There are only five platonic solids. The proof of this is easy. Equilateral triangles angles are each 60 degrees. So you can have three four or five meet at a vertex, but not six as then the angles would sum to 360 degrees and the join would be flat. Similarly you can have at most three squares and at most three regular pentagons meet at a vertex. With hexagons and anythign with more sides, you cannot even have three faces meet at a vertex. So there are at most 5 Platonic solids.
In hyperbolic geometry triangles angles sum to less than 180 degrees so an equilateral triangle’s angles would be less than 60 degrees each. Can you, in hyperbolic space, construct a regular “Platonic solid” in which six equilateral triangles meet at a vertex? How about one where four squares meet at a vertex or three hexagons?
I’m pretty sure that the answer is no. You can prove this using the Euler characteristic. Suppose you have a polygon with V vertices, E edges, and F faces. Suppose further that n edges meet at each vertex and each face has m sides. The Euler characteristic equation says that
V - E + F = 2.
Each edge connects two vertices, and each vertex has n edges meeting at it; this implies that
nV = 2E.
Each edge separates two faces, and each face has m edges; this implies that
mF = 2E.
Combining all of these equations and rearranging, you can show that
1/n + 1/m = 1/2 + 1/E > 1/2.
If n or m get too large, you won’t be able to satisfy this inequality; in fact, the only solutions are {n,m} = {3,3}, {3,4}, {3,5}, {4,3}, or {5,3}. These cases correspond precisely to the five Platonic solids.
You have to be a little careful with that argument: the Euler characteristic of a topological space is an invariant, but an infinite plane isn’t topologically equivalent to a sphere. I think you can still define an Euler characteristic for non-compact manifolds (like the infinite plane) using homology groups, but it’s a lot more complicated than just counting up vertices, edges, and faces.
Take same-size 8 equilateral triangles, and arrange them into a large rhombus.
Warp space so that the opposite sides and vertexes of the rhombus coincide.
This gives us an 8-sided Platonic solid with 4 vertexes and 12 edges, in a non-Euclidean space.
Can’t be done (without putting three holes in the surface). Somewhere, you will get non-triangles or incongruent vertices. In fact, the Euler characteristic is a topological invariant so whether the solid is platonic or not, the argument given by MikeS is solid.
As Hari Seldon mentioned, the Euler characteristic is toplogically invariant, so there’s no change you can make to the topology of the space that will affect it.
I’m having trouble reconciling these statements. In my understanding, “topologically invariant” means that the characteristic doesn’t change, as long as the topology doesn’t change. The Wikipedia seems to agree with me. So if we do change the topology (it takes several cuts and stitches to go from Euclidean space to one that satisfies the conditions I impose), we shouldn’t expect the Euler characteristic to stay the same.
This space is topologically a torus, a genus-1 surface: you’ve periodically identified two parallel pairs of edges. Polyhedra are usually taken to live on a topological sphere, a genus-0 surface. Euler’s formula is V-E+F=2-2g, so your tiling works on the torus but not on a sphere.
Right, Pleonast’s construction is on a torus, a surface topologically distinct from the sphere.
Yes, these three solutions are the only regular solutions for a Euclidean plane, but a hyperbolic plane has many more regular tilings than these, for example.
Well, platonic solids are kind of related to the tessellations of the elliptic plane (if you think of them first as ‘tessellations of the sphere’, and then go from there to plane elliptic geometry by identifying the antipodal points), so there’s something to the idea of relating them to the tessellations of the Euclidean plane, as well – look at it this way: if you want to cover the plane with regular n-gons, k of which meet at each vertex (i.e. a {n,k}-tessellation), the angle between their edges is 360°/k, and thus, the angle sum n*360°/k. Also, the angle sum of a regular n-gon is (n - 2)180°, because you can ‘cut’ it into n-2 triangles (the square in two, the pentagon in three, etc). Thus, we get n360°/k = (n - 2)*180°, which works out to 1/n + 1/k = 1/2, admitting as solutions the tessellations of the plane. However, this was under the tacit assumption of Euclidean geometry – if we don’t require that, the angle sum of each n-gon can be bigger than (n - 2)*180, in elliptic geometry, leaving us with the inequality MikeS derived, which gives us the ‘platonic solids’ as solutions. But of course it can also be smaller, leading to 1/n + 1/k < 1/2, the tessellations of the hyperbolic plane, of which there are infinitely many! These are in construction rather similar to the original platonic solids, it’s just that hyperboloids are not nice, compact things like spheres, and we thus don’t really get any ‘solids’ from that.
Of course, you can also have polyhedrons which aren’t topologically equivalent to a sphere. For instance, you can have a self-intersecting equilateral pentagon with five angles of 36 degrees each (a five-pointed star), and you can then fit a dozen such pentagons together into a self-intersecting regular polyhedron.
If you change the topology, you can do anything. Make it discrete and then there are no edges, faces of vertices. But a platonic solid is generally understood to be topologically a sphere and a solid with 8 faces, 4 vertices and 12 edges will have genus 3 no matter what you do with it. This follows from the formula v - e + f = 2 - 2g. where the four variables are the numbers of vertices, edges, and faces, and the genus, respectively.
