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?
 A: 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 $a$:
$$
(2r-a-x):(x+r)=a:r=x:(2r-x).
$$
A: For simplicity, let us consider the case $r = \frac12$ first.  
Choose a coordinate system so that $OA$ is the $x$-axis and $OC$ is the $y$-axis.
The bigger half circle now has radius $1$. Perform a circle inversion with respect to it, we find:


*

*The bigger half circle get mapped to itself.

*Since the inversion sends $O$ to infinity and the two smaller half circles (red and orange) passes through $O$, they get mapped to two rays (the two dashed lines in red and orange ) parallel to $y$-axis at $x = \pm 1$. 

*Notice circle inversion preserve tangency among curves.  
Since the small circle (green circle with solid boundary) is tangent to the three half circles, 
it get mapped to a circle (green circle with dashed boundary) tangent to the bigger half circle and
the two rays. 
There are only one way to place such a circle. It will have radius $1$ and intersects $y$-axis at $D = (0,1)$ and $E = (0,3)$. 

*Invert it back, the original small circle intersect $y$-axis at $D = (0,1)$ and $E' = (0,\frac13)$.
This means it will have center $C' = (0,\frac23)$ and radius $\frac12(1-\frac13) = \frac13 = \frac23 r$.
By scaling, the expression $\frac23 r$ works for other values of $r$.
$\hspace1in$ 
A: 
By power of a point, $R(R + 2x) = (2x – R)^2$. It gives $R = \dfrac {2x}{3}$.
