Polygon on the cartesian plane

Question

In the Cartesian plane is given a polygon $\mathcal{P}$ whose vertices have integer coordinates and with sides parallel to the coordinate axes. Show that if the length of each edge of $\mathcal{P}$ is an odd integer, then the surface of $\mathcal{P}$ cannot be partitioned into $2 \times 1$ rectangles.

I tried to break $\mathcal{P}$ into disjoint rectangles with odd sides. Now none of the rectangles can covered by those dominoes. But that never works, because the composition of two rectangles might be covered by dominoes. Can someone help me? Thanks.

Answer 1

Csizmadia, György; Czyzowicz, Jurek; Ga̧sieniec, Leszek; Kranakis, Evangelos; Rivera-Campo, Eduardo; Urrutia, Jorge, On tilable orthogonal polygons, Int. J. Pure Appl. Math. 13 (2004), no. 4, 443–459, MR2068717 (2005b:05061).

Summary: "We consider rectangular tilings of orthogonal polygons with vertices located at integer lattice points. Let $G$ be a set of reals closed under the usual addition operation. A $G$-rectangle is a rectangle at least one of whose sides is in $G$. We show that if an orthogonal polygon without holes can be tiled with $G$-rectangles then one of the sides of the polygon must be in $G$. As a special case this solves the conjecture that domino tilable orthogonal polygons must have at least one side of even length. We also explore separately the case of orthogonal polygons placed in a chessboard. We establish a condition which determines the number of black minus white squares of the chessboard occupied by the polygon. This number depends exclusively on the parity sequence of the lengths of the sides of the orthogonal polygon. This approach produces a different proof of the conjecture of the non domino-tilability of orthogonal polygons without even length sides. We also give some generalizations for polygons with holes and polytopes in 3 dimensions.''

What might be an earlier version of this paper is available at http://www.cccg.ca/proceedings/1999/c32.pdf --- Gyorgy Csizmadia, Jurek Czyzowicz, Leszek Gasieniec, Evangelos Kranakis, Jorge Urrutia, Domino Tilings of Orthogonal Polygons, Canadian Conference on Computational Geometry 1999 Proceedings. The 2004 paper is available at http://people.scs.carleton.ca/~kranakis/Papers/Domjou.pdf

Answer 2

To me, this problem is all in the corners. Unless you're counting rectangles of infinitely small volume, the corners would have to be 90 degrees, and otherwise, a triangle is created, which cannot be divided into rectangle. Therefore, P must be a rectangle. Then, we can find that it will have an odd area, as it has two odd side lengths, and your dominoes have an area of 2 square units. You can't divide an odd number by 2 without a remainder, so the surface can't be covered completely. I hope this is what you're looking for, it seems like a simple answer and I may be misinterpreting the question.