Notes
multiple semi-circles iii solution

Solution to the Multiple Semi-Circles III Puzzle

Solution by Circle Areas and Similar Triangles

With the edges labelled as above, the largest semi-circle has area $\frac{1}{8} \pi (a + b)^2$, and the other two have areas $\frac{1}{8} \pi a^2$ and $\frac{1}{8} \pi b^2$. The shaded area therefore has area:

The triangles $A B C$ and $D B A$ are similar, with $A B$ corresponding to $D B$ and $B C$ to $B A$, so $A B : B C = D B : B A$. Hence $\frac{6}{a} = \frac{a + b}{6}$. Rearranging this yields $a(a + b) = 36$ so the shaded area is $9 \pi$.

The use of similar triangles can be replaced by Pythagoras' Theorem. Writing $c$ for the length of $A C$ and $d$ for the length of $A D$, the various right-angled triangles imply the following identities:

Putting this together,

so

Substituting this into the expression for the shaded area gives that that area is $9 \pi$ as before.

Solution by Invariance Principle

The point $C$ can slide on the red semi-circle. As it approaches $B$ then the diagram gets very large, but putting it at $A$ produces a diagram that makes the solution obvious.

The radius of the circle is $3$ so its area is $9 \pi$.