Area of the intersection of four circles of equal radius 
This picture basically shows a rearrangement of four quarters of a circle of radius 1. It asks for the shaded area.
I got the answer to be $\frac{2\pi + 6}{13}$. But then it is incorrect. 
The way I did it is by rearranging the four pieces into respectively a circle a square and a larger square with a inscribed circle. And naming each different pieces in the picture as x y and z.

Note that we could use the first quadrant and then get the final answer by symmetry. We have three quarter-circles in this region with the equations$$\begin{array}{l}{C_1}:{\left( {x - \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\\{C_2}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y - \frac{1}{2}} \right)^2} = 1\\{C_3}:{\left( {x + \frac{1}{2}} \right)^2} + {\left( {y + \frac{1}{2}} \right)^2} = 1\end{array}$$we want to obtain the area beneath ${C_3}$ (to the $x$-axis) but to obtain the endpoint, we have to find the cross point of ${C_2}$ and ${C_3}$, which is $$\begin{array}{l}x = \frac{{\sqrt 3  - 1}}{2}\\y = 0\end{array}$$now, the final answer is $$I = 4\int\limits_0^{\frac{{\sqrt 3  - 1}}{2}} {\left( { - \frac{1}{2} + \sqrt {1 - {{\left( {x + \frac{1}{2}} \right)}^2}} } \right)dx}  = 1 - \sqrt 3  + \frac{\pi }{3}$$
