Fox 337

This one is related to Fox 334, and should be simpler.

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Fox 320 - Solutions

This fox has been discussed extensively. See here...

1. Calculus by Bleaug

He says: "I spotted this problem in a Paul Halmos book "Problems for mathematicians, young and old, 1991", actually in a French translation. He provided the solution below (reformulated by myself) as being proposed by Hugh Montgomery in 1985 in some Math Conference."






2. Checkerboard Solution by Rochberg and Stein:

1. Call sub rectangles as TILEs.


2. Start from the lower left corner of the overall rectangle (let's call this point as the origin)Draw horizontal and vertical lines separated by 1/2=0.5 units starting from the origin.


3. This will create 0.5 x 0.5 squares (and possibly rectangles around 2 edges -top and/or RHS)


4. Color the first square (that has the origin as one of its corners) as black the next one as white, and so on to generate a checkerboard-look.


5. Each TILE will have equal areas of black and white (Why?)


6. Therefore the overall rectangle will have equal areas of black and white.


7. So overall rectangle has at least one integer side.
   

H means the horizontal side is integer, and V means that the vertical side is integer. a1 is the square at the origin. There could have been a smarter combination, but this simple one illustrates the process clearly. This looks like the closest geometric solution we can get -at this time.

When there's hardly no day nor hardly no night
There's things half in shadow and halfway in light

3. Induction by Robinson

1. Assume that each H-tile has a width of 1, and each V-tile has a height of 1. Note that any rectangle can be converted this way without distorting the original problem. (This may increase the number of tiles significantly though)


2. Chose any H-tile, say T(0). (If there is no H-tile, then the result is immediate)


3. If there are H-tiles whose lower border shares a segment with T(0)'s upper border, choose one and call it T(1).


4. Otherwise only V-tiles share this border. In this case, we can expand T(0) upward 1 unit. This does not increase the number of H-tiles. Also, the cut V-tiles still have height 1. (They are still V-tiles)


5. Continue expanding T(0) until either the top of the rectangle is reached or a choice of an adjacent H-tile T(1) is possible.


6. Then repeat the same process from T(1). (Continue upward similarly from T(1) to get T(2), and so on...)


7. This will result in a chain of T(0), T(1), T(2), ... , T(m).


8. Starting from T(0) again, work downward similarly to obtain a bigger chain:
   T(-n), T(1-n), ... , T(0), T(1), ... , T(m-1), T(m) of H-tiles stretching from bottom to top.


9. Remove these tiles and slide the rest together to get a rectangle with fewer H-tiles.


10. Induction applied to this smaller rectangle yields the result for the original rectangle.
    

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Fox 320

Bleaug has a solution for this general case.

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Fox 313

Bleaug submitted the following fox, saying: "a problem which can be expressed in geometric terms and that is elegantly solved using pure algebraic arguments, still giving deep insight."

General case of this problem can be found in literature (with at least 14 proofs :)
But let's try this smaller version here.

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