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1 the Platonic Solids

1 the Platonic Solids

Day! Platonic Solids

1 The Platonic Solids

We take the celebration of Dodecahedron Day as an opportunity embark on a discussion of perhaps the best-known and most celebrated of all polyhedra – the Platonic solids. Before doing so, however, a word about convexity is in order.

A or is said to be convex if, informally, it has no “dents” (see Figure 1). More formally, convexity is described by the property that for any two points in the polygon (or polyhedron), the segment joining these two points lies wholly within the polygon (or polyhedron). The dashed lines in Figure 1 illustrate the nonconvexity of the corresponding – parts of the dashed lines lie outside the figures.

convex polygons nonconvex polygons

Figure 1

An interesting relationship commonly known as “Euler’s formula” is valid for all convex polyhedra – if V denotes the number of vertices of a polyhedron, E the number of edges, and F the number of faces, then V − E + F = 2. (1) Take, for example, a , which has eight vertices (or corners), twelve edges, and six faces. Then we see that V − E + F = 8 − 12 + 6 = 2.

The stage is now set for a discussion of the Platonic solids. A is a polyhedron with the following properties:

(P1) It is convex.

(P2) Its faces are all the same .

(P3) The same number of polygons meet at each of its vertices.

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Note that since a Platonic solid is convex, the polygons referred to in (P2) must also be convex.

Which polyhedra satisfy properties (P1)–(P3)? We provide two different approaches to an- swering this question.

2 A Geometric Enumeration

We take an incremental approach here, and begin by asking, “Which Platonic solids have equilateral triangular faces?” Now the fewest number of which may meet at a is three. It is a simple to construct a polyhedron with three triangles meeting at each vertex – just take a triangular . Four triangles are required (see Figure 2(a)), and hence this polyhedron is called a .

What about four triangles meeting at a vertex? We know from our Egyptian studies that four triangles meet at the apex of a , while two triangles meet at each vertex of the square base. This implies that if we take two square pyramids and join them base- to-base, the “disappear,” leaving 2 + 2 = 4 triangles at each vertex of the interior squares. The result is a convex polyhedron with four equilateral triangles meeting at each vertex. Since two pyramids were used, 2 × 4 = 8 triangles are required (see Figure 2(b)), and hence this polyhedron is called an .

(a) (b)

(c)

Figure 2

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Five triangles at a vertex is a bit trickier. All in all, twenty triangles are required (see Figure 2(c)), and hence this polyhedron is called an (“icosi” is the Greek prefix for “20”). The best way to see how these triangles fit together is to build an icosahedron yourself. (In the final analysis, there is no substitute for hands-on construction.) Should this be momentarily inconvenient, an alternative description follows.

Suppose we try at first to extend our base-to-base square pyramid construction of an octahe- dron to a base-to-base construction of a polyhedron with five triangles meeting at each vertex. Of course five triangles meet at the apex (and the vertex opposite) but, as with the square pyramid, only four triangles meet at each vertex of the “disap- pearing” pentagonal bases (see Figure 3(a)). Because of its construction, the polyhedron in Figure 3(a) is called a pentagonal .

Happily, this situation may be remedied. Consider for a moment the pentagonal “toy drum” of Figure 3(b). The top and bottom are out of phase by 36◦ (see Figure 3(c) for a top view of Figure 3(b)), and the intervening space is filled by a zig-zag of ten equi- lateral triangles. The salient anatomical feature of this “drum” (also called a pentagonal ) is that exactly three triangles (and one ) meet at each vertex. As a result, appending a pentagonal pyramid to both the top and bottom of this antiprism yields a polyhedron with precisely five triangles meeting at each of its twelve vertices (see Figure 4). The contributes ten equilateral triangles, and each of the pentagonal pyramids contributes five, for a total of twenty triangles – hence the name “icosahedron.”

(a)

(b)

(c)

Figure 3

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Our search for Platonic solids with triangular faces ends here, for one is easily convinced that the angles of six equilateral triangles comprise 360◦, and hence any vertex with six equilateral triangles would be “flat.” This, of course, does not result in a convex polyhedron, but rather a tiling of the by equilateral triangles.

(a) (b)

Figure 4

So now we have enumerated all possible Platonic solids with equilateral triangles as faces. What about the next regular polygon, the square? Three squares at a vertex results in our old friend the cube (sometimes called a ), and at four squares we are already flat.

What about regular pentagons? As it happens, there is a Platonic solid with three pentagons at each vertex. As with the icosahedron, the best way to understand this polyhedron is to build it; barring that, we press on...

(a) (b) (6)

Figure 5

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Perhaps the best way to imagine such a solid is to begin with an arrangement of six pentagons as shown in Figure 5(a). Five of these pentagons may be folded up to yield a bowl-like shape; as it happens, two such bowls fit exactly together. The result, as it requires precisely twelve pentagons, is called a dodecahedron (or sometimes a pentagonal dodecahedron).

Now each angle of a regular pentagon had measure 108◦. Hence four such angles have measure 432◦ > 360◦, and hence it is impossible to fashion a vertex of a convex polyhedron with four (or more) pentagons at a vertex.

So on to regular . With three hexagons at each vertex we are already flat, yielding a hexagonal tiling of the plane. As a result, there are no Platonic solids with regular hexagonal faces.

Our search stops here. Since three hexagons result in a flat vertex, three regular polygons with more than six sides, if they met at a vertex, would comprise more than 360◦. So as with the case of four pentagons, no convex polyhedron may be formed.

