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.

Although I did say that in the function f(e), e could be considered to be expressed as a fraction of r, for the sake of clarity, in my solution I kept the r notation rather than substitute the number 1

Thus, I wrote e when I meant the length of e but e/r when I needed the ratio, and I used the expression (r² - e²) instead of (1 - e²), etc.

Posted by TomM on 2002-07-19 02:14:38