Suppose you have a circle that is perfectly inscribed in a rectangle.
A smaller rectangle is placed on top of the first one, such that one corner is on the edge of the circle and the opposite corner matches a corner of the larger rectangle.
If the smaller rectangle is twice as long as it is high, how many of them will fit into of the larger one (without overlapping, of course)?
2
The larger is a square (the only rectangle one can inscribe a circle in). One corner of the smaller rectangle matches the lower left corner of the square. The opposite corner is at the point where the circle touches the square, halfway up the square's height. Its length is then twice its height, and one more can fit in the square right above it.
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Posted by C.B.
on 2003-11-20 16:54:08 |