A polyomino puzzle Is there a polyomino such that it can be glued to an I-shaped pentomino and to a X-shaped pentomino to obtain the same polyomino?  
Or is there simple proof for non-existence of such polyomino? [Edit: See "I-shaped" and "X/(+)-shaped" pentominos below:]
  
 A: This isn't a real answer, but if you allow "infinite polyominos" then you can do it.  Consider the following "infinite polyomino", without the yellow included:

Then adding a either "+" or an "I" pentomino where indicated in yellow will give you the same result up to translation.  In fact, you can add on $n$ "+" pentominos or $n$ "I" pentominos to this so that they'll give you the same answer, for any $n \in \mathbb{Z}$. (Yes $\mathbb{Z}$, as long as you consider "adding a negative pentomino" to mean what I think it should mean :) )
A similar construction works for any pair of finite polyominos.
A: By gluing, if you allow overlapping of a square: e.g. the top square of an "I" shaped "tri-omino" could be glued to the bottom square of the "+" pentomino (vertically aligned), while its center square (of I tri-omino) could be glued orthogonally to the 2nd square from the top of the I pentomino.) This would result in a matching 7-ominos (heptominos). (See my (pitiful) LaTeX attempt at constructing the figure I'm alluding to - just pretend there are no gaps vertically!). 
If overlapping is not allowed, (i.e. gluing must occur from edge to edge of each respective polyomino, then I'm doubting the existence of a polyomino which can be attached to each separate figure with the result a match. But I've no proof, yet.  
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