Subject: Re: Intersections, intersections
Date: Sat, 20 Sep 1997 15:56:08 -0400
From: Alex Bogomolny
denote areas as on the picture: a,b,c.
Combine them in three different ways:
- Circular sector: 3a+2b+c = Pi/4
- Sum of two sectors: the internal lune will be counted twice. Its area is (2a+c). Thus you get 1 + (2a+c) = Pi/2.
- Sum four sectors. Consider the overlaps.
You'll get three equations with three unkonwns. Yours is c.
The second problem can be solved similarly but is clearly more difficult. The answer is sqrt(2)-1 but you'll have to sweat it.
Please do not insist I'll get credit points for you.