Congratulations for your solutions to Hyeyoun Chung, age 17, St Paul's Girls' School, London; Royce Ferguson; Ang Zhi Ping, age 16; Yatir Halevi, age 17, Maccabim and Reut High-School, Israel; and Joe Nielson, Rowan Maclennon-Ryde and Elizabeth Brewster from Madras College, St Andrew's, Scotland.

The radius of the circle OA can be found by using the right triangle formed by $\mathrm{AB}=(1+(\sqrt{2})/2)$ and $\mathrm{BO}=(\sqrt{2})/2)$. Using Pythagorus' theorem, the radius is found to be $\mathrm{OA}=\sqrt{(}2+\sqrt{2})$ units and the area of the circle to be $\pi (2+\sqrt{2})$.

Now, connect the centre of the circle to the 8 points on it's circumference where the white meets red. This divides the white into circular sectors and quadrilaterals.

Using the cosine rule to find $\angle \mathrm{AOD}$ we have $\mathrm{AC}=\mathrm{CD}=1$ and so $\mathrm{AD}=\sqrt{2}$ and hence

${\displaystyle \cos \angle \mathrm{AOD}=\frac{{\mathrm{OA}}^2+{\mathrm{OD}}^2-{\mathrm{AD}}^2}{2\mathrm{OA}.\mathrm{OD}}=\frac{1}{\sqrt{2}}.}$

Hence $\angle \mathrm{AOD}=45$ degrees.

The area of the triangle $\mathrm{AOD}$ is

${\displaystyle \frac{1}{2}\mathrm{OA}\times \mathrm{OD}\times \sin \angle \mathrm{AOD}=\frac{\sqrt{2}}{4}(2+\sqrt{2}).}$

To find the area of the minor segment $\mathrm{AD}$ we subtract the area of triangle $\mathrm{AOD}$ from the area of sector $\mathrm{AOD}$ which gives

${\displaystyle \frac{\pi }{8}(2+\sqrt{2})-\frac{\sqrt{2}}{4}(2+\sqrt{2}).}$

To get the total red shaded area we now add the area of triangle $\mathrm{ACD}$ and multiply by 4 which gives:

${\displaystyle 4[\frac{\pi }{8}(2+\sqrt{2})-\frac{\sqrt{2}}{4}(2+\sqrt{2})+\frac{1}{2}]=\frac{\pi }{2}(2+\sqrt{2})-2\sqrt{2}\approx 2.535\hspace{1em}\mathrm{sq}.\hspace{1em}\mathrm{units}.}$