Notes
subdivided semi-circle solution

Solution to the Subdivided Semi-Circle Puzzle

Solution by Dissection, Lengths in a Regular Hexagon, and the Area of a Triangle

Label the points as in the diagram above, where $O$ is the centre of the semi-circle.

Consider the region formed by the arc $A C$ and line segments $C E$ and $E A$.

As the points are equally spaced around the semi-circle, points $A$, $C$, $E$, and $G$ form part of a regular hexagon. Therefore, $O E$ is parallel to $A C$. This means that triangles $A C E$ and $A C O$ have the same area as both have the same “height” above $A C$. The area of the region formed by the arc $A C$ and line segments $C E$ and $E A$ is therefore the same as the area of the sector $A O C$.

Consider the region formed by the arc $F G$ and line segments $G A$ and $A F$.

The two triangles $O G F$ and $A O F$ have the same length base, as they are both radii of the same circle, and the same height, as it is the height of $F$ above $G A$, so they have the same area. Therefore, the area of triangle $A O F$ is the same as that of triangle $F O E$. So the region formed by the arc $F G$ and line segments $G A$ and $A F$ has the same area as that formed by the arc $F G$ and line segments $G O$, $O E$, and $E F$.

Lastly, the area of the segment cut off by line segment $C D$ is the same as that cut off by line segment $E F$, so combining this with the region above produces the same area as sector $E O G$.

In total, the shaded regions have the same area as two $60^\circ$ sectors of the semi-circle, so their combined area is $\frac{2}{3}$rds of the total.