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Slice a Circle to fit a Rectangle (Posted on 2007-08-22) Difficulty: 4 of 5
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.

No Solution Yet Submitted by Jer    
Rating: 4.1667 (6 votes)

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Solution part 2 | Comment 4 of 10 |

The width will be 2 cos(x) where x is the angle that the line connecting a diameter midpoint to the point of tangency makes with a line normal to the diameter. The length will be 2 + 2 * sin(x).

The angle x cannot exceed the point where one semicircle would extend past the side of the rectangle as determined by the flat portion (diameter) of the other semicircle.  That would occur when x = 60 degrees and the width is 1.

Multiplying the width * length gives 4*cos(x) + 4*sin(x)*cos(x), or 4*cos(x) + 2*sin(2*x). The derivative of this is -4*sin(x) + 4*cos(2*x). Setting this to zero gives cos(2*x) = sin(x), a condition satisfied when x = 30 degrees.  However, this turns out to be a local maximum, rather than minimum.  Graphing the original equation for the area shows this, and also shows that the minimum area is achieved when the width is in fact 1 at x = 60 degrees. At that position, the length is equal to 2 + sqrt(3) ~= 3.732050807568877 for an area of (2 + sqrt(3))/4 ~= .9330127018922193 relative to that of the square.

  Posted by Charlie on 2007-08-22 14:52:28
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