If you were told to draw a rectangle along the lines of a sheet of graph paper such that its area is 40 squares, you could choose rectangles measuring 8x5, 10x4, 20x2 or 40x1.
For two of these, 8x5 or 10x4, you would find that you could draw a diagonal across the rectangle that would pass through exactly 12 squares.
What is the smallest number of squares that could be the area of three different rectangles whose diagonals pass through the same number of squares? How many squares does this diagonal pass through?
(In reply to
Solution by Jonathan Chang)
The answer is Uncertainty.
I too agree with you'll that 2 squares form a rectangle. However, the question mentions three different rectangles and it does not mention that the three different set of rectangle. For instance, if diagonal didn't mean from one corner to the opposite. As the question does not give the assumption that it must be normal diagonal, there is an uncertainty of what fraction of rectangle to be exposed above the diagonal.
Not only that the question does not mention that you must only three rectangles. For instance, if you add more rectangles in the diagram, many rectangles would be different and none of them the same. The answer is still turned up to be uncertainty since some diagonals long and some short and not all diagonal.
The question never mentions the size of the rectangle. For instance, if you take a cubic centimeter rectangle to place it with another, it is too small to see. The answer for this question turns up to be nil since there is no fraction of rectangle appears above the diagonal as a result of smallness.
Subject to the uncertainty of the question, the answer is obvious. Uncertainty!
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Posted by Dustin
on 2005-04-28 16:27:04 |