings of the whole plane. When tiling the plane, the angles of polygons meeting edge-to-edge around each vertex must add to 360°, or two straight angles. If the angles at a vertex add to less than 360°, then we are left with an empty gap and two free edges; and when these two free edges are joined, or glued together, the vertex figure rises out of the plane and becomes a 3-dimensional corner, or solid angle.

To form such a corner we need at least three polygons, or faces - and hence at least three edges and three faces meet around each vertex. For example, three squares fit nicely together in the plane, but leave a 90° gap. When the two spare edges are glued together, the result is to form a corner of a cube, where we have a vertex figure consisting of three regular 4-gons: so we refer to this vertex figure as 4³.

Given a 3-dimensional corner, it may be possible to extend the construction, repeating the same vertex figure at every vertex. The resulting shape may then ‘close up’ to form a convex polyhedron. The assumption that in each vertex figure, the angles meeting at that vertex add to less than 360°, means that all the corners then project outwards - which is roughly what we mean when we say that the polyhedron is “convex”.

A regular polygon is an arrangement of finitely many congruent line segments, with two line segments meeting at each vertex (and never crossing, or meeting internally), and with all vertices alike; a regular polygon can be inscribed in a circle (Problem 36), and so encloses a convex subset of the plane. In the same spirit, a regular polyhedron is an arrangement of finitely many congruent regular polygons, with two polygons meeting at each edge, and with the same number of polygons in a single cycle around every vertex, enclosing a convex subset of 3-dimensional space (i.e. the polyhedron separates the remaining points of 3D into those that lie ‘inside’ and those that lie ‘outside’, and the line segment joining any two points of the polyhedral surface contains no points lying outside the polyhedron).

The important constraints here are the assumptions: that the polygons meet edge-to-edge with exactly two polygons meeting at each edge; that the same number of polygons meet around every vertex; and that the overall number of polygons, or faces, is finite. The assumption that the figure is convex should be seen as a temporary additional constraint, which means that the angles in polygons meeting at each vertex have sum less than 360°.

Problem 186 A vertex figure is to be formed by fitting regular p-gons together, edge-to-edge, for a fixed p. If there are q of these p-gons at a vertex, we denote the vertex figure by p^q. If the angles at each vertex add to less than 360°, prove that the only possible vertex figures are 3³, 3⁴, 3⁵, 4³, 5³.

The vertex figure 4³ is realized by the way the positive axes meet at the vertex (0,0,0), where

• the unit square (0, 0,0), (1, 0,0), (1,1,0), (0,1,0) in the xy-plane (with equation z = 0) meets
• the unit square (0,0,0), (1,0,0), (1,0,1), (0,0,1) in the xz-plane (with equation y = 0), and
• the unit square (0, 0, 0), (0,1,0), (0,1,1), (0, 0,1) in the yz-plane (with equation x = 0).

If we include an eighth vertex (1, 1, 1), and

• the unit square (0, 0, 1), (1, 0, 1), (1, 1, 1), (0, 1, 1) in the plane with equation z = 1,
• the unit square (0,1,0), (1,1,0), (1,1,1), (0,1,1) in the plane with equation y = 1,
• the unit square (1,0,0), (1,1,0), (1,1,1), (1,0,1) in the plane with equation x = 1,

we see that all eight vertices have the same vertex figure 4³. Hence the possible vertex figure 4³ in Problem 186 arises as the vertex figure of a regular polyhedron - namely the cube.

If we select the four vertices whose coordinates have odd sum A = (1, 0, 0), B = (0,1, 0), C = (0, 0,1), D = (1,1,1), then the distance between any two of these vertices is equal to $\sqrt{2}$, so each triple of vertices (such as (1, 0, 0), (0,1,0), (0, 0,1)) defines a regular 3-gon ABC, with three such 3-gons meeting at each vertex of ABCD. Hence the possible vertex figure 3³ in Problem 186 arises as the vertex figure of a regular polyhedron - namely the regular tetrahedron (tetra = four; hedra = faces).

Problem 187 With A = (1,0,0) etc. as above, write down the coordinates of the six midpoints of the edges of the regular tetrahedron ABCD (or equivalently, the six centres of the faces of the original cube). Each edge of the regular tetrahedron meets four other edges of the regular tetrahedron (e.g. AB meets AC and AD at one end, and BC and BD at the other end). Choose an edge AB and its midpoint P. Calculate the distance from P to the midpoints Q, R, S, T of the four edges which AB meets (namely the midpoints of AB, AD, BD, BC respectively). Confirm that the triangles ΔPQR, ΔPRS, ΔPST, ΔPTQ are all regular 3-gons, and that the vertex figure at P is of type 3⁴ . Conclude that the possible vertex figure 3⁴ in Problem 186 arises as the vertex figure of a regular polyhedron PQRSTU - namely the regular octahedron (octa = eight; hedra = faces).

Problem 188

(a) A regular tetrahedron ABCD has edges of length 2, and sits with its base BCD on the table. Find the height of A above the base.

(b) A regular octahedron ABCDEF has four triangles meeting at each vertex.

