Triangle $ABC$ has $AB=9,BC=10,AC=17$. Points $D,E,F$ are midpoints of $\overline{AB},\overline{BC},\overline{AC}$. Let $X\ne E$ be the intersection of the circumcircles of $\triangle BDE,\triangle CEF$. What is $XA+XB+XC$?

1. As in the problem, the circumcenter $O$ of $\triangle ABC$ satisfies $\angle BDO=\angle BEO=90^\circ$ and $\angle CEO=\angle CFO=90^\circ$, so it lies on both circumcircles; hence $X=O$ and $XA+XB+XC=3R$.
2. $s=18$, Area $=\sqrt{18\cdot9\cdot8\cdot1}=36$, $R=\dfrac{abc}{4\cdot\text{Area}}=\dfrac{9\cdot10\cdot17}{144}=\dfrac{1530}{144}=\dfrac{85}8$.
3. $XA+XB+XC=3R=\boxed{\dfrac{255}8}$.