1
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ $n=5$ is possible, but not with regular pentagons. $\endgroup$ – GEdgar Mar 16 '15 at 21:22
  • $\begingroup$ 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...] $\endgroup$ – GEdgar Mar 16 '15 at 21:23
  • $\begingroup$ @GEdgar if you insist upon translational invariance as well, $n=5$ IS ruled out too. $\endgroup$ – user2566092 Mar 16 '15 at 21:23
  • $\begingroup$ You can't mix them, you can only use one type :) $\endgroup$ – Atvin Mar 17 '15 at 6:10
  • $\begingroup$ math.stackexchange.com/questions/91761/… $\endgroup$ – Melquíades Ochoa Mar 3 '16 at 4:31
0
$\begingroup$

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?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.