(i) Let the four triangles which meet at A be ABC, ACD, ADE, AEB. Prove that BCDE must be a square.

(ii) Suppose that all the triangles have edges of length 2, and that the octahedron sits with one face BCF on the table - next to the regular tetrahedron from part (a). Which of these two solids is the taller?

Problem 189 Let O = (0, 0, 0), A = (1,0,0), B = (0,1,0), C = (0,0,1) be four vertices of the cube as described after Problem 186 above. Draw equal and parallel line segments (initially of unknown length 1 — 2a) through the centres of each pair of opposite faces - running in the three directions parallel to OA, or to OB, or to OC

• from $N=\left(a,\frac{1}{2},0\right)$ to $P=\left(1-a,\frac{1}{2},0\right)$ and from $Q=\left(a,\frac{1}{2},1\right)$ to $R=\left(1-a,\frac{1}{2},1\right)$
• from $S=\left(\frac{1}{2},0,a\right)$ to $T=\left(\frac{1}{2},0,1-a\right)$ and from $U=\left(\frac{1}{2},1,a\right)$ to $V=\left(\frac{1}{2},1,1-a\right)$
• from $W=\left(0,a,\frac{1}{2}\right)$ to $X=\left(0,1-a,\frac{1}{2}\right)$ and from $Y=\left(1,a,\frac{1}{2}\right)$ to $Z=\left(1,1-a,\frac{1}{2}\right)$ .

Figure 4: Construction of the regular icosahedron.

These are to form all 12 vertices and six of the 30 edges (of length 1 — 2a) of a polyhedron, see Figure 4. The other 24 edges join each of these 12 vertices to its four natural neighbours on adjacent faces of the cube - to form the 20 triangular faces of the polyhedron: for example,

N joins: to S; to W; to X; and to U.

(i) Prove that NS = NW = NX = NU and calculate the length of NS.

(ii) Choose the value of the parameter a to guarantee that NP = NS, so that the five triangular faces meeting at the vertex N are all equilateral triangles, and each vertex figure of the resulting polyhedron then has vertex figure 3⁵.

The polyhedron is called the regular icosahedron (icosa = twenty, hedra = faces).

In the paragraph before Problem 187 we constructed the dual of the cube by marking the circumcentre of each of the six square faces of the cube, and then joining each circumcentre to its four natural neighbours. We now construct the dual of the regular icosahedron in exactly the same way. Each of the 20 circumcentres of the 20 triangular faces of a regular icosahedron has three natural neighbours (namely the circumcentres of the three neighbouring triangular faces). If we construct the 30 edges joining these 20 circumcentres, the five circumcentres of the five triangles in each vertex figure of the regular icosahedron form a regular pentagon, which becomes a face of the dual polyhedron - so we get 12 regular pentagons (one for each vertex of the regular icosahedron), with three pentagons meeting at each vertex of the dual polyhedron to give a vertex figure 5³ at each of the 20 vertices, which form a regular dodecahedron.

Hence each of the five possible vertex figures in Problem 186 can be realised by a regular polyhedron. These are sometimes called the Platonic solids because Plato (c. 428-347 BC) often used them as illustrative examples in his writings on philosophy.

Constructing the five regular polyhedra is part of the essence of mathematics for everyone. In contrast, what comes next (in Problem 190) may be viewed as “optional” at this stage. The ideas are worth noting, but the details may be best postponed for a rainy day.

Just as you classified semi-regular tilings in Section 5.4, so one can look for semi-regular polyhedra. A polyhedron is semi-regular if all of its faces are regular polygons (possibly with differing numbers of edges), fitting together edge-to-edge, with exactly the same ring of polygons around each vertex - the vertex figure of the polyhedron. Problem 190 uses “the method of analysis” - combining simple arithmetic, inequalities, and a little geometric insight - to achieve a remarkable complete classification of semi-regular polyhedra. There are usually said to be thirteen individual semi-regular polyhedra (excluding the five regular polyhedra); but one of these has a vertex figure that extends to a polyhedron in two different ways - each being the reflection of the other. There are in addition two infinite families - namely

• the n-gonal prisms, which consist of two parallel regular n-gons, with the top one positioned exactly above the bottom one, the two being joined by a belt of n squares (so with vertex figure n . 4²); and
• the n-gonal antiprisms, which consist of two parallel regular n-gons, but with the top n-gon turned through an angle of n radians relative to the bottom one, the two being joined by a belt of 2n equilateral triangles (so with a vertex figure n . 3³).

Notice that the cube can also be interpreted as being a “4-gonal prism”, and the regular octahedron can be interpreted as being a “3-gonal antiprism”. Those interested in regular and semi-regular polyhedra are referred to the classic book Mathematical models by H.M. Cundy and A.P. Rollett.

Problem 190 Find possible combinations of three or more regular polygons whose angles add to less than 360°, and hence derive a complete list of possible vertex figures for a (convex) semi-regular polyhedron. Try to eliminate those putative vertex figures that cannot be extended to a semi-regular polyhedron.