Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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:]

enter image description here enter image description here

share|improve this question
1  
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
1  
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
+100

Here is one:

X
X X
X
X

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
share|improve this answer
    
Nice. Is there anything to be learned from how you arrived at this? –  joriki Jul 1 '11 at 0:51
3  
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
3  
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

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.

$\quad\square$
$\square\square\square$
$\quad\square$
$\quad\square$
$\quad\square$

share|improve this answer
    
The heptomino was all I could think of, too. –  Jack Henahan Jun 29 '11 at 1:36
4  
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

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.