smc

Label the points as in the figure above. Let the side length $AB=AC=s$. Therefore, $BC=s\sqrt{2}$. Since the circumradius of a right triangle is equal to half of the length of the hypotenuse, we have $R=\frac{s\sqrt{2}}{2}$. Now to find the inradius. Notice that $IFAE$ is a square with side length $r$, and also $AD=R$. Therefore, $s=AD=AI+ID=r\sqrt{2}+r$, and so $r=\frac{R}{\sqrt{2}+1}$. Finally, $\frac{R}{r}=1+\sqrt{2},\boxed{1+\sqrt{2}}$.