A square piece of paper $ABCD$ has side length $12$. The paper is folded so that vertex $A$ lands on the midpoint $M$ of side $\overline{CD}$. What is the length of the crease?

1. Place $A=(0,0)$, $B=(12,0)$, $C=(12,12)$, $D=(0,12)$, so $M=(6,12)$. The crease is the perpendicular bisector of $AM$, since folding sends $A$ to $M$.
2. The midpoint of $AM$ is $(3,6)$ and $AM$ has slope $2$, so the crease has slope $-\frac12$: $y-6=-\frac12(x-3)$.
3. It meets $AD$ ($x=0$) at $P=(0,\frac{15}2)$ and $BC$ ($x=12$) at $Q=(12,\frac32)$.
4. $PQ=\sqrt{12^2+6^2}=\boxed{6\sqrt5}$.