What's the distance between two circumcenters This problem was came to a Facebook post of mathigon ......the problem seems trickier than I expected...can you help me to find the distance??


Let $\angle CAB = \angle ABC = \angle BAD = \angle DBA = \theta$
Area of the $\bigtriangleup CAD = \frac{1}{2} 1^2 \sin 2 \theta = \frac{1}{2} \sin 2 \theta$
Area of the sector $CAD$ (consider the right side of the circle on the left) $ = \frac{1}{2} 1^2 2\theta = \frac{1}{2} (2 \theta)$
So the common area between the circles $ = 2 (\frac{1}{2} (2 \theta) - \frac{1}{2} \sin 2 \theta)$
However, it is given that the common area between the circles $= \frac{\pi}{2}$
Hence we need to solve the following equation for $\theta$
$$2 \left(\frac{1}{2} \left(2 \theta \right) - \frac{1}{2} \sin 2 \theta \right) = \frac{\pi}{2}$$
Solving, $\theta = 1.15494$
Finally length of $AB$ is $2 \cos \theta = 0.80794684217$
