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$N$ 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 $E=3,4,6$

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    Is there something more you need, beyond what's in my answer?2011-12-01

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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 $E=3,4,6$.