IBM Research | Ponder This | May 1999 solutions
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May 1999

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First solution:
Set up coordinate axes x,y parallel to the sides of the rectangles. Consider the function . When integrated over each smaller rectangle it gives 0:

which is zero precisely when at least one of (x1 — x0) or (y1 — y0) is an integer -- that is, when the rectangle has at least one integer-length side. So the integral over the large rectangle, being the sum of the integrals over the small rectangles, is also zero, whence the large rectangle has at least one integer-length side.

Second solution:
Again line up coordinates x,y parallel to sides of the rectangles. Select a large prime integer p. Consider lattice points of the form (j/p,k/p) where j,k are integers. Move the coordinates so that no lattice points are on the boundaries of any rectangles. For each small rectangle, the number of lattice points inside the rectangle is divisible by p: it's the product of two integers (the number of points along the bottom edge, times the number along the left edge), at least one of which is divisible by p (by integrality). Therefore the number of lattice points in the large rectangle is divisible by p. Again this is the product of two integers, and since p is prime, one of the integers is divisible by p, whence the length of the corresponding edge is an integer (within an error of 1/p). Since p was arbitrary, that error must in fact be 0.

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