Notes
upside-down semi-circles in a semi-circle solution

Solution to the Upside-Down Semi-Circles in a Semi-Circle Puzzle

Solution by Pythagoras' Theorem

Fix a diagram with $n$ small semi-circles. Let $r$ be the radius of the small semi-circles and $R$ of the outer one. In that diagram, triangle $O B A$ is right-angled. The length of $O A$ is $r$ while the length of $O B$ is $n r$. Applying Pythagoras' theorem shows that:

The shaded area is $n \times \frac{1}{2} \pi r^2$ while the area of the outer semi-circle is $\frac{1}{2} \pi R^2 = (n^2 + 1) \times \frac{1}{2} \pi r^2$. The proportion that is shaded is therefore $\frac{n}{n^2 + 1}$. The first few values of this are: