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Jul 04, 2009 20:39

Happy 4th of July!

This week, I played matchmaker between Erich Friedman and Alexandre Muñiz. (These links contain some necessary background on weightominoes and sumominoes, respectively.) As sort of a thank-you, Alexandre sent me a problem reminiscent of Erich's extension tilings (problem 5) last month.

Is there a n-omino for which all the weightominoes with weight n+1 formed from it can tile a rectangle with equal weight in each square?

To be clear, a basic n-omino has weight 1 in each square cell. Let's define a "weight-extension" as a polyomino where 1 of its cells weighs 2 and the rest of its cells weigh 1. Then there may be up to n weight-extensions of a given n-omino (less if the original is symmetrical), and the problem is to cover a rectangle with these weight-extensions so that every square weighs the same.

Of course, I don't know whether there's an answer for any polyomino (except for the trivial 1x1x2 for the monomino). But my intuition suggests that an asymmetrical octomino with extensions of weight 9 would be a good entry point.

puzzle, open, edited, math

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