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.

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

TopNewestIt 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.

Log in to reply

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 \( 88 \).

Log in to reply

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

Log in to reply

Do you have a diagram at the very least?

Log in to reply

Log in to reply