AMC12 2017 A

Answer (A): Because $\overline{YE}$ and $\overline{EF}$ are parallel to $\overline{AD}$ and $\overline{AC}$, respectively, $\triangle XEY \sim \triangle XAD$ and $\triangle XEF \sim \triangle XAC$. Therefore $$\frac{XY}{XE}=\frac{XD}{XA}\quad\text{and}\quad \frac{XF}{XE}=\frac{XC}{XA}.$$ It follows that $$\frac{XC}{XD}=\frac{XF}{XY}.$$ The Power of a Point Theorem applied to circle $O$ and point $X$ implies that $XC\cdot XG=XD\cdot XB$. Together with the previous equation this implies that $XF\cdot XG=XB\cdot XY$. Let $d=BD$; then $DX=\frac14 d$ and $BY=\frac{11}{36}d$. It follows that $$\begin{array}{rcl} XF\cdot XG&=&XB\cdot XY=(BD-DX)\cdot(BD-DX-BY)\\[4pt] &=&\left(d-\frac14 d\right)\left(d-\frac14 d-\frac{11}{36}d\right)\\[6pt] &=&\frac34\cdot\frac49 d^2=\frac{d^2}{3}. \end{array}$$ To determine $d$, note that because $ABCD$ is a cyclic quadrilateral it follows that $\alpha=\angle BAD=\pi-\angle DCB$. Applying the Law of Cosines to $\triangle ABD$ and $\triangle BCD$ yields $$\cos\alpha=\frac{AB^2+AD^2-BD^2}{2\cdot AB\cdot AD} =\frac{3^2+8^2-d^2}{2\cdot3\cdot8} =\frac{73-d^2}{48},$$ and $$-\cos\alpha=\cos(\pi-\alpha) =\frac{CB^2+CD^2-BD^2}{2\cdot CB\cdot CD} =\frac{2^2+6^2-d^2}{2\cdot2\cdot6} =\frac{40-d^2}{24}.$$ Therefore $$\frac{73-d^2}{48}=\frac{d^2-40}{24},$$ and solving for $d^2$ gives $d^2=51$. Hence $XF\cdot XG=\frac13 d^2=17$.