6. Given that the three sides of triangle $A B C$ are all positive integers, if $\angle A=2 \angle B$, and $C A=9$, then the minimum possible value of $B C$ is $\qquad$ .
By the Law of Sines, $\frac{a}{\sin 2 B}=\frac{9}{\sin B}=\frac{c}{\sin 3 B} \Rightarrow \frac{a}{2 \cos B}=\frac{c}{3-4 \sin ^{2} B}=9$ $\Rightarrow\left\{\begin{array}{l}a=18 \cos B, \\ c=36 \cos ^{2} B-9\end{array} \Rightarrow \cos B>\frac{1}{2}\right.$ and the denominator of $\cos B$ is one of $2,3,6$, so $\cos B=\frac{2}{3}$ or $\frac{5}{6}$. Therefore, the minimum possible value of $B C$ is 12.