# Tiling rectangles with V 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 U 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 V 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 V-pentomino will tile. Examples shown, with the $1\times 1$, $1\times 2$ or $2\times 2$, you can tile a $3\times 3$ as follows: Now we don't need to consider $1\times 1$, $1\times 2$, or $2\times 2$ any longer as we have found the smallest rectangle tilable with copies of V plus copies of each of those three.

There are at least 20 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 are (most of) the remaining ones. An easy one for 1x5:

and a more difficult one for 1x6:

1x7 takes a lot more:

1x8:

1x9:

1x10:

1x12:

2x3:

2x5:

2x6:

2x7:

2x8:

2x9:

3x4:

3x5:

3x7:

3x8:

4x5:

4x6:

5x6:

I assume the number of solutions here is infinite (probably in both directions), I'll post more when I have them.

• The 1x5 into 5x6 is minimal, your 1x6 into 6x10 is nice, but there's a smaller one – theonetruepath Apr 21 '18 at 0:19
• @theonetruepath ah, of course. I think I found it. – Glorfindel Apr 21 '18 at 10:37
• Yup that's minimal – theonetruepath Apr 21 '18 at 11:10
• 1x5, 1x6, 1x7, 1x8, 1x9 all minimal. At least 11 more to find. – theonetruepath May 9 '18 at 10:00
• I think I got (almost) all of them now. – Glorfindel May 21 '18 at 18:17

20's a lot, but here's a few to get it started.

$3\times 6$ tiled with a $2\times 4$

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

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

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$6\times 6$ tiled with a $4\times 4$

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• Yup all four are optimal – theonetruepath Apr 17 '18 at 3:42