Saturday, February 16, 2019

geometry - Polyomino T hexomino and rectangle packing into rectangle


Let's pack some (one or more) T hexominoes together with some (one or more) small $a\times b$ rectangles into some bigger $m\times n$ rectangle without holes and overlapping pieces.


For example, I can pack two T hexominoes and eight $1\times1$ rectangles into a $5\times4$ rectangle:



Task


Find as many as you can different integer pairs $\{a,b\}$ so that one or more rectangles of size $a\times b$ can be packed together with T hexominoes into a $m\times n$ rectangle without holes and overlapping pieces.


Provide images of solutions if you claim that some combination is possible. Please, put your images in spoiler tags so that other users can try find themselves.


If some answers will have same number of pairs $\{a,b\}$, then answer with smaller total area of outer rectangles on example images is preferable. If area also equal, then earlier posted answer is preferable.


EDIT: Please, add to your answers total area of outer rectangles. If one tiling implicitly includes several $a\times b$ rectangles, then multiply area by number of combinations it includes.



Notes


I can guarantee that there exist solutions following integer pairs:
$1\times1$, $1\times2$, $1\times3$, $1\times4$, $1\times5$, $1\times6$, $1\times8$, $1\times9$,
$2\times2$, $2\times3$, $2\times4$, $2\times5$, $2\times6$


I'm wondering why a $1\times7$ rectangle solution is hard to find (maybe it's impossible?) when there exist solutions for with $1\times8$ and $1\times9$ rectangles.


1 week after posting, I'll put up my own answers for unfound combinations.



Answer



2x2 - area 108 - optimal



2x2




2x3 - area 72 - optimal



2x3 2x3 2x3



2x4 - area 108 - optimal



2x4



3x4 - area 84 - optimal




3x4



1x5 - area 54 - optimal



1x5



2x5 - area 304 - optimal



2x5




3x5 - area 576 - optimal



3x5



2x6 - area 240 - optimal



2x6



1x7 - area 1034




1x7



1x8 - area 432 - optimal



1x8



3x8 - area 5880



3x8




1x9 - area 585 - optimal



1x9



Note that by subdividing the yellow rectangles:
2x3 indirectly solves 1x1, 1x2, 1x3
2x4 indirectly solves 1x4 and 2x2
2x5 indirectly solves 1x5
2x6 indirectly solves 1x6



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