At 10:24 AM 4/11/04, Erich Friedman wrote:
i have a rug in my bathroom in the shape of a rectangle, surrounded by a border of constant width, surrounded by another border of constant width. i have often wondered whether any such rug can have integer dimensions with the three areas being equal.
in other words, if the center rectangle is L by W, the inner border has width A, and the outer border has width B (with L, W, A, and B all being positive integers), can we have
2LW = (L+2A)(W+2A) and 3LW = (L+2A+2B)(W+2A+2B) ?
i looked for small solutions with a computer and couldn't find any. did i not search far enough, or is there some number theoretic reason solutions do not exist?
Thanks to Fermat, I can show that there are no solutions. Given L and W, we can solve the first quadratic for A. For A to be an integer it is necessary that the discriminant, L^2 + 6LW + W^2, be a square. Having solved for A, we can then express the discriminant of the second quadratic in B in terms of L and W. After some algebra, we get L^2 + 10LW + W^2, which must also be a square if B is an integer. Now, L^2 - 2LW + W^2 and L^2 + 2LW + W^2 are certainly squares, so that would amount to a total of four squares in arithmetic progression. But Fermat showed that no such squares exist. A proof can be found at http://www.mathpages.com/home/kmath044.htm . -- Fred W. Helenius <fredh@ix.netcom.com>