A sphere is inscribed in a truncated right circular cone (tangent to both bases and the lateral surface). If the volume of the truncated cone is $3$ times that of the sphere, what is the ratio of the larger base radius to the smaller base radius?
With the same tangency relations $\rho=\sqrt{r_1r_2}$, $h=2\rho$, set the frustum volume equal to $3$ times the sphere volume: $\frac13\pi(2\rho)(r_1^2+r_1r_2+r_2^2)=3\cdot\frac43\pi\rho^3$.
Writing $t=r_1/r_2$ and using $\rho^2=r_1r_2$ to clear square roots, this reduces to $t^2-5t+1=0$.
The larger-to-smaller ratio is the bigger root, $t=\dfrac{5+\sqrt{21}}2$.
