You know what a cube is: a solid three-dimensional object with six identical sides, each of which is a square. The cube is one of the five Platonic solids: highly symmetrical shapes known at least as far back as the ancient Greeks. In addition to the six-sided cube, we have the four-sided tetrahedron, the eight-sided octahedron, the twelve-sided dodecahedron, and the twenty-sided icosahedron. Their common feature is that all their sides are identical. For example, all four sides of the tetrahedron are equilateral triangles, and all twelve sides of the dodecahedron are regular pentagons. Since all of their faces are identical, they are all well suited to use as dice. Many games use cubes (common six-sided dice) since they are easy to manufacturer and ship, but the popular (at least when I was growing up, before computer games appeared) fantasy role-playing game Dungeons and Dragons uses all five kinds, as shown below.
It has long been known that these five are the only shapes with this sort of complete symmetry. Today I want to discuss the question "Why are there exactly five Platonic solids?" Why not more, or fewer? Why these particular five?
First, let's be clear about the definition of a Platonic solid. It is a geometric shape like the ones pictured above, all of whose faces are identical regular polygons. What's a "polygon", and what is "regular"?
Well, start by drawing some dots on a piece of paper. We will refer to them as vertices. Now connect the dots with straight lines (called edges) that do not cross, to form a closed shape. That's a polygon. For example, a triangle has 3 vertices and 3 edges.
A regular polygon is one for which all the edges are the same length, and all the angles between the edges are the same. For instance, a triangle with all three edges of equal length is regular; we call it an equilateral triangle. If a four sided polygon is regular, we call it a square. If a five sided one is regular, we call it a regular pentagon.
So, to make a Platonic solid, you pick a regular polygon: either an equilateral triangle, a square, or a regular pentagon. You make a bunch of copies of the polygon; we call these the faces. Finally, you glue them together along adjacent edges to form a solid three-dimensional shape.
However, if you try this with regular hexagons (6-sided polygons), you will find that you cannot glue them into a solid "dice-like" shape. It just does not work: the angles are not right. This provides a clue to figuring out why there are only five possibilities.
Imagine we draw three squares together as shown in (A) below, then cut out the L-shape and fold on the edges between colors. The three faces meet at a common vertex, forming half of a cube. But if we had included the fourth square, as in (B), we could not achieve this: any attempt to fold the edges down would not fit together.
Let's see if we can generalize this observation, so we can characterize which shapes will and will not fit together.
If we draw several identical regular polygons around a common vertex, like the three squares above, we have to leave a gap, like in (A), if we want to be able to fold it into a three-dimensional shape. Since squares have 90 degree angles between adjacent edges, four squares take up all 360 degrees: no gap. We are limited to using three squares, and we wind up creating the cube.
Two squares are insufficient, since we need three to form a three dimensional corner. So the cube is the only possible Platonic solid built from squares.
Equilateral triangles have an interior angle of 60 degrees, so six of them completely fill the space around a vertex, just like four squares do. So, we are limited to using 3, 4, or 5 triangles around a vertex, if we want to build a Platonic solid. These give the tetrahedron, octahedron, and icosahedron, respectively.
With just three triangles around each vertex, the tetrahedron is very "spiky": the points are sharp, because we have to fold the paper tightly to glue the edges together. Try it yourself with a piece of paper: print the picture above, cut out the three-triangle strip, fold where the colors change, and try to bend it to make a three dimensional corner. In contrast, when we make the icosahedron, with five triangles around each vertex we do not need to fold the paper as sharply, so the corners are not as sharp. In fact, the icosahedron is much more like a sphere (smooth all over), in comparison to the tetrahedron.
We can also use regular pentagons. The angle between adjacent edges is 3*180/5 = 108 degrees, so we can fit three together, leaving a gap, much like the three squares in (A), but four will not fit. So there is only one possible Platonic solid made from pentagons, namely the dodecahedron.
Finally, if we try something with 6 or more sides, like a hexagon, the interior angles will be 120 degrees (for hexagons) or more (for 7 or more sided shapes), so even three will fill or overfill the 360 degrees available, with no room for a gap. So there are no Platonic solids with faces of six or more sides. That means the five we found are all of them: there are no others.
