platonic solids are regular bounded bodies, with plane surfaces
and straight edges, whose fces are all the same, edges are all the same and
corners are all the same. So if you've studied the details of one face, one
edge and one vertex, you know all about one. Since the body is bounded, it must
have some face that faces out into space away from the body; and all other faces
are just the same as this one, so must do the same. The ancient greeks studied
these and established that there were: tetrahedron (triangular-based pyramid),
cube, octahedron (with eight triangular faces), dodecahedron (12 pentagonal faces)
and the icosahedron (20 triangular faces). Forget that for a moment and let's
see what we can work out for ourselves.
Let's first just consider the network of vertices, edges and faces; since all faces are outward-facing, they must form the outer surface of a lump, with no holes through it. Any network of edges between nodes can be built up by successively adding edges and nodes. Starting with a blank lump and drawing a loop on its surface, connecting one node back to itself, we have one node and one edge splitting the surface into two regions. Keeping our network of edges and nodes connected, we can add a node to an existing edge, splitting it, or we can add an edge out of an existing node; in the latter case, the other end of the edge is either an existing node or a new node. When it ends in a new node, the region it extended into isn't split, so we make no new region but we add an edge and a node. When we split an existing edge, we likewise make no new region but add an edge (by splitting one into two) and a node. The other case has our new edge out of an existing node go to another existing node; since we keep our network connected, these two nodes were already connected, so the new edge completes a loop via its two nodes, thereby splitting the region into which it intruded. So we add an edge and a region; but we reused existing nodes, so didn't add any of them. Thus each step in building up our network adds an edge and either a node or a region. So suppose we have E edges, V vertices (nodes) and R regions (faces); the values of V +R and E +2 are equal in our initial loop and all growth of the network adds one to each, so we have V +R = E +2. Since we can build up any network of nodes and edges over the surface of a blob this way, the nodes and edges of our platonic solid satisfy this equation.
OK, so we have V vertices, E edges and R regions, satisfying V +R = E +2. Now, each edge has two sides and two ends; each corner is the same as all others, so the number of edges of which each vertex is an end is the same, let's call it k (for corner); likewise, each face is the same, so has the same number of edges, call it p (for polygon). The corners of faces happen at vertices and are delimited by edges, so there's as many of them at each vertex as there are edges, i.e. k; and each region has exactly as many corners as edges, since it has one between each pair of adjacent edges. So the number of corners is (from the vertices) k.V and is also (from the faces) p.R; so k.V = p.R. Counting the sides of edges, each has one side in each of two regions; couting via the edges we get 2.E sides; counting via the faces we get p.R, so p.R = 2.E. Counting ends of edges, by the edges we get 2.E again; by the vertices at which they end, k end at each vertex, so we have k.V ends and k.V = 2.E. Thus p.R = 2.E = k.V. In particular, at least one of p and R must be even; as must at least one of k and V. Substituting E = V +R −2 into this new equation, we get p.R = 2.V +2.R −4 = k.V, so
multiply the first by k −2 to get the second into the form
whose start and end we can rearrange as
By their specifications, k, p and R are whole numbers greater than two. The fact that all three are positive makes the last factor, 2.k +2.p −k.p, also need to be positive, so we get (k −2).(p −2) = k.p +4 −2.k −2.p < 4, which requires at least one of k and p to be 3, so that its −2 factor is 1, while the other is < 6 (to make its factor, hence the product, < 4). So we don't have many possibilities to try; one of k and p is 3, the other is 3, 4 or 5.
Let's start by considering the case p = 3; this is the case of a platonic solid whose faces are triangles. We have 4.k = R.(2.k +6 −3.k) = R.(6 −k), so 6 −k is a positive divisor of 4.k; if k is 5 it's 1, so divides everything and we get 20 = R (the icosahedron); 4 makes it 2, giving 16 = 2.R or R = 8 (the octahedron); 3 makes it 3 which definitely does divide 4.k, giving us 12 = 3.R and R = 4 (the tetrahedron); and k can't be less than 3, so that's our last possibility for p = 3. The other case to consider is k = 3, so three faces meet at each corner. This gives us 12 = R.(6 −p); we've already seen that p = 3 works, so how about p = 4 and p = 5 ? With p = 4, we get 12 = 2.R so R = 6; a solid with six square faces, the cube. With p = 5, we get R = 12 (the dodecahedron).
