Gluing Polygons Consider all polygons whose vertices are lattice points and edges are parallel to the axes such that no more than two edges meet at a vertex. For two polygons A and B, define A+B be to the set of polygons which can be partitioned into two poygons congruent to A and B.
Given two polygons A and B of same area. Do there always exist a polygon P such that , there is a polygon X in A+P and Y in B+P such that X is congruent to Y? What can we say about area of P? 
Edit: I wanted to ask a more general question. If such a P exists, how will one go on constructing it?,And how to characterize the pairs for which such a P does not exist? Is this a known problem?? etc
 A: Define an inclusion of a polygon $A$ to be a subset of the complement of the polygon whose boundary minus $A$ consists of a single edge.  Inclusions sometimes contain other inclusions.  The congruence class of an inclusion is the congruence class of the pair (polygon, free edge).
Define the inclusion vector $Iv(A)$  to be the nonngegative linear combination of congruence classes of inclusions of $A$ describing all inclusions up to congruence.
Notice that for any $X$ in $A + P$, as long as neither $A$ nor $P$ fit inside an inclusion of the other, then $Iv(X) \ge Iv(A) + Iv(B)$ 
Let's say that a polygons $X$  and $Y$ bind to each other with strength $k$ if they can be arranged up to congruence so that their intersection is a connected set of $k > 2$ edges.  Then any inclusions $J$ of an element $X$ of $A + P$ that are not inclusions of $A$ or of $P$ must be polygons that bind to  $A$ and $P$ with total strength at least the (length of  the perimeter of $J$) - 1.
With the picture established by all these definitions, it's easy to make counterexamples.  For $A$ and $B$, start with large rectangles with crenelated boundaries (every other square along the boundary removed). There are necessarily 4 straight segments of length 2 on the boundary at the 4 corners, but that is all.  Now bore out at least 5 maximal inclusions from $A$ and $B$ in the form of $S$-patterned double spirals that do not bind to $A$ or $B$ anywhere except at the four straight stretches.  Make sure the maximal inclusions removed from $A$ and $B$ are not congruent, and that the resulting polygons $A'$ and $B'$ have the same area.    
The difference between the vectors $Iv(A)$ and $Iv(B)$ is sufficiently large that adding a polygon $P$ can never create enough inclusions to bridge the difference.
