Notes
five semi-circles solution

Solution to the Five Semi-Circles Puzzle

Solution by Pythagoras' Theorem

With the lengths labelled as in the above diagram, Pythagoras' theorem gives the following identities:

Putting some of these together leads to:

Combining these gets:

The area of a semi-circle of diameter $D$ is:

And so the areas of the semi-circles add to $10 \pi$.

Solution by the Intersecting Chords Theorem

(With a smidgeon of Pythagoras' Theorem)

In the diagram above, the line segment $I H$ is a diameter of the outer circle. As $E$ is on the middle circle, length $g = 8$ and $f = 4$. The intersecting chords theorem then says that $a d = f g = 32$. Applying Pythagoras' theorem to the right-angled triangle $E J B$ yields the identity $b^2 = e^2 + d^2$, and applying it to $A J B$ produces:

And so $a^2 + b^2 = 144 - 64 = 80$ as before.

(Note: This method was inspired by the cosine rule and the invariance principle since if $a^2 + b^2$ is constant then when applying the cosine rule to triangle $A E B$ the term $2 a b cos(C)$ has to be constant. In the above diagram, this is $a d$.)

Solution by the Parallelogram Law

In the above diagram, $K$ is the point on the inner middle circle diametrically opposite to $E$, so that $A K$ is congruent to $E B$ and $K B$ to $A E$. So $A E B K$ is a parallelogram. According to the parallelogram law,

Since $A B = 12$ and $E K = 4$, this means that (with $a$ and $b$ as in above sections):

and the solution follows as before.

Solution by the Invariance principles

The point $E$ can move around on the circumference of the middle circle, and this leads to two special configurations.

The left-hand semi-circle has radius $4$ and the right-hand one has radius $2$. So their areas are:

In this configuration, the diameter of each semi-circle is the hypotenuse of a triangle with other sides $6$ and $2$, so its square is $6^2 + 2^2 = 40$. The sum of the squares of the diameters is therefore $80$ which leads to the area of the semi-circles being $10\pi$ as before.