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

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 $\phantom{\rule{thinmathspace}{0ex}}n=3\phantom{\rule{thinmathspace}{0ex}}$ vertices, because $6\cdot {60}^{\circ }={360}^{\circ }$.
For $\phantom{\rule{thinmathspace}{0ex}}n=4\phantom{\rule{thinmathspace}{0ex}}$ it works too, because $4\cdot {90}^{\circ }={360}^{\circ }$.
I think that $\phantom{\rule{thinmathspace}{0ex}}n=6\phantom{\rule{thinmathspace}{0ex}}$ is good too, but how do I prove, that other than that, it isn't possible to do that?
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stangeix
Step 1
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?
Step 2
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?
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imire37
Explanation:
It is possible to cover the plane with convex polygons with more than 6 vetices, if you do not put a bound on their size. Take a tiling of the hyperbolic plane by heptagons in the disc model and apply a map $\left(r,\theta \right)↦\left(\frac{r}{1-r},\theta \right)$ to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).