Saturday, December 1, 2007

Solutions and Discussion re Squares problem

Since I only received one comment from the squares problem, I guess readers were either bored by this question or are awaiting my solution(s)!

Solution I: Divide side AB into segments of lengths 2+x, 2-x. Here, x can be any real between -2 and 2. Similarly, divide BC into segments of lengths 2-y and 2+y. One can demonstrate that the order here is irrelevant. Then the product of the areas of either pair of non-adjacent rectangles can be expressed (after rearrangement of factors) as
(2+x)(2-x)(2+y)(2-y) = (4-x2)(4-y2).
From the restrictions on x and y, it follows that x2 is greater than or equal to zero and less than 4. Similarly for y.
Therefore, (4-x2)(4-y2) ≤ 4⋅4 or 16.
This also demonstrates that the maximum product occurs when x=0 and y=0! QED

Solution II (using the Arithmetic-Geometric Mean Inequality): We will prove a general result for squares of side m. This will be forthcoming and there may be a visual surprise! Stay tuned!

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