Re: A Putnam Geometry Problem
Was this really on the Putnam? [seems too easy]
Answer:
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Such a rectangle must be able to hold two squares of side
1/sqrt(2), so it has to have length AT LEAST twice that, or
sqrt(2). The width must be 1 to hold a square with side
1-epsilon for all epsilon>0, i.e., alpha = sqrt(2).
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