Is there another solution to this problem? Take a half circle with radius 2r with the center of O. Now take the half points from the radiuses from O forming the border of the half circle and draw two more half circles with radius r. Finally draw a circle touching the original half circle from the inside and the smaller half circles from the outside. What's the radius of this circle?
So let x be the radius in question, the centre of one of the smaller half circles be A, the centre of the circle in question C, then OC is 2r-x, AC is x+r and OA is r. With the Pythagorean theorem, $(r+x)^2=r^2+(2r-x)^2$, $x=2/3r$.

http://ggbm.at/v6XZM74R
Is there a way to prove this without the Pythagorean theorem? Similar triangles? Alternatively, how would you do a compass-and-straightedge construction?

Let $T$ be the point on $AC$ where the circles touch. Draw from $T$ the common tangent to intersect $OC$ at $P$. Let $OP=PT=a$, so that $PC=2r-x-a$. From the similarity of $AOC$ and $PTC$ you get two equations and can solve for $x$ and $a$:
$$
(2r-a-x):(x+r)=a:r=x:(2r-x).
$$
