# Semicircles, semi mysterious

This puzzle by Catriona Shearer has been driving me nuts today:

I stared at a while, didn’t come up with anything good, and finally gave up and started looking at the solutions people were posting. They fell into three categories: Answers with no real attempt to show reasoning; algebraic solutions that were kind of opaque and didn’t really give any insight; and geometric solutions that depended on assertions I couldn’t see a justification for.

I still haven’t seen a solution that makes me particularly happy.

The geometric solutions in particular mostly looked like they assumed the left small semicircle’s base makes a 45° with the large semicircle’s base (or something equivalent). Which it does, but how do you know it does? In particular, it appeared people were assuming the square diagonal and the perpendicular bisector shown below are the same; in this diagram they’re not, because the large “semicircle” is really a semiellipse, but how do you know they do coincide when it’s a semicircle?

The clearest solution posted which I felt didn’t make any unjustified assumptions was Bilal Aydin’s. Here’s my version of it.

Let the radii of the small and large semicircles be $r$ and $R$. In the following diagram $D$ is the center of the large semicircle, $B$ is the center of the small one on the left, and $E$ is the center of the small one on the right. $C$ is the point where the base of the large semicircle is tangent to the angled semicircle, and radius $BC$ is perpendicular to that base. Since the base of the angled semicircle is a chord of the large one, its perpendicular bisector passes through $B$ and $D$.

Then $BE=2r$ and $BC = r$ so $CE = \sqrt{3}r$. Let $DE=x=R-r$. Now $(BD)^2 = r^2+(\sqrt{3}r-x)^2$. $AB=r$, so $AD^2 = r^2+r^2+(\sqrt{3}r-x)^2$. But $AD = R$. So

$r^2+r^2+(\sqrt{3}r-x)^2 = (r+x)^2$

$2r^2+3r^2-2\sqrt{3}rx+x^2 = r^2+2rx+x^2$

$4r^2= 2rx(1+\sqrt{3})$

$x = 2r/(1+\sqrt{3}) = r(\sqrt{3}-1)$

And then $R = x+r = \sqrt{3}r$. So the large semicircle is three times the area of each small one; the shaded area is $2/3$ of the whole.

That works, but it’s very mysterious. Stuff cancels on both sides for no obvious reason. And if you try to figure out $CD = \sqrt{3}r-x$ you find it’s just $r$, which means triangle $BCD$ is isosceles and the angled semicircle’s base is 45° from the large semicircle’s base. Those seem like things that should be easy to prove geometrically, and yet… how?