# Pairs and pieces

• Sep 19th 2011, 06:51 AM
GGPaltrow
Pairs and pieces
Find all such pairs of positive integers (m,n), for which a rectangle with dimensions of m x n can be built from such pieces created from 6 squares marked as "O" (note that you can both rotate and flip the pieces):
Code:

O O OOOO
(if it's not clear, it's a F-type shape)

OK. So we now that we can create a 3x4 rectangle by simply putting together this shape and the one rotated by 180^:
Code:

OOOO  O O
As we can create 3x4, we can also create 4a x 3a for a positive integer a. But what else? I'm kind of clueless...
• Sep 19th 2011, 10:47 AM
Soroban
Re: Pairs and pieces
Hello, GGPaltrow!

Quote:

Find all such pairs of positive integers $(m,n)$,
for which a rectangle with dimensions $m\!\times\!n$ can be built
from pieces created from 6 squares as shown.
(Note that you can both rotate and flip the pieces):

. . $\begin{array}{cc}\square \;\,\square \;\; \\ [-2mm]\square\! \square\! \square\! \square \end{array}$

OK, so we know that we can create a 3x4 rectangle
. . by simply putting together this shape and the one rotated by $180^o.$

. . $\begin{array}{c}\blacksquare\! \blacksquare\!\blacksquare\! \blacksquare \\ [-2mm] \square \! \blacksquare\! \square\! \blacksquare \\ [-2mm] \square \! \square \! \square \! \square \end{array}$

We can also create $4a\times 3b$ rectangles for positive integers $a$ and $b.$
But what else? .I'm kind of clueless ...

Me too . . .

I've tried to find other possible rectangles, but have failed so far.
• Sep 20th 2011, 06:27 AM
GGPaltrow
Re: Pairs and pieces
Hmm.. maybe sth like this:

Suppose we have:
Code:

O O OOOO
in which the four squares form an edge. Then, let's try to put sth beside it (as putting above leads to a 3x4 rectangle):

Code:

  OO     O O OOO OOOOO
This can't be transformed into a rectangle as our "F" can't be put inside it in any way. The same goes with this flipped:
Code:

    OO     O O O OO OOOOO
Which is even more wrong as we leave our black F laying and blocked from the "open" side so we can only complete it to a 4x3.

Making any F lie on its "back" beside the black one doesn't make sense cause it leads to completing each of them to a 4x3 which means we eveantully end with 4ax3b.