 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  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 One Solution | Comment 2 of 10 | `Let A, B, C, D, E, and F (in that order) lie ona horizontal line such that AE is the diameterof the upper semicircle with center C and BF isthe diameter of the lower semicircle with centerD. Let G be the point on the upper semicirclesuch that GF is tangent to the upper semicircle.The sides of the enclosing rectangle are paralleland perpendicular to the line GF.`

Let x = |CD| and t the measure of angle CFG. Then

A = [2 + x cos(t)][2 - x sin(t)]             (1)

where

1
sin(t) = -------                             (2)
1 + x

Plugging x from (2) into (1) gives

cos(t)^3 + 2 sin(t)[1 + sin(t)]
A = ---------------------------------        (3)
sin(t)

Setting dA/dt = 0 gives

2 cos(t)^3 - 2 cos(t)^2 - 3 cos(t) + 2 = 0   (4)

Of the three real roots of this equation, only
one lies in [-1,1]. Therefore,

cos(t) ~= 0.57318274451645

t ~= 55.02753223576 degrees

x ~= 0.22036411444

Plugging these values into (1) or (3) gives
the ratio

A/4 ~= 0.967166262

This value agrees with what I get using Sketchpad.

 Posted by Bractals on 2007-08-22 12:42:24 Please log in:

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