Twelve semicircles line the inside of a regular hexagon with side length $4$ as shown: each side is divided into two equal parts, and each part is the diameter of a semicircle. What is the radius of the circle tangent to all of these semicircles?

1. Let $A$ be the center of the hexagon. By symmetry the circle is centered at $A$ and touches all twelve semicircles in the same way, so it is enough to look at one of them. Take a side with midpoint $M$ and let $B$ be the center of one of the semicircles on that side.
2. $AM$ is the apothem of the hexagon: $AM=\dfrac{\sqrt3}2\cdot4=2\sqrt3$, and $AM\perp$ the side. The side has length $4$, so $B$ (a quarter point of the side) is at distance $MB=1$ from $M$, and the semicircle has radius $1$.
3. In the right triangle $AMB$, $AB=\sqrt{(2\sqrt3)^2+1^2}=\sqrt{13}$.
4. The circle is externally tangent to the semicircle, along the line $AB$, so $AB=r+1$. Therefore $r=\boxed{\sqrt{13}-1}\approx2.61$.