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 Slice a Circle to fit a Rectangle (Posted on 2007-08-22)
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)

Comments: ( Back to comment list | You must be logged in to post comments.)
 Part Three Analysis Comment 10 of 10 |
` `
`Let A(-cos(t),sin(t)) and B(cos(t),sin(t))be the endpoints of the spliting chord.`

The rotated and translated segment is
tangent to the large segment at point
C(cos(p),-sin(p)) with endpoints A' and B'
on lines y = sin(t) and y = -1 respectively.

Both t and p in (0,90). The angle mentioned
in the results post is t+p.

The edges of the enclosing rectangle are
y = sin(t), y = -1, x = -1, and x = f(t,p)

where

1+sin(t)
cos(p) = ----------
2*cos(t)

and

2-sin(p)*(1-sin(t))
f(t,p) = 1 + --------------------- - sin(t)*cos(p)
2*cos(p)

The area of the enclosing rectangle is

Area = cos(t)*{ 2-sin(p)*[1-sin(t)]

+ 2*cos(p)*[2-sin(t)*cos(p)] }

The Perl program finds the minimum for this area.

I tried to set d(Area)/dt = 0 to find the minimum,
but finally gave up after filling page after page
with calculations.

Edited on August 24, 2007, 4:44 pm
 Posted by Bractals on 2007-08-24 16:43:17

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