Here is a problem which arose from an earlier question. I'll change the terminology but not the question: A polyomino is a region with a connected interior made by joining one or more unit squares edge to edge. We will allow translating, reflecting and rotating polyominos but always with corners at integer points (and sides horizontal and vertical.)

It seems obvious that any two polyominoes, neither a single square, can be positioned so that their interiors are disjoint but they share at least two edges. Find a proof, preferably an elegant one, or perhaps a counter-example.

If we only insisted on being connected at corners this would not be true.

Here is a partial argument which is perhaps not that elegant and perhaps not that easy to finish. Let us say that one is red and the other blue. Every polyomino has a minimal bounding rectangle. I won't illustrate, but clearly if each has two consecutive filled squares on the boundary of the bounnding rectangle then those can be put in contact. So assume that the blue one does not have two consecutive colored squares on the boundary. This means that all four corners of the bounding rectangle are unoccupied. If the same is true of the red one then we can position them as shown to the left below (only a small part of each polyomino is indicated). White squares are definitely empty. The solid colored squares are definitely filled and include the *rightmost* blue square of the top or of that polyomino and the *leftmost* red square at the bottom of the red polyomino. The lightly colored squares with question marks might be empty or full. The green lines are not crossed by either polyomino in their current positions. The red polyomino can be moved down by one and there are either two edges of contact or else we can move it down another one and definitely have two edges. To the right is a similar situation where the red poyomino has a square of the bounding rectangle filled.

It remains to consider the case where the blue polyomino does not have two consecutive filled squares on the boundry of the bounding rectangle but the red one does, but not at any corner.