The goal of this problem is to slice a unit circle into two pieces which can be fit into a rectangle of minimal area.
Consider these three methods:

One: Slice the circle across its diameter and slide the pieces along each other a little.
What distance between the radii of the two semicircles gives the smallest rectangle? This rectangle has a smaller area than square that circumscribes the original circle. What is the ratio of rectangle to square?

Two: Slice the circle along a diameter and place these into a rectangle so the straight edges of each semicircle are along opposite edges of the rectangle.
What is the ratio of rectangle to square this time?

Three: Slice a segment off of the circle and place this segment to the side. The large piece's straight edge is along one side of the rectangle. The straight edge of the segment is tangent to the large piece but not necessarily parallel to a side of the rectangle.
What chord length minimizes the rectangle? What is the ratio?

Note: Part Three is may be particularly difficult to find an exact solution for. If you have a method of finding a good approximation feel free to share your results.

(In reply to re: computer exploration of part 3 by Charlie)

If the .4125 thickness of the segment is taken as exact (it seems to be), then the chord length is 2*sqrt(1-.5875^2)~=1.618448330963951. 

The minimum occurs when the rightmost point on the segment in its new position is directly over the rightmost part of the circle.

The angle I've called alpha, the angular measure from the top of the circle (taking the segment from the left side), is 90 minus twice the arctan of half the chord length, or about 12.03875807124709 deg.

The height of the rectangle is about 2.403427777711721, with the top being tangent to the segment. The width is of course 2-.4125 = 1+.5875 = 1.5875, for an area of 3.815441597117356, or .9538603992793391 of the area of the original square.

The leftmost point on the segment does not quite reach the left side of the rectangle (defined by the chord). It's close enough that it makes one think that the true solution might have this gap closed, but then again, maybe not--maybe this is just a coincidence that it lies so close to the left edge.

Posted by Charlie on 2007-08-23 16:21:41