Ten circles of radius $1$ are arranged in a ring so that each is tangent to the next, and all ten are internally tangent to a larger circle $\omega$. What is the radius of $\omega$?

1. Let $O$ be the center of $\omega$. By symmetry the centers of the ten unit circles lie on a circle about $O$ of some radius $d$, and adjacent centers are $2$ apart (the circles are tangent).
2. Adjacent centers subtend an angle of $36^\circ$ at $O$, so $d\sin18^\circ=1$. Since $\sin18^\circ=\dfrac{\sqrt5-1}4$,
   $$d=\frac4{\sqrt5-1}=\sqrt5+1 .$$
3. Each unit circle is internally tangent to $\omega$, so the radius of $\omega$ is $d+1=\boxed{\sqrt5+2}$.