Let $\triangle ABC$ be a right triangle with right angle at $C.$ Let $D$ and $E$ be points on $\overline{AB}$ with $D$ between $A$ and $E$ such that $\overline{CD}$ and $\overline{CE}$ trisect $\angle C.$ If $\frac{DE}{BE} = \frac{8}{15},$ then $\tan B$ can be written as $\frac{m \sqrt{p}}{n},$ where $m$ and $n$ are relatively prime positive integers, and $p$ is a positive integer not divisible by the square of any prime. Find $m+n+p.$
We have $\angle BCE = \angle ECD = \angle ECA = \tfrac 13 \cdot 90^\circ = 30^\circ$. Drop the altitude from $D$ to $CB$ and call the foot $F$. 


Let $CD = 8a$. Using angle bisector theorem on $\triangle CDB$, we get $CB = 15a$. Now $CDF$ is a $30$-$60$-$90$ triangle, so $CF = 4a$, $FD = 4a\sqrt{3}$, and $FB = 11a$.  Finally, $\tan{B} = \tfrac{DF}{FB}=\tfrac{4\sqrt{3}a}{11a} = \tfrac{4\sqrt{3}}{11}$. Our final answer is $4 + 3 + 11 = \boxed{018}$.