Let the point where the ladder touches the wall be $A$, the point where the ladder touches the ground be $B$, the point where the wall and ground intersect be $C$, the midpoint of the ladder be $M$, and the points on the ground and wall closest to $M$ be $F$ and $E$, respectively. Also, let $ME=x$. Then, $MF=2x$. $\triangle AME\cong\triangle MFB$ by ASA, so $AE=2x$ and $BF=x$. $MECF$ is a rectangle, so $EC=2x$ and $CF=x$. Thus, $AC=4x$ and $BC=2x$. By the Pythagorean Theorem, $16x^2+4x^2=20x^2=36$ $\implies x=\frac{3\sqrt5}5$ Thus $AC=4x=\boxed{\frac{12\sqrt5}5}$.