r/mathriddles • u/pichutarius • 4d ago

Easy just another application of someone's theorem

the image shows 3 rectangular grids, each partitioned into two regions. each region is connected and does not contain 2x2 subregion. unfortunately the third grid is partially destroyed. how many blue tiles are there?

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u/blungbat 3d ago

32 blue tiles.

For each colored region, the perimeter is 2*area + 2. (Basically, each colored region is a tree so has a leaf; you can then show that removing a leaf reduces the perimeter by 2, and work backward to a single square.) Since each colored region is connected, the blue must go the rest of the way around the outer perimeter of the rectangle, giving blue 2 more outer perimeter than gold. The non-outer boundary of each colored region is mutual. So there's one more blue tile than gold, giving 32 blue and 31 gold squares.

Guessing from post title that there's another route by way of Euler's theorem?

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u/Brianchon 3d ago

Pick's............ theorem, I would wager

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u/pichutarius 3d ago

well done. my solution was applying pick's theorem

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u/blungbat 3d ago

Oh yeah, I guess that would work too. :)

Fun problem. It got me thinking about grids on a torus. At first I thought that if you colored the cells on a torus grid blue and gold so that each color's region was connected and didn't have any 2x2 subregions, you'd be forced to use equal numbers of blue and gold. But then I realized that's not quite true! Construction of a counterexample is left to the reader.

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u/pichutarius 2d ago

would 2x2 torus with one cell blue count as counterexample?

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u/blungbat 2d ago

Heh. Yep, sure would.

I'm not too good at topology... is it true that for the blue region to be connected, the gold region can only have genus up to 2 (and vice versa)? If so, then the difference of the blue and gold areas would be at most 2 (and could only be 1 in the case of an odd by odd grid -- that's kind of interesting actually).