2014 AMC 10B Problem 22

1. Size of the semicircles. Each side of length $2$ holds two semicircles side by side, so each has diameter $1$ and radius $\frac12$.
2. Coordinates. Put the center of the square at $(0, 0)$, so its corners are $(\pm 1, \pm 1)$. On the top side, the two semicircles are centered at the middles of the two halves: $\left(-\frac12, 1\right)$ and $\left(\frac12, 1\right)$. The other sides are the same picture turned.
3. Where is the middle circle? The picture looks the same after a quarter turn, so the circle touching all eight semicircles must be centered at $(0, 0)$. Call its radius $r$.
4. Distance between centers. Use the semicircle centered at $\left(\frac12, 1\right)$:
$$\sqrt{\left(\tfrac12\right)^2 + 1^2} = \sqrt{\tfrac14 + 1} = \sqrt{\tfrac54} = \frac{\sqrt5}{2}.$$
5. Touching circles. When two circles touch from the outside, the distance between their centers is the sum of the radii. (The touching point lies on the line joining the centers. That line points from $\left(\frac12, 1\right)$ down into the square, so it does meet the curved part of the semicircle.) So
$$r + \frac12 = \frac{\sqrt5}{2}, \qquad r = \frac{\sqrt5}{2} - \frac12 = \frac{\sqrt5 - 1}{2}.$$
6. Sense check: $\sqrt5 \approx 2.24$, so $r \approx 0.62$, a reasonable size for a circle inside a square of side $2$.

The answer is $\boxed{\textbf{(B)}\ \frac{\sqrt5-1}{2}}$.

Two circles that touch are really a fact about their centers: the distance between centers equals the sum of the radii (touching from outside) or the difference (one inside the other). Symmetry tells you where the unknown center is, coordinates give the distance, and one equation finishes the job. With any group of touching circles, join the centers first.