3 An Algebraic Enumeration

The foregoing is not the only approach to an enumeration of the Platonic solids. We now embark on an algebraic description, making use of Euler’s formula (see (1)). Our attack is to translate (P1)–(P3) into algebraic analogues. To begin, suppose that we have a Platonic solid before us, with the number of vertices, edges, and faces being denoted by V , E, and F , respectively. We see from (P2) that all faces of this solid are the same regular polygon. Now suppose that this polygon has p sides.

Then pF is simply the total number of sides on all faces of the Platonic solid. Because each of the solid is the meeting place of exactly two faces (and hence two of the sides among the pF ), we find that our count includes every edge of the solid exactly twice. Thus, we have

0 (P2) pF = 2E.

(If you have trouble visualizing this, take hold of the nearest Platonic solid and work through the previous paragraph.)

From (P3), we see that the solid has the same number of polygons meeting at each vertex; this number shall be denoted by q. Then there must also be exactly q edges meeting at each vertex as well, so that qV counts the total number of edges incident at all vertices of the polyhedron. But each edge is incident at exactly two vertices, so that qV counts each edge of our solid exactly twice. Thus, we have

0 (P3) qV = 2E.

Finally, because our Platonic solid is convex, we replace (P1) with Euler’s formula:

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0 (P1) V − E + F = 2.

0 0 0 We can now solve for V and F from (P3) and (P2) and substitute into (P1), yielding 2E 2E − E + = 2. q p A little algebra yields 1 1 1 1 + = + , (2) p q 2 E which must be valid for any Platonic solid.

Now p and q are integers 3 or greater, and E is a positive integer. So if p ≥ 4 and q ≥ 4, we would have 1/p + 1/q ≤ 1/2, making (2) impossible. As a consequence, we must have p = 3 or q = 3 (or possibly both).

Assume for the moment that p = 3. Then (2) becomes 1 1 1 = + . (3) q 6 E

Since E is a positive integer, this means that 1/q > 1/6, or equivalently, q < 6. Since at least three polygons must meet at the vertex of a polyhedron, the only possibilities are q = 3, q = 4, or q = 5. With each choice of q, E may be determined from (3). V may then be 0 0 determined from (P3), and F from (P2).

Since (2) is symmetric in the occurrence of “p” and “q” (each playing precisely the same role), the reader is invited to make an analogous argument with the assumption that q = 3. When this is done, Table 1 below is obtained, which includes all possibilities for p and q as described in the previous few paragraphs.

An alternate way to obtain Table 1 is to begin again with (2), and observe that since E > 0, we may write 1 1 1 + > . p q 2 Multiplying through by 2pq gives 2q + 2p > pq, which may be rearranged as (p − 2)(q − 2) < 4. This gives the same set of solutions for p and q; although it is somewhat simpler to solve, one benefit of the initial approach was that we were able to solve for E as well.

As a final note, the polyhedron with regular faces of p sides, where q are assembled at each vertex, is sometimes denoted by {p, q}. This notation for referring to a polyhedron is called a Schl¨aflisymbol.

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p q E V F Platonic Solid {p, q}

3 3 6 4 4 Tetrahedron {3, 3} 3 4 12 6 8 Octahedron {3, 4} 4 3 12 8 6 Cube {4, 3} 3 5 30 12 20 Icosahedron {3, 5} 5 3 30 20 12 Dodecahedron {5, 3}

Table 1

Notice that there are exactly five possibilities, each corresponding to one of the Platonic solids described in the previous section. As expected, an algebraic approach yields the same set of Platonic solids as an incremental geometric approach.

4 Exercises

1. Build the five Platonic solids.

2. Find all possible nets for the cube. In other words, find all arrangements of six con- tiguous squares in the plane which may be folded to obtain a cube. (Two nets are considered the same if one may be obtained from the other by a rotation and/or a reflection.)

3. Find all possible nets for the octahedron.

4. Color the faces of an octahedron with four colors so that all four colors are incident at each vertex. Then color the faces of an icosahedron with five colors so that all five colors are incident at each vertex.

5. Number the vertices of a dodecahedron with the numbers 1 through 5 so that each pentagonal has five differently numbered vertices.

6. Color the edges of an icosahedron with three different colors so that each face of the icosahedron has three differently colored edges.

7. Build two square pyramids and then arrange them base-to-base to form an octahedron.

8. Construct an icosahedron by constructing two pentagonal pyramids and a pentagonal antiprism, and then arranging them appropriately.

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9. (a) Build four tetrahedra and one octahedron so that all of the polyhedra have the same edge length. Arrange them to form a larger tetrahedron. Use this con- struction to find the ratio of the of an octahedron to the volume of a tetrahedron with the same edge length as the octahedron. (b) Build eight tetrahedra and one octahedron so that all polyhedra have the same edge length. On each face of the octahedron, affix a tetrahedron. What Platonic solid do the exposed vertices form? The resulting figure is called a stella octan- gula, and may be thought of as two large tetrahedra intersecting in a common, smaller octahedron. 10. A tetrahedron may be cut into two congruent parts by a plane parallel to and midway between a pair of opposite edges of the tetrahedron (see Figure 6). Build two of these pieces and arrange them to form a tetrahedron.

Figure 6

11. A cube may be cut into two congruent parts by a plane perpendicularly bisecting a long diagonal of the cube (see Figure 7). Build two of these pieces and arrange them to form a cube.

Figure 7 Figure 8

12. A cube may be inscribed in a dodecahedron as in Figure 8. Thus, we see that a dodecahedron may be formed by affixing six “roofs” on the faces of a cube (one such roof is shown in a darker orange). Build a cube and six roofs, and arrange them to form a dodecahedron.

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