fill a rectangle with hexagons Suppose I have a rectangle and I want to fill it with hexagons without having any white space. The hexagon doesn't need to be in the regular hexagon shape so the only thing that matters is that it needs to have 6 sides. Each hexagon can have maximum one common side with another hexagon.
I have been trying to solve it and I think it is impossible but I cannot prove it:S.
Can somebody help me filling it out or proving that it is impossible?
 A: Assume there is such a partition of the given rectangle into hexagons.
Denote by $v$ the total number of vertices and by $v_b$ the number of vertices on the boundary of the rectangle, among them the the $4$ corners of the rectangle. Let there be $e$ edges and $f$ hexagons. All in all we have a polyhedral partition of $S^2$ with the exterior of the rectangle as an extra facet. 
By Euler's formula we have
$$e+2=v+f+1\ .\tag{1}$$
The edges can be counted in two ways:
$$2e = 6f+v_b\ ,\tag{2}$$
since every edge  belongs to two hexagons or one hexagon and the extra facet, and
$$2e\geq 3(v-4) +2\cdot 4=3v-4\ ,\tag{3}$$
since all vertices apart from the corners of the rectangle have degree $\geq3$.
When we multiply $(1)$ by $6$ and subtract from it $(2)$ we obtain together with  with $(3)$ that
$$6v-6-v_b=4e\geq 6v-8\ ,$$
or $v_b\leq2$. But the latter contradicts $v_b\geq4$.
A: As there are a few details missing from the question about what is and is not allowed, I'll offer what I believe is a proof that such a tiling by hexagons is not allowed. First, suppose that we are tiling the unit square because it's just easier to work with.
$(1)$ Now, suppose that a tiling $T$ has to have each tile a non-degenerate hexagon and each intersection of two distinct tiles is equal to either the emptyset or a single shared edge (not a segment of an edge), or possibly either of these also union a disjoint set of vertices not belonging the that edge. A tile may not 'share an edge' with itself. That is each tile has exactly six distinct edges.
$(2)$ Next, assume that each edge of the square belongs to the edge-set of exactly one tile, and that no tile has two or more edges of the square belonging to its edge-set (this is a rather strong condition, but it's one I can't seem to find a proof without).
Consider a second copy of a square tiled by $T$ and glue this square to the original square along corresponding edges so that we now have a spherical tiling by hexagons which still meets the above conditions in $(1)$ on the tile intersections (this is why we needed the conditions on tiles at the boundary of the square). Now consider the 'dual graph' to this tiling given by marking a single vertex in each tile and drawing an edge between two vertices if the corresponding tiles intersect in an edge. This graph is embedded in a sphere and so can also be embedded in the plane. This graph also has the property that each vertxes has degree exactly equal to six because each edge of a tile is shared by exactly one other distinct tile. By the theorem linked to by hardmath in the comments, this is a contradiction as all planar graphs have at least one edge with degree less than or equal to five.
