Eight semicircles line the inside of a square with side length 2 as shown. What is the radius of the circle tangent to all of these semicircles? $\text{(A) } \dfrac{1+\sqrt2}4 $\quad$ \text{(B) } \dfrac{\sqrt5-1}2 $\quad$ \text{(C) } \dfrac{\sqrt3+1}4 $\quad$ \text{(D) } \dfrac{2\sqrt3}5 $\quad$ \text{(E) } \dfrac{\sqrt5}3$
Solution
Let $A,B$ be the centers of two semicircles on opposite sides (radius $\frac12$), placed as in the figure so that segment $AB$ passes through the center of the square. Then $AB=\sqrt{2^2+1^2}=\sqrt5$, and along $AB$ we pass through a semicircle radius, the diameter of the circle, and another semicircle radius: $\sqrt5=\frac12+2r+\frac12=1+2r$.
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?
$\text{(A) } \dfrac{1+\sqrt2}4 $\quad$ \text{(B) } \dfrac{\sqrt5-1}2 $\quad$ \text{(C) } \dfrac{\sqrt3+1}4 $\quad$ \text{(D) } \dfrac{2\sqrt3}5 $\quad$ \text{(E) } \dfrac{\sqrt5}3$ 
