Why a tesselation of the plane by a convex polygon of 7 or more sides is not possible? I read in several places, including Wikipedia, that a tessellation of the plane by a single, convex, $n-$sided polygon is not possible for $n\geq7$. I was not able to locate a proof, or a paper that discusses this. Maybe it is too trivial, but I am not able to figure it out myself. 
So I would like to ask you for help. Where can I find a proof for this?
Also, I am interested in the mathematical background behind plane tiling (especially periodic), If you can suggest papers on the topic it would really help.
 A: There is a way to cover the entire plane with convex heptagons. The one in the center is regular, then a ring of seven of them are almost regular. The next ring out are already pretty long and narrow, and so on.
From the point of view of your question, the key fact is that these are not all congruent. I'm not sure the word "tessellation" would be correct here.
This construction has been known for a very long time, it is in page 77 in Mathematical Snapshots by H. Steinhaus, probably plenty of other books and web pages. STEINHAUS 
Evidently it is also discussed on pages 248-249 of GARDNER 
I'm afraid I have been unable to find an image on line. The closest I have found in in $\mathbb H^2,$ see the tiling called (7,3) at HYPERBOLIC  and imagine what would happen if the whole thing were stretched radially, in polar coordinates $(r,\theta) \rightarrow \left( \tan \frac{\pi r}{2},\; \theta  \right).$ I should admit that I do not really know whether this gives a good approximation of the covering of $\mathbb R^2$ beyond the first three or four rings depicted. Probably not.
A: As a supplement to Will Jagy's answer, permit me to include two figures.
The first image below is from
Jay Kappraff's 1990 book, Connections: the geometric bridge between art and science.
The second image is from a 1983 paper by 
Danzer, Grünbaum, Shephard, "Does Every Type of Polyhedron Tile Three-Space?"
in Structural Topology 8 (perhaps the source for Kappraff?).

          


          


A: Here's an almost complete answer to the claim: No convex 7+ sided polygon can tessellate the plane. We don't need to compute interior angles or averages. We only need to count the number of interior angles in two ways.
We'll use contradiction of course. Let our polygon $P$ have $n$ sides and let it have area 1. Take a circle of area $A$. Keep every copy of $P$ that lies at least partially in the circle.
I will assume, without careful proof, that the number of polygons in our circle, which can be considered faces in the context of the Euler characteristic of the resulting planar graph, is $F=A+O(A^{1/2})$.
Now I will also assume, without careful proof, that the number of edges in our resultant graph is $E=nF/2+O(A^{1/2})$, as every edge is shared by two polygons except for the $O(A^{1/2})$ on the boundary. Thus $E=nA/2 + O(A^{1/2})$.
Using the Euler characteristic equation for a planar graph, we can compute
$$V = E-F+1 = \left(\frac{n}{2}-1\right)A + O(A^{1/2}).$$
Now we can count the number of interior angles in two ways. First, it is easily seen that the number must be $nF$. But also,
$$nF = \sum_{\mbox{v in all vertices}} \mbox{(# of angles meeting at v)}
\\ = \sum_{\mbox{v in interior vertices}} \mbox{(# of angles meeting at v)} + \sum_{\mbox{v in edge vertices}} \mbox{(# of angles meeting at v)}.$$
The number of edge vertices $V_e$ is assumed to be $O(A^{1/2})$ so that the number of interior vertices $V_i$ is assumed to be the same as $V=(n/2-1)A+O(A^{1/2})$
Because the number of angles meeting at a vertex must be bounded above given any polygon $P$, the second term in our split sum must be $O(A^{1/2})$.
Now using convexity, we note that for interior angles, the number of interior angles meeting at any interior vertex must be greater than or equal to 3. So,
$$nF \geq 3V_i + O(A^{1/2}) = 3V + O(A^{1/2}).$$
Substituting gives
$$nA \geq 3(n/2-1) A + O(A^{1/2}),$$
which cannot hold in the limit as $A \to \infty$ when $n > 6$. This gives us our desired contradiction.
A: In a tessellation of the entire plane by convex tiles at least three tiles will meet in each corner of each tile. Therefore the average of inner angles of the tile cannot be higher than 120 degrees. The sum of the inner angles of an $n$-gon is $(n-2)180$. If $n>6$, then their average $180(n-2)/n=180(1-\frac2n)>180(1-\frac26)=120$.
A: If you mean a regular polygon, then consider what happens at a vertex. If $k$ polygons meet at a vertex, then $2 \pi = k \alpha$, where $\alpha$ is the internal angle of the polygon, $\alpha = (n-2)\pi/n$. Then $2n = k(n-2)$. This equation has positive integer solutions only when $n\le6$. Indeed
$$
k = \frac{2n}{n-2} = 2 - \frac{4}{n-2}
$$
and so $n-2$ is a divisor of $4$, i.e., $n-2=1,2,4$, which gives $n=3,4,6$.
