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)?

Firstly, since the circle is inscribed in a rectangle the rectangle is a square.

Assume the circle is of radius 1.

The coordinates of the corner of the smaller rectangle is ((1-2h),(1-h)) where h is the height of the smaller rectangle.

Using the equation of a unit circle, x^2 + y^2 = 1, and putting in the above coordinates we get the equation (1-2h)^2 + (1-h)^2 = 1.

This simplifies to 5h^2 - 6h + 1 = 0,
which is the same as (5h-1)(h-1) = 0.
Therefore, h=(1/5),1 but h cannot =1.

So the height of the rectangle is 1/5 the radius, and the length is 2/5 the radius. We can therefore fit 10 of these inside the larger rectangle heightwise, and 5 lenghtwise. Which means that we can fit 50 of the smaller rectangles inside of the larger rectangle, regardless of the size of the circle.

50 smaller rectangles fit in the larger rectangle.

Posted by wonshot on 2003-11-20 15:16:34