Notes
two semi-circles in a circle solution

Solution to the Two Semi-Circles in a Circle Puzzle

Solution by Congruent Triangles and Pythagoras' Theorem

In the above diagram, the points $E$ and $F$ are midpoints of those chords and the line through $E$ and $F$ is the perpendicular bisector through those chords so is a diameter of the circle.

The length of $E F$ is the sum of the lengths of $B E$ and $A F$. Let $O$ be the point so that $E O$ has the same length as $A F$, and so $F O$ has the same length as $B E$. Then $O E B$ and $A F O$ are right-angled triangles with the same side lengths adjacent to the right-angles, so are congruent. This means that the lengths of $O B$ and $O A$ are the same. As $A$ and $B$ are on the circumference of the circle and $O$ is on a diameter, it is therefore the centre of the circle.

This means that $O A$ is a radius and $F O$ has the same length as $B E$, so with $a$ as the length of $A F$, $b$ as the length of $B E$, and $r$ as the length of $F O$, Pythagoras' theorem applied to the triangle $A F O$ shows that:

The shaded areas are $\frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi r^2$ and the whole circle has area $\pi r^2$, so the shaded area is $\frac{1}{2}$ of the total circle.

Solution by Midpoints of Chords and Pythagoras' Theorem

In the above diagram, $M$ is the midpoint of the chord $A B$, $E$ of $B C$, and $F$ of $A D$. The line through $E$ and $F$ is the perpendicular bisector of each of the chords $B C$ and $A D$ and so is a diameter of the circle. The point $O$, which lies on this diameter, is the centre of the circle.

The points $G$, $H$, and $I$ are so that $G M$ is perpendicular to $E F$, $B H$ is perpendicular to $A F$, and $I M$ is also perpendicular to $A F$.

As $M$ is the midpoint of $A B$, $G$ is the midpoint of $E F$, and so the length $G F$ is half of the sum of the radii of the two semi-circles. Similarly, $I$ is the midpoint of $A H$ and so the length of $I F$ is the average of the radii of the two semi-circles. Hence $M G$ and $M I$ have the same length. Since angles $G \hat{M} I$ and $O \hat{M} A$ are both right-angles, angles $G \hat{M} O$ and $I \hat{M} A$ are equal. Therefore, triangles $M G O$ and $M I A$ are congruent.

The length of $O G$ is therefore the same as that of $A I$, which is half the difference of the radii of the smaller circles. Since the length of $F G$ is half the sum of these radii, $F O$ is the same length as the smaller radius, namely the length of $B E$.

Applying Pythagoras' theorem to triangle $A F O$ then shows that, with $a$ and $b$ the radii of the semi-circles and $r$ of the outer circle,

As above, then, the shaded areas are $\frac{1}{2} \pi a^2 + \frac{1}{2} \pi b^2 = \frac{1}{2} \pi r^2$ and the whole circle has area $\pi r^2$, so the shaded area is $\frac{1}{2}$ of the total circle.

Solution by Invariance Principle

There are two special configurations of this puzzle. In the first, one of the circles is as large as can be and the second is non-existent. In the second, the two circles are of equal size.

In this case, it is clear that the shaded area is half the area of the full circle.

In this case, the diameters of the smaller circles form two sides of a square and the diagonal of this square is then a diameter of the larger circle. Since the length of the diagonal is $\sqrt{2}$ times the length of a side, the area of the large circle is $2$ times the area of the two semi-circles made into a circle.