Notes
two semi-circles and two quarter circles solution

Solution to the Two Semi-Circles and Two Quarter Circles Puzzle

Solution by Pythagoras' Theorem and the Area of a Circle

In the above diagram, $O$ is the centre of the two semi-circles and $D$ is such that angle $C \hat{D} O$ is a right-angle.

By including the white regions to both the yellow and blue, the comparison is between the two quarter circles and the two semi-circles.

Let $R$ be the radius of the larger blue semi-circle and $r$ of the smaller. Let $b$ be the radius of the left quarter circle and $a$ of the right. So the sides of the blue triangle are $a$, $b$, and $2 r$. Let $d$ be the length of the line segment $O D$ and $h$ of $C D$.

There are three right-angled triangles, namely $C D B$, $C D A$, and $C D O$. Applying Pythagoras' theorem to all three yields:

Adding the first two equations together gives:

And then substituting in the third shows that $a^2 + b^2 = 2 r^2 + 2 R^2$.

Multiplying by $\frac{1}{4} \pi$ gives:

and so the blue and yellow regions have the same area.

Solution by the Cosine Rule and Parallelogram Identity

The above can be shortened by using either the cosine rule or the parallelogram identity.

Using the cosine rule avoids needing point $D$. Applied to triangles $C O B$ and $C O A$ it yields:

Then, since $\cos(B \hat{O} C) = - \cos(C \hat{O} A)$, adding the equations gives:

This is also the result of applying the parallelogram identity to the parallelogram obtained by duplicating triangle $C A B$ and rotating it about $O$ (so that $A B$ as a diameter).

Solution by Invariance Principle

The positions of point $C$ on the circumference and $A$ and $B$ on the diameter can be varied, leading to two positions where the result is obvious. In both, $C$ is positioned at the midpoint of the arc.

Here, $A$ and $B$ are at the ends of the diameter, meaning that the blue and green is a full circle. The triangle is an isosceles right-angled triangle so its sides are in the ratio $1 : 1: \sqrt{2}$. So the yellow and green regions comprise half a circle of twice the area of the blue and green.

Here, $A$ and $B$ are at the centre of the blue semi-circle, and the yellow quarter circles combine to be a semi-circle of exactly the same size.