I have known this fellow for years on my Cadd forum. He is a wild guy. Dave check out his angle on this.
You two should hang out Dave, you both give me a headache, Ha-ha just kidding. I still do not see this as proof though. Because we can only check it using real inputs.
William -- Let me see if I have this correct:
You have a circle of R1 and a non-diametral chord within that circle. That chord has a length of 2B. There is an inner circle, concentric with the first that is tangent to the chord. Therefore, this is a circle of radius R2 = {R1² - B²)^1/2. Right?
The area of the outer hole is A1 = pi * R1². The area of the inner hole is A2 = pi * (R1² - B²). So, assuming that 2B is a constant, A2 is not a constant.
It dawned on me later that the Area of the outer annulus is: A1 - A2 = pi * (R1² - (R1² - B²)) = pi * B² -- which is constant!
Denrep thanks for posting this, it is causing all kinds of fun everywhere. If only the painter had hooked the inside of the hole in the merry go round deck, with his tape measure, and swung an arc with the tape measure over the outer edge of the merry go round, until he found the smallest distance possible we would know everything about the circle using the formula aevald posted to solve the circular segment.
Sincerely,
William McCormick