Let $\triangle ABC$ be a triangle with integer side lengths and the property that $BC=9$ and $\angle B=2\angle A$. What is the sum of all possible perimeters of such triangles?

1. Let $a=BC=9$, $b=CA$, $c=AB$. As in the standard argument, the law of sines gives $\cos A=\dfrac b{2a}$ and combining with the law of cosines gives $b^2=a(a+c)$ (the case $c=a$ would force $B=90^\circ$ and $b^2=2a^2$, impossible).
2. Since $A<60^\circ$, $\frac12<\cos A<1$, so $a
3. $b^2=9(9+c)$ means $9\mid b^2$, so $b=3t$ with $c=t^2-9$. The condition $9<3t<18$ gives $t=4$ or $t=5$.
4. $t=4$: $(a,b,c)=(9,12,7)$, perimeter $28$. $t=5$: $(a,b,c)=(9,15,16)$, perimeter $40$. Both are triangles, and the angle condition holds (for example $\cos A=\frac56$, $\cos B=\frac7{18}=2\cdot\frac{25}{36}-1$ for the second).
5. The sum of all possible perimeters is $28+40=\boxed{68}$.