Notes
circles in rhombii sequel solution

Solution to the Circles in Rhombii Sequel Puzzle

Solution by Trigonometry

Consider a rhombus as in the diagram above with $\alpha$ as angle $O \hat{A} C$. For simplicities’ sake, set the length $O A$ as $1$ unit. Then $O A$ is the hypotenuse of the right-angled triangle $O A B$ so the length of $O B$ is $\sin(\alpha)$.

The point labelled $Q$ is the centre of the upper small circle. The point labelled $D$ is where the circle touches the line $E C$, so $Q D$ and $O Q$ are radii of this circle and so of equal length. Therefore, triangles $Q O E$ and $Q D E$ are congruent, so angles $Q \hat{E} O$ and $D \hat{E} Q$ are equal and therefore both equal to $\frac{\alpha}{2}$. Since $O E$ has length $1$, this means that $O Q$ has length $\tan\left(\frac{\alpha}{2}\right)$.

The goal is for the large circle to have the same length as the two smaller circles combined, so the area of the large circle is double that of one of the smaller circles, and so the radius of the larger circle is $\sqrt{2}$ times that of the smaller. This implies:

The half-angle formulae show that with $t = tan\left(\frac{\alpha}{2}\right)$ then $\sin(\alpha) = \frac{2 t}{1 + t^2}$. Substituting in to the above gives:

which rearranges to $1 + t^2 = \sqrt{2}$ and so $t = \sqrt{\sqrt{2} - 1}$ (taking the positive square root as $\alpha$ is acute).

This gives a value of $\alpha$ as $65.53$ to $4$ significant figures.