So a fairly simple analysis using uncomplicated whole number algebra has told us there are at most five possibilities. Now let's work out what V and E are for each of the solutions, using p.R = 2.E = k.V, and make a table of the results:
Pause to check: yes, we do get E +2 = V +R in each case. Now, notice that swapping the values of k and p switches you to a row with the same E and swaps R with V; in the case of the tetrahedron, this is the same row; otherwise, we interchange the cube and octahedron or the dodecahedron and icosahedron.
So let's look at a cube and an octahedron. The cube has three pairs of square faces; in each pair, one face is opposite the other; there are twelve edges, four bounding each of an opposed pair of faces and the other four connecting up a corner of each to a corner of the other. Each face has four edges, at which it meets the four faces not opposite to it. Mark the centre-point of each face and, inside the cube, imagine lines connecting each face's centre-point to the four centre-points of the non-opposite faces. Since each of these lines connects the centres of two faces sharing a common edge, we can associate each line with an edge of the cube; and, indeed, each edge of the cube is where two faces meet, so there is such a line associated with it. For each corner of our cube, look at the three lines associated with the edges of the cube into the given corner: these form a triangle among the centre-points of the three faces that meet at our corner. There's one such triangle for each corner of the cube; make faces of these eight triangles, meeting along the lines we constructed as edges, with the cube's faces' centres as vertices and the figure they form is the octahedron. So inside every cube is an octahedron waiting to be noticed, whose edges correspond to those of the cube while its faces correspond to corners of the cube and its corners correspond to faces of the cube.
Start with the octahedron: take centre-points of its faces, connect each to
that of each face with a shared edge; as each face of the octahedron has three
sides, each centre-point is connected to three others. Each vertex of the
octahedron has four edges out of it, whose associated new lines connect four
centre-points in a square. There are six vertices, so we get six square faces
for the figure on the octahedron's face-centres; we're back at the cube. The
cube and the octahedron are thus described as mutually
dual in the sense
that this transformation turns each into the other. The dodecahedron and
icosahedron are likewise mutually dual. When we do the same with the
tetrahedron, each face is a triangle and each vertex has three edges meeting at
it; so the dual likewise has three edges meeting at each vertex and three edges
bounding each face; sure enough, the tetrahedron is self-dual.
While I'm describing such constructions, consider a cube and pick one corner; colour it green and each corner with which it shares an edge red. There are three of these corners, each diagonally opposite each of the other two across one of the squares that met in our original corner; no two red vertices are joined by an edge. Each has one edge back to our green vertex and two more to the corners diagonally opposite that green vertex on one of the faces that meet at it; colour these vertices green. The only remaining vertex of our cube is the one opposite where we started; its three edges go to the three vertices we've just coloured green, so colour it red. Each edge of our cube now connects a red vertex to a green vertex. If you now select either the red vertices or the green vertices, you have four of them; and they lie at the corners of a tetrahedron inscribed inside the cube. It has six edges, one in each face of the cube; pick a face F of the cube and a perpendicular edge S. Consider the plane, through the diagonal of F that's an edge of our tetrahedron, whose other direction is that of S. This has two vertices of our tetrahedron in its intersection with F; the other two vertices lie in the face opposite F and are not on our plane, but are equal distances from it on opposite sides of it. Shearing, with our plane invariant, to move one of those vertices through the displacement along S shall bring that vertex into our face F; the remaining vertex shall move through the same distance in the opposite direction, putting it twice as far from F as it was. We now have three of our tetrahedron's vertices in F; shear again, with F as invariant plane, to move the remaining vertex paralle to F's diagonal that isn't an edge of the original tetrahedron. This moves it to the point twice S's displacement away from the vertex we pulled into F, along the edge of F this vertex moved along. Shrink the sheared tetrahedron perpendicular to F by a factor of two, without changing its scale in directions parallel to F. Neither shear changed its volume and this shrink has halved it. The result is a tetrahedron formed by one corner of our cube and the three corners of the cube linked to it by edges; our original tetrahedron cut off four of these corner-pieces from the cube, whose total volume is thus four halves of that of the original tetrahedron; adding the original tetrahedron gets us three times its volue, equal to the whole cube; so the tetrahedron we initially inscribed within the cube has a third of its volume.
So, now you know why there are at most five platonic solids; with
any luck you can imagine the tetrahedron (like a cone or pyramid, but with a
triangular base rather than a circle or a square), cube (lots of things come
packed in cardboard boxes that are cuboid; just imagine one with all sides of
equal length) and octahedron (two square pyriamids with their square faces stuck
together). To convince you that the dodecahedron and icosohedron exist, I'd
need to do some actual geometry (the reasoning above is all topological), such
as showing you one – if you know anyone that plays Dungeons and Dragons,
ask them to show you a
d12 and a