Problem 13
Circle O is incribed in a square with side length 1. Circle O' is inscribed in the lower right corner of the square so that O' is tangent to O and the two sides of the square (see figure below). What is the radius of O'?
Solution
Let P be the point where the circles touch and Q be the lower right corner of the square. Then PQ = (sqrt(2) - 1)/2. Let r be the radius of the small circle.
Then (r + (sqrt(2)*r)) = PQ. Solving for r we get the radius of O' to be (3 - 2*sqrt(s))/2 or for those that put the answer in decimal form, about 0.085786437626904951198311275790302
