# Tiling rectangles with T pentomino plus rectangles

Also in this series: Tiling rectangles with F pentomino plus rectangles

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with U pentomino plus rectangles

Tiling rectangles with V pentomino plus rectangles

Tiling rectangles with W pentomino plus rectangles

Tiling rectangles with X pentomino plus rectangles

The goal is to tile rectangles as small as possible with the T pentomino. Of course this is impossible, so we allow the addition of copies of a rectangle. For each rectangle $a\times b$, find the smallest area larger rectangle that copies of $a\times b$ plus at least one T-pentomino will tile. Examples shown, with the $1\times 1$ or the $1\times 2$, you can tile a $3\times 3$ as follows: Now we don't need to consider $1\times 1$ or $1\times 2$ any longer as we have found the smallest rectangle tilable with copies of T plus copies of $1\times 1$ and $1\times 2$.

There are at least 10 more solutions. I tagged it 'computer-puzzle' but you can certainly work some of these out by hand. The larger ones might be a bit challenging.

I assume these are the remaining ones you found as well; for 2x3

a simple 7x5 = 35 one: for 2x2

for 1x5:

for 2x4:

for 2x5:

for 3x4:

and finally a large one for 3x5

which uses the same central figure formed by the Ts as the 1x5.
28x40 = 1120 • Nice. Will have to wait for confirmation of minimality until I get back to Australia as my net connection there seems to be down. May 4 '18 at 9:36
• I got a friend to power cycle my router, back online now. 2x3 into 5x7 is minimal. So are 2x2 into 8x10 and 1x5 into 10x11. May 6 '18 at 5:30
• @theonetruepath I think I found the remaining ones, and I suspect there are none left. Forming long straight boundaries with just Ts is hard. May 16 '18 at 19:15
• Yup all minimal except I can't confirm the 3x5 as my program hasn't got there yet. When I re-ran it after reconfiguring to avoid the bug which made me miss solutions occasionally, it found the 12x15 but I forgot to remove the 16x15 for the 2x5 which it had previously found instead. I think it's time to award the answer on this one... May 17 '18 at 0:16
• @theonetruepath thanks. If there are 10 more solutions (other than 1x1 and 1x2) and your program hasn't found the 3x5 yet, we're still missing one as Riley found 3 and I found 7. May 17 '18 at 6:33

Here's three of them

$3\times 6$ tiled with $1\times 4$

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$4\times 4$ tiled with $1\times 3$

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$8\times 8$ tiled with $1\times 6$. Not sure if optimal.

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• All optimal yes Apr 17 '18 at 4:17