Notes
three overlapping semi-circles solution

Solution to the Three Overlapping Semi-Circles Puzzle

Solution by Angle in a Semi-Circle and Pythagoras' Theorem

In the diagram above, $D$ is the point where the yellow semi-circle intersects the diameter of the blue. As $A C$ is a diameter of the yellow semi-circle, using angle in a semi-circle, angle $A \hat{D} C$ is a right-angle and so triangle $A D C$ is a right-angled triangle.

The point $O$ is the centre of the blue circle, so $O A$ and $O B$ are both radii and so are of length $2$. Therefore, $D$ is the midpoint of $O B$. Since $C$ is directly above $D$, it is equidistant from $O$ and $B$ and so triangle $O C B$ is isosceles with $O C$ and $C B$ of the same length. Since also $O C$ and $O B$ have the same length, $O C B$ is therefore equilateral.

Point $F$ is both the centre of the red semi-circle and the midpoint of $A C$. Triangle $A O C$ is an isosceles triangle with angle $A \hat{O} C = 120^\circ$, since triangle $O C B$ is equilateral. Cutting along $O F$ creates two right-angled triangles that can be reassembled into an equilateral triangle, demonstrating that $O F$ is half of $O A$. Therefore also $E F$ is the same as $O F$.

As the triangle $A O F$ is a right-angled triangle, applying Pythagoras' theorem to it results in

The areas of the red and yellow semi-circles are given by

Putting these together, and using the fact that $A O$ has length $2$, the shaded area is

Note: $O F$ is half of $O E$ which has length $2$ so $O F$ has length $1$, it is also possible to show that $A F^2$ is $3$ without using Pythagoras’ theorem, see equilateral triangle for the details, so the shaded area is $\frac{1}{2} \pi 1^2 + \frac{1}{2} \pi 3 = 2 \pi$ and this does not use Pythagoras’ theorem.