jmc

Let $D$ be the midpoint of $\overline{BC}$. Then by SAS Congruence, $△ABD\cong △ACD$, so $\angle ADB=\angle ADC=90^o$. Now let $BD=y$, $AB=x$, and $\angle IBD=\frac{\angle ABD}{2}=\theta$. Then $\cos (\theta )=\frac{y}{8}$ and $\cos (2\theta )=\frac{y}{x}=2\mathrm{co}{\mathrm{s}}^2(\theta )-1=\frac{y^2-32}{32}$. Cross-multiplying yields $32y=x(y^2-32)$. Since $x,y>0$, $y^2-32$ must be positive, so $y>5.5$. Additionally, since $△IBD$ has hypotenuse $\overline{IB}$ of length $8$, $BD=y<8$. Therefore, given that $BC=2y$ is an integer, the only possible values for $y$ are $6$, $6.5$, $7$, and $7.5$. However, only one of these values, $y=6$, yields an integral value for $AB=x$, so we conclude that $y=6$ and $x=\frac{32(6)}{(6{)}^2-32}=48$. Thus the perimeter of $△ABC$ must be $2(x+y)=\boxed{108}$.