imc

We connect the centers of the circle and one of the semicircles, then draw the perpendicular from the center of the middle circle to that side, as shown. We will start by creating an equation by the Pythagorean theorem: $$\sqrt{1^2+{\left(\frac{1}{2}\right)}^2}=\sqrt{\frac{5}{4}}=\frac{\sqrt{5}}{2}.$$ Let's call $r$ as the radius of the circle that we want to find. We see that the hypotenuse of the bold right triangle is $\frac{1}{2}+r$, and thus $r$ is $\boxed{\frac{\sqrt{5}-1}{2}}$.