counting edges in tesselations of a torus Here's another one that no one's rushing to answer on stackexchange.
Tesselate a torus with finitely many simply connected polygons.  Do not allow four or more of them to meet at a point.  Count the edges by viewing them as edges of a graph, not as edges of a polygon; i.e. don't count a "straight line" as just one edge if it's the boundary between polygons A and B until you reach a point after which it's the boundary between A and C; at that point one edge ends and the next starts.  (For example, if the torus lies flat on a table, you can divide it into (1) the upper north quadrant; (2) the upper south quadrant; (3) the lower east quadrant; and (4) the lower west quandrant.  In one sense, each of these is a rectangle with four edges, but two of the edges get interrupted halfway through by a vertex, so we'll count six graph edges.)
Then: The average number of edges of the tesselating polygons is exactly 6.
Proof: $V-E+F=0$, then massage.
The question: Is the statement after "then" in citable literature somewhere?
 A: Yeah, this works, book is Topological Graph Theory by Gross and Tucker, mostly pages 216-217. First we have, on page 4, Theorem 1.1.1 by Euler, the sum of the valences is twice the number of edges. So the average valence is $\frac{2 E}{V}.$
Very much in the same vein, the sum over all the faces is twice the number of edges, so the average edges per face is $\frac{2 E}{F}.$
Your hypothesis is that the average valence is exactly three, so
$$ 3 V = 2 E.   $$
The Euler characteristic says $V - E + F = 0,$ or $F = E - V,$ or 
$$3 F = 3 E - 3 V = 3 E - 2 E = E,$$ so
$6 F = 2 E$ and 
$$ \frac{2 E}{F} = 6. $$ 
There is a figure of $K_7$ on the torus on page 137, the dual would be seven hexagons. So that is the prototype, but of course you can take any large section of hexagon wallpaper and carefully place a rectangle on it to get another example. I don't know what happens if you start with a pentagon and go from there. There is also the one with octagons and squares.
Indeed, we can do this with any doubly periodic tiling, perhaps convex polygons if that can be arranged. Then, for any vertex with valence larger $n >3,$ replace the vertex with a little $n$-gon with vertices along each of the $n$ edges. For doubly periodic, this is a finite set of replacements, so we can make all sorts of examples on the torus that satisfy the valence three requirement. In particular, I have in mind one made up of pentagons, each pentagon with three vertices of valence 3 but two vertices of valence 4. Replacing these with little squares, we get a tiling of the torus with squares and heptagons. Who knew?
