Imagine you have a rectangular piece of fabric of dimension 3x4. I want to cut a circular piece out, and obviously the largest diameter possible is 3. But if I want a larger circle I can make a straight cut through the rectangle and then sew the pieces back together in any orientation and position (the pieces cannot be joined at points and cannot overlap). Now with this 1 cut and sewing operation what is the largest circle I can make? What about with n (finite) operations? 24.255.30.187 (talk) 05:44, 2 June 2013 (UTC)[reply]

Intuitively, I'd expect the best config for one cut to be like so (forgive the crude pic):

+        +------+
|\      /___    |
| \____//   \   |
|      /     \  |
+-----+       +-+

That is, the two semi-circles should be cut out from opposite sides, and be tangent to each other. StuRat (talk) 06:05, 2 June 2013 (UTC)[reply]

Each cut has to be a single straight line. 24.255.30.187 (talk) 06:08, 2 June 2013 (UTC)[reply]

The trivial upper bound for the diameter is ${\displaystyle {\sqrt {2\times 3\times {4 \over \pi }}}}$which is 3.90882 . Bo Jacoby (talk) 06:16, 2 June 2013 (UTC).[reply]

You can do StuRat solution with a single cut by cutting along the tangent of the two circles this gives a radius of approximately 1.56. Cutting along the diameter of the circle gives a circle with a radius of about 1.75.--Salix (talk): 08:42, 2 June 2013 (UTC)[reply]

How do you get 1.56? The diagonal of the rectangle (length 5) should pass through the point of tangency, so the radius should be 1.25.--80.109.106.49 (talk) 09:10, 2 June 2013 (UTC)[reply]

I've got my numbers experimentally. For the first case if the radius is r and the angle of the tanget is θ then for the semi circles to fit we require ${\displaystyle 2r+2r\sin \theta =4}$and ${\displaystyle 2r\cos \theta =3}$. Getting expression for cos and sin and using the pythagorean identity gives an equation in r with exact solution 25/16 ≈ 1.5625. The second is harder to give an explicit solution for, an aproximate solution has an angle of 45.4334° and r ≈ 1.75189.--Salix (talk): 10:31, 2 June 2013 (UTC)[reply]

I just realized my mistake. The diagonal isn't a diameter.--80.109.106.49 (talk) 13:06, 2 June 2013 (UTC)[reply]

Here is r=1.75
However, if the grey stripe was a trapezoid or even a triangle instead of a rectangle, the circle might get a bit bigger. --CiaPan (talk) 09:09, 3 June 2013 (UTC)[reply]

Should Bo Jacoby's upper bound not be ${\displaystyle {\sqrt {4\times 3\times {4 \over \pi }}}}$? A quick back of the envelope estimate (by assuming that you can make the whole area of the sheet into the circle) puts that at ${\displaystyle 4\times {\sqrt {3 \over \pi }}}$, which seems to match the numerical value given MChesterMC (talk) 15:13, 3 June 2013 (UTC)[reply]

You are right. Thanks. Bo Jacoby (talk) 19:45, 3 June 2013 (UTC).[reply]