# How to solve the system of inequalities when using topological proof to show that there are exactly 5 platonic solids? (Beginner)

I have seen other answers explaining the topological proof up until the point of

$$1/p + 1/q > 1/2$$ and $$p$$, $$q$$ are greater than or equal to three

Then they proceed to say that the 5 platonic solids have the only values that satisfy these conditions. My question is how do I solve for these values? By 'graphing systems of linear inequalities with two variables'? I'm a high schooler so I see a system of inequalities and I try to graph it but the graph I get is weird... Below I show what I get when I just input "$$1/p + 1/q > 1/2$$": enter image description here Keep in mind that in that picture, I made '$$p$$' into '$$x$$', and '$$q$$' into '$$y$$'. Then I solved for y and put the inequality into Geogebra. This is because inputing '$$1/x + 1/y > 1/2$$' does not result in a graph for some reason, while '$$y > (2x)/(x-2)$$' does (even more confusion).

Am I totally wrong? How do you solve these inequalities and arrive to the desired numbers for the 5 platonic solids? Any help would be greatly appreciated! Please keep in mind that I'm a high schooler, though. I also haven't done any calculus yet. If the method is too advanced kindly just tell me the name, no need to explain. I'll remember it for when I'm older. Thank you.

We're looking for integer values of $$p$$ and $$q$$ that satisfy the three given equations $$p \geq 3$$, $$q \geq 3$$ and $$\frac 1p + \frac 1q > \frac 12$$.

Suppose both $$p$$ and $$q$$ are not equal to $$3$$. Since they are integers, they must both be greater or equal to $$4$$. But if that's the case, then $$\frac 1p$$ and $$\frac 1q$$ are both less or equal to $$\frac 14$$, meaning that $$\frac 1p + \frac 1q \leq \frac 14 + \frac 14 = \frac 12$$, contradicting one of the equations.

So to solve the equation either $$p$$ or $$q$$ must be equal to $$3$$.

Let's consider only the case where $$p$$ is equal to $$3$$. Then we have $$\frac 13 + \frac 1q > \frac 12$$. Subtracting $$\frac 13$$ from both sides yields $$\frac 1q > \frac 16$$, which implies $$q < 6$$.

So if $$p$$ is equal to $$3$$, $$q$$ must be an integer with $$3 \leq q < 6$$, so $$q$$ must be either $$3$$ or $$4$$ or $$5$$.

By a symmetrical argument, if $$q$$ is equal to $$3$$, $$p$$ must be equal to $$3$$ or $$4$$ or $$5$$.

So we have five cases remaining:

$$(p = 3, q = 3), (p = 3, q = 4), (p = 3, q = 5), (p = 4, q = 3), (p = 5, q = 3).$$

It is now easy (and necessary) to check that all these five cases satisfy the three given equations.

There isn't really a general method to do this kind of proof, it just requires finding a bunch of solutions, then a sufficient bunch of good arguments for why there aren't any more solutions.

• +1 It's worth noting that this proves there are at most five regular polyhedra. It does not guarantee that each possibility actually exists. That's easy to prove for the tetrahedron, cube and octahedron, harder for the dodecahedron and its dual the icosahedron. – Ethan Bolker Jul 19 '19 at 20:30
• It makes so much sense when reading it (thanks by the way :D) but how does one think of these things? What can you do to identify problems that you need to solve in this way? I don't think I could've ever gotten that on my own :( How do you know it's not something you should graph, for example? – Lolo123 Jul 20 '19 at 8:10
• There's not really an interesting class of problems I use this method to solve, as I made up the method myself. To learn how to build your own proofs to solve problems, you could have a look at contest mathematics. The problems there are all designed to be solved using improvised logical reasoning like this, and each has their own solution methods (usually several possible approaches) that don't fit any other problem. – Magma Jul 20 '19 at 8:20
• Thank you! I'll check it out. – Lolo123 Jul 21 '19 at 15:29

Multiply both sides of the inequality by $$2pq$$ to get $$2q+2p > pq$$, which is equivalent to $$(p-2)(q-2) < 4$$. Now $$p-2$$ and $$q-2$$ are integers $$\ge 1$$, so it is easy to list the five possibilities: $$(p-2,q-2) \in \{(1,1),(1,2),(1,3),(2,1),(3,1)\},$$ which implies that $$(p,q) \in \{(3,3),(3,4),(3,5),(4,3),(5,3)\}.$$