Tiling a Jordan polygon I saw this problem some years ago, don't remember the source:

Let $P$ be a Jordan polygon (i.e. the only points of the plane belonging to two edges are the polygon vertices) that can be tiled with parallelograms.

*

*Does it follow that the sum of the areas of all the rectangles among those parallelograms is the same independent of the tiling?

*Does it follow that the sum of the areas of all parallelograms with angles $\alpha$ and $\beta$ is the same independent of the tiling?


I couldn't find a solution, but I believe the answer to both questions is positive.
Does anyone know how this problem can be solved?
 A: When you tile a simple polygon by parallelograms, the tiling can be partitioned into "zones" of rectangles with parallel sides, meeting end-to-end, and forming a path from one edge of the polygon to a parallel edge of the opposite orientation. Two parallel zones cannot cross, so the pairing of opposite edges into zones is uniquely determined. Additionally, two non-parallel zones cross either zero times or one time, according to whether the edges at the ends of the zones appear in nested or alternating order around the polygon. That is, the number of crossings is again uniquely determined by the polygon.
You get exactly one rectangle for each two zones that are determined by perpendicular edges of the polygons and that cross each other. The shape of this rectangle is just the Cartesian product of the two perpendicular edges, so it is also determined. So the shapes of all rectangles in the tiling, and not just the sum of their areas, is equal for all tilings. The same goes for parallelograms with other angles.
A: Not every simple polygon can be tiled with parallelograms.

Kannan, Sampath, and Danny Soroker. "Tiling polygons with parallelograms." Discrete & Computational Geometry 7, no. 2 (1992): 175-188.
DOI.


Kenyon, Richard. "Tiling a polygon with parallelograms." Algorithmica 9, no. 4 (1993): 382-397.
DOI.

Here are two untilable examples from the Kannan et al. paper:
     
