Suppose you had a square intersecting a circle like in Circular Logic, where one vertex is inside the circle and three are outside.

If the vertex inside the circle was not located at the center, but anywhere inside the circle, such that segments ending at that vertex were of length a and b and the circle had radius r, what would be the area of their intersection be?

Note: A geometric solution is expected, without using calculus.

(In reply to re: can this be simplified? -- spoiler by Bractals)

hmmmm...  that would be a whole different calculation. I can't imagine one formula would cover such different situations.

Also, Gamer didn't include a measure for the side of the square in the formulation of the puzzle.

Posted by Charlie on 2006-08-09 13:46:54