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, $(r+x)^2=r^2+(2r-x)^2$, $x=2/3r$. ![enter image description here]( < Is there a way to prove this without the Pythagorean theorem? Similar triangles? Alternatively, how would you do a compass-and-straightedge construction?

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By power of a point, $R(R + 2x) = (2x – R)^2$. It gives $R = \dfrac {2x}{3}$.