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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Covering the plane with convex polygons?I have got the following task here:Prove, that you can&#039;t cover the &quot;Plane&quot; 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&#039;t possible to do that?

stangeix · Accepted Answer

Step 1I&#039;m going to assume you mean tiling the plane with copies of the same regular n-gon; so we can&#039;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 2What the corresponding angles be, if we consider a regular n-gon, where   n  &amp;gt;  6? Would angles like that make sense, in a tiling situation?

imire37 · Accepted Answer

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).

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

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