Problem in understanding a proof there are five Platonic solids.

Thanks to several comments by Gerry Myerson, it is now clear that I wasn't clear, up to a state where I seriously confused myself. In a renewed attempt:

Recently, I've been thinking about Platonic solids (in preparation of a maths camp next summer) and about proofs that there are only five of these. There is one step in what seems to be the main proof that causes problems.

To avoid problems, polyhedra may contain infinitely many faces and Platonic solids consist of finitely many faces, all of these the same regular polygon, such that at every vertex the same number of polygons meet.

The proof I'm talking about considers the possibilities for $k$ regular $n$-gons meeting at each vertex ($k\geq3$) and considers when the sum of angles at each vertex (i.e. $k$ times the angle in a regular $n$-gon) is strictly less than $360^\circ$. This yields exactly five options and all five options can then be realised.

From considering counterexamples to several subtly wrong definitions of Platonic solids (in this setting), it seems that the requirement that $k$ times the angle in a regular $n$-gon) is strictly less than $360^\circ$ corresponds to the finiteness of the number of faces. I, however, fail to see why the finiteness of the number of faces implies this. (the implication in the other direction would be equally unclear, but irrelevant). Does anyone have suggestions how to (easily) prove that the finiteness of the number of faces implies that sum of angles at each vertex has to be strictly less than $360^\circ$? I still guess there's some obvious thing I'm missing.

• I think a polyhedron, by definition, has only finitely many faces. I don't see how you get one polyhedron, much less infinitely many, where the sum of the angles at one vertex, much less the sum of the angles at each vertex, is 360 degrees. Nov 7 '13 at 11:52
• If you don't require polyhedra to be finite, then you have more than five Platonic polyhedra, don't you? The infinite square lattice, the infinite hexagonal lattice, the infinite triangular lattice? Nov 7 '13 at 11:57
• A hexagonal lattice is made of hexagons; a triangular lattice is made of triangles. But you are really confusing me. You say that, to you, a Platonic solid has only finitely many faces; you say that if we restrict to finite polyhedra then there is no problem; it seems to me that it follows that you have no problem (and moreover you have no polyhedron with angles at a vertex summing to 360 degrees). Nov 7 '13 at 12:10
• I managed to get myself confused. Thanks for pointing it out. The problem is quite clear in my head but damn hard to write down. I'm edited the question accordingly now. Hopefully it is a bit clearer this way.
– HSN
Nov 7 '13 at 12:30
• If the sum of angles of identical regular polygons intersecting a vertex add up to 360 degrees, then the polygons must lie on a plane. Is this what's confusing you? Nov 7 '13 at 14:05

I am still not too clear on what you are asking regarding the infinite number of faces, but let me say a little on what I think you are getting at.

Suppose we have a convex polyhedron. This means that Euler's polyhedral formula holds and we have $$V - E + F = 2$$ where $V$, $E$ and $F$ are the number of vertices, edges and faces respectively. We define the vertex angle at each vertex $\theta_v$ as the sum of all the face angles meeting at that vertex. We then define the angular deficit as $$\delta_v = 2\pi - \theta_v$$ We now want to compute the total angular deficit $\sum_v\delta_v$ over all vertices of the polyhedron. Clearly, we have $$\sum_v\delta_v = 2\pi V - \sum_v\theta_v$$ The latter summation is the sum of all face angles across the polyhedron. Since each face of a convex polyhedron is a convex polygon, we can write $$\sum_v\theta_v = \sum_f\phi_f = \pi\sum_f (E_f-2)$$ where we write $\phi_f$ as the net interior angle of face $f$, and $E_f$ as the number of edges comprising face $f$. This gives us the net result $$\sum_v\delta_v = 2\pi V - \pi\sum_f (E_f-2) = 2\pi V + 2\pi F - \pi \sum_f E_f$$ The latter sums over the edges of all faces, and since each edge belongs to exactly two faces, the sum counts each edge twice. Therefore we finally have $$\sum_v\delta_v = 2\pi V + 2\pi F - \pi \sum_f E_f = 2\pi(V - E + F) = 4\pi$$ This result is known as Descarte's Theorem. The take away for this is that the net angular deficit is always positive. In a platonic solid, by definition we require the vertex angles to be the same across all vertices. This means that the excess at each vertex must be given by $$\delta_v = \frac{4\pi}{V}$$ and that the vertex angle is strictly less than $2\pi$.

I will close off with a remark that convexity is somewhat necessary. It is not true for non-convex polyhedra that the vertex angles are less than $2\pi$.

• @MertAktaş I have no idea, I don't think this site allows that feature.
– EuYu
Jul 3 '15 at 3:40

Let the regular polyhedron consist of regular p-polygons ($p \ge 3$) meeting k ways ($k \ge 3$) at every vertex. Then for the included angle at each polygon vertex

$\boldsymbol{\alpha = \dfrac{(p-2)\pi}{p}} \tag{1}$
so $\boldsymbol{\dfrac{\pi}{3} \le \alpha < \pi} \tag{2}$

Using the concept of angular deficit at each vertex ($\delta_v = \frac{4\pi}{V}$), noting that $V \ge 4$ for a polyhedron: $0 < \delta_v \le \pi$.

But by equation (1): $\delta_v = 2\pi - k\alpha = 2\pi - k \dfrac{(p-2)\pi}{p}$, so $\pi \le 2\pi - k \dfrac{(p-2)\pi}{p} < 2\pi$. Then

$\mathbf{1 \le k(1 - \dfrac{2}{p}) < 2} \tag{3}$

From $k\alpha = 2\pi - \delta_v < 2\pi$, the restrictions on $p,k$, and inequality (2), we get

$\mathbf{3 \le k < 6}$ and $\mathbf{3 \le p < 6}$

By checking all possible combinations of $(p,k)$ against inequality (3), solutions are:

$(p,k) \in \{(3,3),(3,4),(3,5),(4,3),(5,3)\}$