# Tiling rectangles with U pentomino plus rectangles

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

Tiling rectangles with N pentomino plus rectangles

Tiling rectangles with T 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 U 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 U-pentomino will tile. Example shown, with the $1\times 1$, you can tile a $2\times 3$ as follows:

Now we don't need to consider $1\times 1$ any longer as we have found the smallest rectangle tilable with copies of U plus copies of $1\times 1$.

There are at least 6 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.

Here is a way to tile a

6x13 = 78

rectangle with U pentominoes and 1x4 rectangles, which is an improvement over @athin's 9x10 solution:

As a bonus, here are two suboptimal solutions, one of which is asymmetric:

For 1x5:

12x20 = 240

for 1x6:

14x24 = 336

and for 3x4:

19x40 = 760

• Yup that's optimal – theonetruepath Apr 20 '18 at 6:21
• @theonetruepath I̶ (my program) found two more, I believe we've reached the "6 more solutions" now. – Glorfindel May 17 '18 at 10:08
• ... make that three, actually. So I guess this question counts as solved as well. – Glorfindel May 17 '18 at 11:14
• Yup I haven't reached 19x40 yet, all others minimal. – theonetruepath May 17 '18 at 12:00
• FYI, my program takes a fixed rectangle size (e.g. 3x4) and then enumerates all 'boards' in increasing order. Larger rectangle sizes are evaluated much faster, and for this particular pentomino + rectangles with width > 1 there's another trick: you know that the U pentominos always come in pairs. – Glorfindel May 17 '18 at 12:04

Beside Riley's 2 solutions:

$1 \times 4$

Area: $10 \times 9 = 90$

$2 \times 3$

Area: $10 \times 7 = 70$

• Both nice solutions, the $2\times 3$ is optimal but there is a smaller area for $1\times 4$ – theonetruepath Apr 18 '18 at 18:05

Here's two to get this one started

$3\times 4$ using a $1\times 2$

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$7\times 4$ using a $1\times 3$

.

• Both optimal yes – theonetruepath Apr 17 '18 at 3:43