Polyomino question

[Dr. Gordon Hamilton]

Jack,

The best that can be done is 10. There are 11 different solutions.
The 9-solution below isn't right because it does not include the L 
polyomino. I have been presenting this problem to elementary school students 
in Calgary, Canada for the last year. You can see the problem at: 
www.galileo.org/math/puzzles.html under the name "Babushka Squares".

Gord




Subject: Re: Polyomino question


>
>> For n = 5, the best I can do is 9:
>>   O
>> OOOOO
>> OOO
>> I don't know if this is unique; I would be very surprised if 8 is 
>> possible.
>
> It's not unique.  See also the somewhat similar:
>
>     O
>  OOOOO
>   OOO
>
> Eight is not possible.  First you need to have five in a row:
>
>  OOOOO
>
> Then consider the "X" pentomino:
>
>   O
>  OOO
>   O
>
>
> In order to overlay this onto the
> five-in-a-row without exceeding a total area of 8, you have three
> basic possibilities (ignoring reflections and rotations):
>
>   O        O         O
>  OOOOOO   OOOOO    OOOOO
>   O        O         O
>
> Now consider the "W" pentomino:
>
>    O
>    OO
>     OO
>
> No matter how this is oriented with respect to any of the three
> basic possibilities, it can only overlap two squares of the central
> row and at most one square not on the central row -- meaning that
> two of its five squares must be added to the enclosing polyomino.
> This pushes the minimum area to 9, and since that has already been
> shown to be possible, the answer for n = 5 is shown to be exactly 9.
>
>
>   Jack
>
>
>
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