# Tiling rectangles with Hexomino plus rectangle #2

Previous puzzle in this series: Tiling rectangles with Hexomino plus rectangle #1

Next puzzle in this series: Tiling rectangles with Hexomino plus rectangle #3

The goal is to tile rectangles as small as possible with the given hexomino, in this case number 2 of the 25 hexominoes which cannot tile a rectangle alone. 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 of the given hexomino will tile.

Example shown, with the $1\times 1$ (or the $1\times 2$) you can tile a $2\times 4$ as follows:

Now we don't need to consider $1\times 1$ (or $1\times 2$) further as we have found the smallest rectangle tilable with copies of the hexomino plus copies of $1\times 1$ (or $1\times 2$).

I have tilings for another nine rectangles: Hexomino plus $1\times 3$, $1\times 4$, $1\times 5$, $1\times 6$, $1\times 7$, $1\times 8$, $2\times 2$, $2\times 3$, $2\times 4$

No Computer All of them could be tiled by hand (with significant effort in some cases), so I'm making this a no-computer puzzle. This also means please don't look up answers on the web... if you post an answer it should be because you found it 'by hand'. This does not preclude you from for example using an image program to manipulate shapes on the screen, just from using a computer to search for or automate the arrangement.

Here are the first few in the order I found them.

$2\times2$

$4\times4=16$ square

$1\times3$

$6\times6=36$ square

$2\times3$

$10\times12=120$ rectangle

$2\times4$

$10\times14=140$ rectangle

$1\times4$ (Updated)

$7\times10=70$ rectangle

$1\times7$

$10\times10=100$ square

• Nice work. All optimal except the 2x3. – theonetruepath Jun 9 '18 at 7:22

Here's a smaller solution for $2 \times 3$:

8x12=96

and Jaap's $1 \times 7$ solution was a great help in finding one for $1 \times 6$:

10x12=120

Here's a (computer-assisted, I admit) generalizable solution for $1 \times n = 4k$. By subdividing the rectangles, you can find solutions for $n$ not divisible by $4$. In those cases, the 'padding' (i.e. areas with just rectangles) don't need to be so large.

When $n$ is divisible by $4$, $1 \times n$ fits in a $2n \times 2n+1$ (left)
When $n$ is even but not divisible by $4$, $1 \times n$ fits in a $2n \times 3n+1$ (right)
When $n$ is odd, $1 \times n$ fits in a $2n \times 5n+1$

• If I can't find anymore today, I'll run my computer program to see if there are solutions for more rectangle sizes. – Glorfindel Jun 9 '18 at 7:51
• Yup 2x3 and 1x6 both optimal – theonetruepath Jun 9 '18 at 8:40
• The (2n)x(2n+1) when n is divisible by 4 is optimal for n=8 (size 16x17) but not n=4 (already found as 7x10=70 rather than 8x9=72). I might do a special run for n=12, I hadn't got that far. The right hand one is probably never optimal as you already have a better 1x6. I stopped before finding the 1x10. – theonetruepath Jun 9 '18 at 21:04
• ...which means that from my list above, only 1x5 remains unsolved... – theonetruepath Jun 9 '18 at 21:06
• Let's leave the 1x5 for somebody without a tiling program :) – Glorfindel Jun 9 '18 at 21:30