# Covering the plane with convex polygons?

I have got the following task here:

Prove, that you can't cover the "Plane" with convex polygons, which have more than $\,6\,$ vertices!

The answer is pretty obvious for $\,n=3\,$ vertices, because $6\cdot 60^\circ = 360^\circ$.

For $\,n=4\,$ it works too, because $4\cdot 90^\circ = 360^\circ$.

I think that $\,n=6\,$ is good too, but how do I prove, that other than that, it isn't possible to do that?

• $n=5$ is possible, but not with regular pentagons. – GEdgar Mar 16 '15 at 21:22
• Are we allowed to mix $7$-, $8$-, and $9$-gons? Are we allowed to have different (non-congruent) $7$-gons? [pjs36 ruled all these interesting cases out...] – GEdgar Mar 16 '15 at 21:23
• @GEdgar if you insist upon translational invariance as well, $n=5$ IS ruled out too. – user2566092 Mar 16 '15 at 21:23
• You can't mix them, you can only use one type :) – Atvin Mar 17 '15 at 6:10
• math.stackexchange.com/questions/91761/… – Melquíades Ochoa Mar 3 '16 at 4:31

I'm going to assume you mean tiling the plane with copies of the same regular $n$-gon; so we can't mix squares and hexagons, for example.
Why are $60^\circ, 90^\circ,$ and $120^\circ$ (for a hexagon) important, feature of the polygon do they measure?
What the corresponding angles be, if we consider a regular $n$-gon, where $n > 6$? Would angles like that make sense, in a tiling situation?