# Tiling rectangles with Heptomino plus rectangle #6

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Next puzzle in this series Tiling rectangles with Heptomino plus rectangle #7

The goal is to tile rectangles as small as possible with the given heptomino, in this case number 6 of the 108 heptominoes. 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 heptomino will tile.

Example with the $1\times 1$ you can tile a $2\times 5$ as follows: Now we don't need to consider $1\times 1$ further as we have found the smallest rectangle tilable with copies of the heptomino plus copies of $1\times 1$.

I found 14 more. I considered component rectangles of width 1 through 11 and length to 31 but my search was not complete.

List of known sizes:

• Width 1: Lengths 1 to 8, 10 to 12
• Width 2: Lengths 2, 3, 5
• Width 3: Length 5

Most of these could be tiled by hand using logic rather than just trial and error.

## 2 Answers

Finally, a more interesting heptomino :) (in the sense that previous ones all had generalizable solutions who looked very much like this hexomino)

Here's the minimal solution for $1 \times 2$:

$3 \times 6 = 18$ and for $2 \times 2$:

$6 \times 13 = 78$ For $3 \times 5$:

$19 \times 22 = 418$ My program found another one for $2 \times 7$:

$21 \times 30 = 630$ a very narrow one for $1 \times 10$:

$6 \times 31 = 186$ another one for $1 \times 11$:

$12 \times 32 = 384$ and another one for $1 \times 12$:

$12 \times 26 = 312$ This is probably the $1 \times 8$ solution you're looking for:

$17 \times 22 = 374$ I like how this one and Jaap's attempt are fundamentally different; this one is 'chaos' and the other one 'order'. It's asymmetric but it can be turned in a symmetric one; there are two ways to tile the irregular shape formed (twice) by the darker shaded polyominos. If you use the same one for both, you get a symmetric solution.

Here is the minimal solution for $1 \times 9$:

$19 \times 22 = 418$ • Yup 1x1=3x6, 2x2=6x13 both minimal. From 1x1 to 1x7 all would make nice hand tiling puzzles. – theonetruepath Jun 19 '18 at 7:07
• 3x5=19x22 is optimal yes. – theonetruepath Jun 20 '18 at 7:58
• got an extra dollar sign in the last one – SilverCookies Jun 20 '18 at 7:58
• I love the symmetry of all of these - including the asyymetric 2x7! – Chris Jun 21 '18 at 16:58
• Your net area just shot up to 2400, time to award I think – theonetruepath Jun 22 '18 at 10:20

Here are a few more solutions.

$1\times3$

$1\times4$

$1\times5$

$1\times6$

$1\times7$

$2\times3$

$2\times5$

Edit:

Here is a $1\times8$ solution that is surely non-optimal.

• Yup 1x3,4,5,6,7; 2x3,5 all optimal – theonetruepath Jun 20 '18 at 3:50
• @theonetruepath I added a 1x8 solution that is surely not optimal. – Jaap Scherphuis Jun 20 '18 at 6:17
• The 1x8 is a thing of beauty... but not optimal, as you say. My optimal solution is asymmetric, I haven't searched all solutions to see if there's a symmetric one. – theonetruepath Jun 20 '18 at 7:56