Posted by J. Holmer on November 4, 2007, 4:44 pm, in reply to "coverings by rectangles and coverings by cubes"
Board Administrator
I think it would be good to provide a solution to #15 as part of your solution to #7, but don't get carried away with precision in the solution to #15; restrict to the case of two dimensions if you want. A rectangle with sidelengths c_1 and c_2 can be covered with \epsilon x \epsilon cubes such that the sum of the areas of the cubes is less than (c_1+\epsilon)(c_2+\epsilon). You'll need to appeal to Lemma 1.1 and Lemma 1.2.
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