Three one-inch squares are placed with their bases on a line. The center square is lifted out and rotated 45 degrees, as shown. Then it is centered and lowered into its original location until it touches both of the adjoining squares. How many inches is the point $B$ from the line on which the bases of the original squares were placed?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ \sqrt{2}\qquad\textbf{(C)}\ \frac{3}{2}\qquad\textbf{(D)}\ \sqrt{2}+\frac{1}{2}\qquad\textbf{(E)}\ 2$
Consider the rotated middle square shown in the figure. It will drop until length $DE$ is 1 inch. Then, because $DEC$ is a $45^{\circ}-45^{\circ}-90^{\circ}$ triangle, $EC=\frac{\sqrt{2}}{2}$, and $FC=\frac{1}{2}$. We know that $BC=\sqrt{2}$, so the distance from $B$ to the line is 
$BC-FC+1=\sqrt{2}-\frac{1}{2}+1=\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.
[AMC10200519Sol.png](https://artofproblemsolving.com/wiki/index.php/File:AMC10200519Sol.png)


Note
(Refer to Diagram Above)
After deducing that $BC=\sqrt{2}$, we can observe that the length from $C$ to the baseline is $\frac{1}{2}$. This can be obtained by subtracting $FC$ from the side length of the square(s), which is $1$.
Adding these up, we see that our answer is $\boxed{\textbf{(D) }\sqrt{2}+\dfrac{1}{2}}$.
- sdk652