Notes
multiple semi-circles solution

Solution to the Multiple Semi-Circles Puzzle

Solution by Angle in a Semi-circle

In the above diagram, let $a$ be the length of $A B$, $b$ of $B C$, and $c$ of $C D$. From the formula for the area of a circle, the area of a semi-circle can be found from its diameter as $\frac{1}{2} \pi \left( \frac{1}{2} d\right)^2 = \frac{1}{8} \pi d^2$.

So the area of the purple semi-circle is $\frac{1}{8} \pi a^2$. The second largest semi-circle has area $\frac{1}{8} \pi (a + b)^2$, so the cyan region has area:

Since the purple and cyan regions have the same area, $\frac{1}{8} \pi a^2 = \frac{1}{4} \pi a b$ so $a = 2 b$.

The diameter of the full semi-circle is $a + b + c = 3 b + c$. So the area of the yellow region is:

This has the same area as the cyan region, so $\frac{3}{4} \pi b c = \frac{1}{4} \pi a b$ which means that $3 c = a$.

The diameter of the full semi-circle is therefore $a + \frac{1}{2} a + \frac{1}{3} a = \frac{11}{3} a$ so its area is $\frac{1}{8} \pi \left( \frac{11}{3} a \right)^2 = \frac{121}{72} \pi a^2$.

The unshaded regions have diameters $\frac{1}{2} a$ and $\frac{1}{3} a$ so have areas $\frac{1}{8} \pi \left( \frac{1}{2} a \right)^2 + \frac{1}{8} \pi \left( \frac{1}{3} a \right)^2 = \frac{13}{288} \pi a^2$.

The fraction that is shaded is therefore: