Given: There is a square, ABCD, with side lengths l, a constant. Point E lies somewhere on side CD. Points B and E both lie on the perimeter of a circle, with center at point F, and with radius y. The length of line segment CE is labeled x. Finally, the space of overlap between circle F and square ABCD has a fixed area, equal to Z.
The goal: Find a function y=f(x) which gives the radius of circle F based upon the length of line segment CE. Use in the function, if necessary, the positive real constants l and Z.
For any curious, I was staring at some interesting tiling while in a bathroom at a university, and, being very bored, made up this problem. I'm not sure, however, how to solve it, and thought maybe someone else would like to.
Moving circle on a square with fixed overlap
Moderators: gmalivuk, Moderators General, Prelates

 Posts: 2
 Joined: Wed May 16, 2012 11:46 pm UTC

 Posts: 131
 Joined: Sat Mar 21, 2009 1:40 am UTC
 Contact:
Re: Moving circle on a square with fixed overlap
Unless I'm mistaken, we need a bit more information, because there are infinitely many circles with points B and E on their perimeters. Do you mean the one such that BE is a diameter? Or the one such that F lies on BC? Or some other circle?
 gmalivuk
 GNU Terry Pratchett
 Posts: 26818
 Joined: Wed Feb 28, 2007 6:02 pm UTC
 Location: Here and There
 Contact:
Re: Moving circle on a square with fixed overlap
There are, but that only means that the function will depend on more than just x. There aren't, after all, infinitely many such circles that would enclose a given area Z of the square.

 Posts: 2
 Joined: Wed May 16, 2012 11:46 pm UTC
Re: Moving circle on a square with fixed overlap
I believe there should only be one circle that meets all of the requirements for any given length of CE. If it helps, I was thinking chiefly of situations where the overlap was less than half the area of the square (in other words, where point F lies on the same side of line BD as point C).
Who is online
Users browsing this forum: No registered users and 5 guests