TMUA practice: Straight Lines, problem 1

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} .