This is an original theorem of mine.

Let P be an n-sided polygon. Define an m-covering to be a set of convex polygons, each of which has at least 4 but at most m sides, such that the union of the regions enclosed by these polygons is the region enclosed by P, and each polygon in the set shares exactly one side with P.

For example, the following are two failed attempts at 4-coverings. The figure on the left shows a dark green polygon that has been almost covered by four convex quadrilaterals, colored blue, light green, red, and yellow. But it is not a 4-covering because the dark green triangular regions at the left and the right have not been covered. The figure on the right shows a triangle that has been completely covered by four quadrilaterals, colored blue, red, and yellow. But it is also not a 4-covering because the blue and yellow quadrilaterals are not convex.

Theorem

No m-coverings are possible, for any m >= 4.

Lemma

First, I shall prove the weaker theorem that no 4-coverings are possible.

In any such covering, each quadrilateral must share 1 side with the region outside of the polygon and it must share part of 1 side with at least 3 other quadrilaterals. This is true because each quadrilateral has 3 sides that are inside the polygon, and no other quadrilateral may cover more than 1 of those sides, and each of the sides must be completely covered.

Now if we draw this as a graph, we have n points Q_1...Q_n representing the quadrilaterals and each of those points has an edge going to a point P that represents the outside of the polygon. Each of the Q_i must then have 3 other edges going to other Q_i. And of course, the whole graph must be planar. The diagram illustrates the situation.

Now we simply must check that this planar graph is not possible. Suppose otherwise and let Q_i be any of the quadrilaterals. Now, since Q_i must connect to three other quadrilaterals, it must have an edge to at least one quadrilateral Q_j that it is not adjacent to. Therefore the cycle Q_i -> Q_j -> N -> Q_i encloses a region R containing at least one other quadrilateral.
Now choose Q_i and Q_j, without loss of generality, such that R encloses a minimal nonzero number of quadrilaterals m. Let Q_k be in R. Q_k must have an edge to at least one quadrilateral Q_l that it is not adjacent to (Note: it is possible that Q_l is equal to Q_i or Q_j). But then the region L enclosed by Q_k -> Q_l -> N -> Q_k lies entirely within R, yet does not contain Q_k. Therefore L encloses a nonzero number of quadrilaterals that is smaller than m, contradicting the choice of Q_i and Q_j. Therefore, the graph cannot be planar, so no n-sided polygon may be covered by quadrilaterals, such that each quadrilateral shares exactly one side with the polygon.

Generalization

The result of the lemma can be generalized from 4-coverings to m-coverings by splitting up each convex polygon in the m-covering into convex quadrilaterals. In other words, suppose for contradiction that there is an m-covering. Then subdivide each convex polygon as shown into convex quadrilaterals (in the diagram, the blue hexagon, part of a supposed tiling of the green polygon, is subdivided into quadrilaterals along the black lines). Thus we could obtain a 4-covering, therefore the m-covering was impossible as well.