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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:]

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What do you mean by "the same polyomino" ? How could it stay the same if you had stuff to it ? –  Joel Cohen Jun 28 '11 at 22:42
oh sorry, what i wanted to say was can the polyomino obtained after gluing some polyomino to the I shaped pentomino and the polyomino obtained after gluing the same polyomino to the X shaped one be the same –  bleh Jun 28 '11 at 22:51
I'm wondering, whether we can use the fact that, on a chess board, the I-shape covers 3 squares of one color and 2 of the opposite, whereas the cross shape splits 4+1. This won't do it by itself, because translation by a single square switches the colors, but may be we can rule out some cases? –  Jyrki Lahtonen Jun 29 '11 at 9:52
An immediate consequence of that is that the color imbalance of the third polyomino must be 1 or 2. I don't see this leading anywhere now. –  Jyrki Lahtonen Jun 29 '11 at 10:13
@Jyrki: that was my first take on it - in fact, I think you can say that the color imbalance must be 2, since the equations aren't quite consistent otherwise, but simple parity doesn't say anything beyond that. –  Steven Stadnicki Jun 29 '11 at 21:57

3 Answers 3

up vote 14 down vote accepted

Here is one:


You get these two congruent polyominoes:

  X          Y
Y X X      Y Y Y
Y X        X Y
Y X        X X
Y          X
Y          X
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Nice. Is there anything to be learned from how you arrived at this? –  joriki Jul 1 '11 at 0:51
I arrived at it by proving its impossibility. Then I started to write up my proof here, and saw my mistake. –  TonyK Jul 1 '11 at 0:53
Specifically, I knew that the glued-together polyomino must have a column of five squares on the left, and a column of a single square on the right (or vice versa). I thought this led to a contradiction, but it turned out to lead to the solution. –  TonyK Jul 1 '11 at 0:55

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:

infinite poly

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.

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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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The heptomino was all I could think of, too. –  Jack Henahan Jun 29 '11 at 1:36
hi,thnx for the answer,but this puzzle is trivial if overlapping is allowed as we can glue a very large polyomino to the top of any pentomino to obtain the same large polyomino! –  bleh Jun 29 '11 at 6:37
I suppose that putting the polyominoes on a cylinder of girth 3 (or other changes in the topology) would also amount to bending the rules :-) –  Jyrki Lahtonen Jun 29 '11 at 18:14

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