1 the Platonic Solids

Dodecahedron 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 polygon or polyhedron 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 polygons { 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 cube, which has eight vertices (or corners), twelve edges, and six square 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 Platonic solid is a polyhedron with the following properties: (P1) It is convex. (P2) Its faces are all the same regular polygon. (P3) The same number of polygons meet at each of its vertices. Matsko 1 2011 Dodecahedron Day! Platonic Solids 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 triangles which may meet at a vertex is three. It is a simple matter to construct a polyhedron with three triangles meeting at each vertex { just take a triangular pyramid. Four triangles are required (see Figure 2(a)), and hence this polyhedron is called a tetrahedron. What about four triangles meeting at a vertex? We know from our Egyptian studies that four triangles meet at the apex of a square pyramid, 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 squares \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 octahedron. (a) (b) (c) Figure 2 Matsko 2 2011 Dodecahedron Day! Platonic Solids 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 icosahedron (\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 octahedron to a base-to-base pentagonal pyramid 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 bipyramid. Happily, this situation may be remedied. Consider for a moment the pentagonal \toy drum" of Figure 3(b). The top and bottom pentagons 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 equilateral triangles. The salient anatomical feature of this \drum" (also called a pentagonal antiprism) is that exactly three triangles (and one pentagon) 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 pentagonal antiprism 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 Matsko 3 2011 Dodecahedron Day! Platonic Solids 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 plane 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 hexahedron), 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 Matsko 4 2011 Dodecahedron Day! Platonic Solids 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 hexagons. 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 edge 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: Matsko 5 2011 Dodecahedron Day! Platonic Solids 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.

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