Strange tiles

Consider a $10 \times 11$ rectangular room which is divided into $110$ unit squares in the natural way by lines parallel to the sides of the room. We also have a collection of strange tiles. Each tile consists of six unit squares connected together in the following shape:

What is the maximal possible number of non-overlapping tiles we can pack into the room such that each tile is covering exactly six unit squares of the room? Give proof and diagram of the tiling.

#Combinatorics #Sharky

Easy Math Editor

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`2 \times 3`	$2 \times 3$
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`\sum_{i=1}^3`	$\sum_{i=1}^3$
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Comments

It is easy to pack 14:

I can also prove that this number is at most 15: the corner squares can't be used, and if a square on the edge is used then the two adjacent squares on the edge aren't used either, which totals to $2 \times 5 = 10$ squares from the long sides, $2 \times 4 = 8$ squares from the short side, and $8 \times 9 = 72$ squares from the middle, for a total of $10+8+72 = 90$ squares or $\left\lfloor \frac{90}{6} \right\rfloor = 15$ pieces.

Since the packing for 14 looks solid, I believe there shouldn't be any packing for 15, but I haven't proved it.

Ivan Koswara - 5 years, 7 months ago

I think you can lower the upper bound from 15 to 14 by looking at the squares adjacent to the corners.

If both of them are filled, then on at least one side, at least two squares will be empty. This should bring down the number of usable squares to $88$ .

Siddhartha Srivastava - 5 years, 7 months ago

My guess is 13, but I have no proof it is the best

Julian Poon - 5 years, 7 months ago

Do you have a diagram at the very least?

Sharky Kesa - 5 years, 7 months ago

Julian Poon - 5 years, 7 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$