In qball's Circular Logic, we were asked to determine the Area of the region common to a circle and a square, one vertex of which was at the center of the circle. The figure was drawn to show that e (the edge of the square) was greater than r (the radius of the circle), and the Area resolved to 25% that of the circle, or πrČ/4.

But what if e were less than r? The second and third verteces would be inside the circle, and the Area becomes a function of e (expressed as a fraction of r). When e is less than r/(√2), the fourth vertex is within the circle, and the area of the common region becomes the area of the square eČ

But what is the function f(x) such that A=f(e) for r ≥ e ≥ r/(√2) ?

Note: we already know that f(r) = πrČ/4, and f(r/(√2)) = rČ/2, because although the derivative of the greater function becomes discontinuous at these points, the function itself does not.

It was a comment by Manolo on the page for the original Circular Logic problem that suggested this one to me.

(In reply to A solution by Nick Reed)

Nick, just a minor correction: I believe that the area of each of the right angled triangles is half that: or (er * Sin a)/2. Other than that you have a very tight proof. Mine was full of messy Arctangents - yuk.

Posted by Eric on 2003-09-07 10:01:59