AMC12 2012 B

Construct the midpoints $E=(4,0)$ and $F=(10,0)$ and triangle $\triangle EMF$ as in the diagram, where $M$ is the center of square $PQRS$. Also construct points $G$ and $H$ as in the diagram so that $BG\parallel PQ$ and $CH\parallel QR$. Observe that $\triangle AGB\sim\triangle CHD$ while $PQRS$ being a square implies that $GB=CH$. Furthermore, $CD=6=3\cdot AB$, so $\triangle CHD$ is 3 times bigger than $\triangle AGB$. Therefore, $HD=3\cdot GB=3\cdot HC$. In other words, the longer leg is 3 times the shorter leg in any triangle similar to $\triangle AGB$. Let $K$ be the foot of the perpendicular from $M$ to $EF$, and let $x=EK$. Triangles $\triangle EKM$ and $\triangle MKF$, being similar to $\triangle AGB$, also have legs in a 1:3 ratio, therefore, $MK=3x$ and $KF=9x$, so $10x=EF=6$. It follows that $EK=0.6$ and $MK=1.8$, so the coordinates of $M$ are $(4+0.6,1.8)=(4.6,1.8)$ and so our answer is $4.6+1.8 = 6.4 =$ $\boxed{\mathbf{(C)}\ 32/5}$.