Triangle $A B C$ has sides $A B=14, B C=13$, and $C A=15$. It is inscribed in circle $\Gamma$, which has center $O$. Let $M$ be the midpoint of $A B$, let $B^{\prime}$ be the point on $\Gamma$ diametrically opposite $B$, and let $X$ be the intersection of $A O$ and $M B^{\prime}$. Find the length of $A X$.
Since $B^{\prime} B$ is a diameter, $\angle B^{\prime} A B=90^{\circ}$, so $B^{\prime} A \| O M$, so $\frac{O M}{B^{\prime} A}=\frac{B M}{B A}=\frac{1}{2}$. Thus $\frac{A X}{X O}=\frac{B^{\prime} A}{O M}=2$, so $A X=\frac{2}{3} R$, where $R=\frac{a b c}{4 A}=\frac{(13)(14)(15)}{4(84)}=\frac{65}{8}$ is the circumradius of $A B C$. Putting it all together gives $A X=\frac{65}{12}$. $\fbox{65 / 12}$.