Notes
two triangles and three concentric semi-circles solution

Solution to the Two Triangles and Three Concentric Semi-Circles Puzzle

Solution by Similar Triangles

In the above diagram, point $O$ is the centre of the concentric semi-circles.

Consider the smaller triangle $A D E$ and the two smaller circles. Point $F$ is where the side $A E$ touches the circle, so angle $O \hat{F} E$ is a right-angle as it is the angle between a radius and tangent. Triangles $O D E$, $O F E$, and $O FA$ are all right-angled triangles with $O D$ and $O F$ of the same length, and $O E$ and $O A$ of the same length. They are therefore congruent. They are also similar to triangle $A D E$ as triangles $O F A$ and $E D A$ are both right-angled triangles sharing the angle at $A$. The congruency shows that the length of $E A$ is twice that of $E D$, and so by similarity $O E$ has double the length of $O D$. Therefore the middle semi-circle has double the radius of the smaller one.

Applying the same argument to the triangle $A B C$ shows that the outer semi-circle has double the radius of the middle semi-circle, and so four times that of the smaller semi-circle.

Since area scales with the square of the length scale factor, the area of the outer semi-circle is $16$ times that of the smallest, while the middle semi-circle has area $4$ times that of the smallest. The white region therefore has area $4 - 1 = 3$ times that of the smallest and so the fraction that is shaded is $\frac{13}{16}$.

Solution by Equilateral Triangles

Reflecting the diagram to create whole circles shows that the original triangles are half of a triangle with the property that its circumcircle and in-circle are concentric. Such a triangle must be equilateral, and the radius of its circumcircle is double that of its in-circle. The result then follows as above.