# Classify the regular polygons that fit about a common vetex

I know this can be solved using a (quasi-)symmetric groups approach from crystallography, but I wish to solve it with a more simple approach, number theory motivated.

I wish to classify the regular polygons for which several copies can fit around a common vertex.
The inner verices angle of an $$n$$ polygon is $$\pi-\frac {2\pi}{n}$$, and we wish it to be equal to a fraction of a circle $$=\frac {2\pi}{m}$$. A quick algebra brings us to $$mn=2m+2n$$ or $$m(n-2)=2n$$. So this may hold for any $$n$$ s.t. $$n-2|2n$$. This means that $$n-2=(n-2,2n)=(n-2,n-2+n+2)|n+2$$. Now $$(n+2,n-2)=1$$ for odd $$n$$'s and $$(n+2,n-2)=2$$ for even ones. This means that only $$n=3,4$$ are possible, indeed 4 squares and 6 triangles fit around a common vertex. But three hexagons can also fit, why did we miss this option?

Thanks a lot.

It is not the case that $$(n+2,n-2)=2$$ for even $$n$$. If $$n\equiv 2\pmod{4}$$, then $$4$$ divides both $$n+2$$ and $$n-2$$.
Note that you can simplify the reasoning a bit by just using the fact that if $$a|b$$ and $$a|c$$, then $$a|b+c$$ rather than the more general fact about GCDs. In particular, you know $$n-2|2n.$$ It is also clear that $$n-2|-2n+4.$$ Therefore, $$n-2|4.$$ This takes you immediately to your answer without needing to break into cases.
• Right! That's a better way to go. As with my approach - if $n$ is even than indeed $(n+2,n-2)$ is either 2 or 4, not just 2, which also solves it. Thanks! – galra Jan 6 at 16:53