# Regular polygons meeting at a point

$N$ regular polygons with $E$ edges each meet at a point with no intervening space. Show that $$N = \frac{2E}{E-2}\qquad E=\frac{2N}{N-2}$$ I did this part considering that the internal angles must add up to $2\pi$ , i.e $\left( 1 -\dfrac2E \right)\pi\times N =2\pi$. I am unable to explain the following:

Using the result given above, show that the only possibilities are $E=3,4,6$

-
For $E=5$: have you seen a dodecahedron? A proof of your claim is often used as part of the proof of why there are only five Platonic solids. –  Ｊ. Ｍ. Nov 28 '11 at 23:47
@J.M. But we're talking about polygons not polyhedrons. –  kuch nahi Nov 29 '11 at 0:51
I said your result is a component of the proof of there only being five Platonic solids. Try searching for a proof of that theorem; you'll see what I mean. –  Ｊ. Ｍ. Nov 29 '11 at 1:27
Is there something more you need, beyond what's in my answer? –  Gerry Myerson Dec 1 '11 at 6:33

If $E\gt6$ then your second equation tells you something impossible about $N$.

EDIT: Perhaps this was too subtle, so here are the details.

Suppose $E\gt6$. Then by the second equation in the question,

$2N/(N-2)\gt6$, $2N\gt6N-12$, $4N\lt12$, $N\lt3$.

But if $N$ regular polygons meet at a point with no intervening space, then $N\ge3$, so we have a contradiction to our assumption that $E$ was greater than 6.

So, $E\le6$.

Also, there are no polygons with fewer than 3 edges, so $E\ge3$.

Well, that doesn't leave very many values of $E$, does it? And $E=5$ leads to $N=10/3$, which is absurd.

So, the only possibilities are $E=3,4,6$.

-