Geometry Riddle
Posted by Rob Herman at June 27th, 2006
All right. Weekends seem to be better for most people. Since Origins is this weekend, we’re going to shoot for 1KBWC next weekend, so that’s Sunday 7/8. If you are reading this, you are almost certainly invited. Please let me know so I know how many snacks to acquire.
Now for the riddle:
Due to their use in dice, most gamers will be familiar with the 5 convex regular polyhedra: the tetrahedron, cube, octahedron, dodecahedron, and icosahedron. The riddle: Why are these 5 the only regular polyhedra that can exist?
Simultaneous Limitation and Hint: We’re only talking about Euclidean space here.
[Edit: Comments contain a solution. It’s involved enough that a quick glance probably won’t spoil everything; just so you know.]
Bother.
I am more fully confident of the completeness of my arguments now that I have typed them out, incidentally.
The issue is the generation of vertices. A vertex must be a meeting of a specific whole number, 3 or greater, of identical regular polygons, in such a way as to maintain convexivity. Therefore, the internal angles of the vertices of the component polygons at the polyhedral vertex must sum to less than 360*. Thus, we are restricted to polygons of vertex angle less than 120*, ie the triangle, square and pentagon.
Now consider the number of polygons that meet at each vertex. For pentagonal and square faces, this can be no greater than 3, and thus must be exactly 3, generating the cube and the dodecahedron. For triangular faces, this can be no greater than 5, generating the tetrahedron, octahedron and icosahedron for 3, 4 and 5 triangles, respectively.
Anything else would be uncivilized. QED.
Ok, apparently this text box truncates posts on a less-than sign (shift+,). I blame Rob.
Yeah, Alatar got it. :)
Apparently WordPress’s comment field is paranoid about anything that looks like HTML tags.
You also do a topological proof based on Euler’s Formula.
F - E + V = 2
or, the number of faces minus the number of edges plus the number of vertices equals 2. Euler’s formula is derived by observing that each edge joins two vertices, and has two adjacent faces, so
p F = 2 E = q V,
for positive integers p, q
p = number of sides of each face
q = number of faces meeting at each vertex.
therefore
2E/p - E + 2E / q = 2, so
1/p + 1/q = 1/2 + 1/E, and since E is positive
1/p + 1/q lessthan 1/2. Since p and q are at least 3, there are only five possibilities,
{3,3}, {4,3}, {3,4}, {5,3}, and {3,5}, QED.
(Thanks to my algebriac topography textbook. The proof probably dates to the mid 18th century, when Euler laid the framework for algebriac topography, but I can’t find a reference. The first proof, provided by alatar, is as old as Euclid. yay math.)
Plato attempt this proof long before Euclid. His argument kind of glosses over a step though so it is not a solid proof. It goes something like this.
1. If you connect the centers of all the faces of a regular polyhedron then another polyhedron is created inside of it.
(this can be proven with some relatively simple geometry)
2. By repeating step 1 with the inner polyhedron you get a smaller version of the outer polyhedron. (this makes sense since the centers of the faces of the inner polyhedra are directly between the center and a vertx of the outer polyhedron). This creates the following pairings of polyhedra where performing step 1 on one polyhedron in the pair creates the other.
(tertrahedron, tetrahedron) (cube, octohedron) (dodecahedron, icosahedron)
3. Each of these pairings must include a polyhedron that has a triangle for a face (He kind of glosses over this as common sense. Ah the power of the dialogue)
4. Since due to conclavity the only polyhedrons that can be formed with triangles are tetraherdon, octohedron, and icosahedron there are no more pairings and therefore no more regular polyhedrons QED (except not)
It sounds like he’s giving a roundabout version of the proof Alatar indicated. Each of the pairings must include a polyhedron that has a triangle for a face because otherwise you would have to fit four (or more) n-gons where n GT 3, and there aren’t enough degrees for that.