Notes
three quarter circles ii solution

Solution to the Three Quarter Circles II Puzzle

Solution by Pythagoras' Theorem and Lengths in an Isosceles Right-angled Triangle

Let the points be labelled as above, where $D$ is the point on $C E$ such that $F D$ is perpendicular to $C E$ and $B$ on $A C$ likewise. Let $F D$ have length $a$ and $F B$ have length $b$. As $A C E$ is a quarter-circle?, angle $A \hat{C} E$ is a right angle, and by construction so are $F \hat{D} C$ and $C \hat{B} F$, so $B C D F$ is a rectangle. Therefore $C D$ also has length $b$. Applying Pythagoras' theorem to triangle $F D C$ we have:

As $A C E$ is a quarter circle, angle $A \hat{E} C$ is $45^\circ$, as is angle $C \hat{A} E$. Therefore, triangles $F D E$ and $A B F$ are isosceles right-angled triangles. Hence $E F$ has length $\sqrt{2} a$ and $A F$ has length $\sqrt{2} b$. By the area of a circle, the total shaded area is therefore:

Solution by Invariance Principle

The two shaded quarter circles can be drawn any size relative to each other. There are two special configurations.

In this configuration, the quarter circles are the same size making a semi-circle. The radius of this semi-circle is $8$, so the area is:

In this version, one quarter circle is shrunk to a point. The radius of the remaining quarter circle is $8 \sqrt{2}$, making its area: