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)?
In thinking rectangles, not squares, if the inscribed circle had its centre at the intersection of the diagonals of a rectangle whose length was 3 times the diameter of the circle, then 6 smaller rectangles, as defined, would cover the larger one; the width of the smaller rectangle would be the radius of the circle.
Considering Tristan's '2 solutions', two rectangles with a width of the diameter of the circle would cover a rectangle of length 4 times the circle's diameter.
I am wondering if further cases of this type exist, including a mix of horizontal and vertical rectangles.

Posted by brianjn
on 20031123 23:06:08 