8. [35] Let $A B C$ be an acute triangle with circumcenter $O$ such that $A B=4, A C=5$, and $B C=6$. Let $D$ be the foot of the altitude from $A$ to $B C$, and $E$ be the intersection of $A O$ with $B C$. Suppose that $X$ is on $B C$ between $D$ and $E$ such that there is a point $Y$ on $A D$ satisfying $X Y \| A O$ and $Y O \perp A X$. Determine the length of $B X$.
Answer: $96 / 41$ Let $A X$ intersect the circumcircle of $\triangle A B C$ again at $K$. Let $O Y$ intersect $A K$ and $B C$ at $T$ and $L$, respectively. We have $\angle L O A=\angle O Y X=\angle T D X=\angle L A K$, so $A L$ is tangent to the circumcircle. Furthermore, $O L \perp A K$, so $\triangle A L K$ is isosceles with $A L=A K$, so $A K$ is also tangent to the circumcircle. Since $B C$ and the tangents to the circumcircle at $A$ and $K$ all intersect at the same point $L, C L$ is a symmedian of $\triangle A C K$. Then $A K$ is a symmedian of $\triangle A B C$. Then we can use $\frac{B X}{X C}=\frac{(A B)^{2}}{(A C)^{2}}$ to compute $B X=\frac{96}{41}$.