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 e²
But what is the function f(x) such that A=f(e) for r ≥ e ≥ r/(√2) ?
Note: we already know that f(r) = πr²/4, and f(r/(√2)) = r²/2, 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.
