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A polyhedron is a solid bounded by F plane faces, which meet in E edges and V vertices. You may assume Euler’s formula, that $V − E + F = 2$.

In a regular polyhedron the faces are equal regular m-sided polygons, n of which meet at each vertex. Show that

$F =\frac{4n}{h}$ , where $h = 4 − (n − 2)(m − 2)$

By considering the possible values of h, or otherwise, prove that there are only five regular polyhedra, and find V , E and F for each.

I'm stuck on the second part of this question. I've tried bounding h by saying that the number of faces meeting at a point can be a maximum of $\frac{2\pi}{\theta}$ where $\theta$ is the interior angle. But I realise now that obviously doesn't show anything.

I was wondering if I could get a good hint as to a first step as I'm not sure how it wants me to look at the possible values of h.

Thanks in advance :)

Jake Stone
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  • I've now solved it :) I just needed to observe that $h$ must be positive and that limits the values of $n$ and so you can just check a couple cases. – Jake Stone Mar 07 '23 at 17:50
  • I was just about to post an answer... :( – D S Mar 07 '23 at 17:51
  • Post it anyway I'd be interested to see how you structured it :) – Jake Stone Mar 07 '23 at 17:57
  • sorry I'd gone to sleep after posting that comment. Basically I figured that $(n-2) \cdot (m-2) \le 3$, and $n > 2$ and $m > 2$. Letting $a = n-2$ and $b = m-2$, we have the following pairs $(a,b)$: $(1,1), (2,1), (3,1), (1,2), (1,3)$. – D S Mar 08 '23 at 12:24

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