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

Matonya 2022-08-18 Answered
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 60 = 360 .
For n = 4 it works too, because 4 90 = 360 .
I think that n = 6 is good too, but how do I prove, that other than that, it isn't possible to do that?
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Answers (2)

stangeix
Answered 2022-08-19 Author has 10 answers
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 , 90 , and 120 (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
Answered 2022-08-20 Author has 8 answers
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 ( r , θ ) ( r 1 r , θ ) to yield a tiling of the whole plane (if we straighten out the edges after applying the transformation).
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