It turns out that there is an entirely different approach we could have used. The analysis of angles we discussed above was known to Euclid around 300 B.C., but this second approach is much more recent. Various mathematicians, including Descartes in 1639, Euler in 1751, and Cauchy in 1811, contributed to the discovery that for any "sphere-like polyhedron" (i.e. any geometric solid built from polygons, without holes, such as the Platonic solids), there is a relationship between the number of vertices V, edges E, and faces F, often called Euler's formula:
V - E + F = 2
Start with a cube, for example. It has V=8 corners, E=12 edges, and F=6 faces; 8-12+6 = 2 as claimed. Now imagine we take one of the square faces of the cube and cut it into 2 triangles, as illustrated below:
What does this do? The number of vertices does not change, but the number of faces goes up by one, as does the number of edges. So the overall value of V-E+F does not change, since the extra face cancels the extra edge. A similar argument shows that no matter how you add or remove vertices, faces and edges, the value of V-E+F does not change, provided you do not cut holes through the solid (i.e. turn it into a doughnut or inner-tube shape).
Using Euler's formula, we can classify the Platonic solids without having to calculate with angles. Suppose the Platonic solid is made from n-sided polygons, and suppose k of them meet at every vertex. For instance, the cube has k=3 polygons around each vertex, and each of the polygons is a square, with n=4 edges.
If there are F faces, then before we glue them together to build the solid, we have nF edges and nF vertices. We glue edges in pairs, so there will be E = nF/2 edges in the final solid. We glue k vertices together at each corner, so there will be V = nF/k overall in the final solid. By Euler's formula, we have
nF/k - nF/2 + F = 2.
Dividing by twice the number of edges (2E) gives
1/k + 1/n = 1/E + 1/2.
Since E>0, we know 1/k + 1/n must exceed 1/2. We also know that n is at least 3 (triangles have 3 sides, other polygons have more) and that k is at least 3 (we need at least three faces per vertex to get a three dimensional solid).
What whole numbers n and k, each at least 3, satisfy the constraint that 1/k + 1/n must exceed 1/2? We can list them out.
When k=3, 1/n must exceed 1/6, so n can only be 3, 4, or 5.
When k=4, 1/n must exceed 1/4, so n can only be 3.
When k = 5, 1/n must exceed 0.3, so n can only be 3.
And if k is 6 or more, 1/n must exceed 1/3, so there are no solutions for n (since n must be at least 3).
So once again, we have exactly five solutions, no more and no less, and once again they correspond exactly to the known Platonic solids. The tetrahedron has k=3 triangles (n=3) meeting at each vertex. The cube has k=3 squares (n=4) and the dodecahedron has k=3 pentagons (n=5) meeting at each vertex. The octahedron has k=4 triangles (n=3) at each vertex, and the icosahedron has k=5 triangles (n=3) at each vertex.
There are other interesting questions you can ask about geometric shapes. For instance, someday I intend to do an article on "tessellation", which is the process of covering a sheet of paper with a repeating grid of shapes (not just squares, but triangles and hexagons work too). One can also ask more exotic questions, such as: how many four-dimensional Platonic "hypersolids" are there? But that requires being able to visualize geometry in 4 dimensions, which is too tricky to delve into today.
If you enjoyed this article, you might also like Algorithmic Art. Or perhaps Counting and Number Systems.
I hope you found this interesting. You can click the "M" button below to email this post to a friend, or the "t" button to Tweet it, or the "f" button to share it on Facebook, and so on. As usual, please post questions, comments and other suggestions using the box below, or G-mail me directly at the address mentioned in the Welcome post. Remember that you can sign up for email alerts about new posts by entering your address in the widget on the sidebar. If you prefer, you can follow @ingThruMath on Twitter to get a 'tweet' for each new post. The Contents page has a complete list of previous articles in historical order. You may also want to use the 'Topic' and 'Search' widgets in the side-bar to find other articles of related interest. See you next time!
