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)?
The problem states that "one corner is on the edge of the circle" but not that "only one corner is on the edge of the circle". If two corners are on the edge of the circle they lie at the ends of the diameter that connects opposite sides of the square, that is at the points of tangency. As the smaller rectangle then also has two vertices that coincide with two vertices of the square, it takes up half the square, and two such rectangles fill the square.
This corresponds to the other solution of the quadratic.
Posted by Charlie
on 2003-11-20 15:45:07