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)?
Well based on pd's picture lets make a unit circle, and a unit square on the origin. The point where the corner of the small rectangle to a point on the circle is point ((1-2x),(1-x)). To find the point, make a quadratic formula adapted from x^2+y^2= 1: (1-2x)^2 + (1-x)^2 = 1
1 - 4x + 4x^2 + 1 - 2x + x^2 = 1
5x^2 - 6x + 2 = 1
5x^2 - 6x + 1 = 0
Using the quadratic formula, you get roots 1 and .2. 1 cant be a root so x = 0.2.
Thus the corner is (0.6 , 0.8).
Eventually, you get that the 4 corners of the small rectangle is (1,1), (0.6,1), (1,0.8), (0.6, 0.8). You find that it is a 0.4 x 0.2 rectangle. Thus, 5 rectangles can fit the length of the large square, and 10 rectangles can fit the width of the large square. 5 x 10= 50.
50
solution
