2014 AMC 12B Problem 19

Let the top radius be $1,$ the bottom radius $r,$ and the sphere radius $a.$ The sphere touches both bases, so the frustum height is $2a.$ In an axial cross-section put the sphere center at $(0,a)$; the right slanted side through $(r,0)$ and $(1,2a)$ has equation $2ax+(r-1)y-2ar=0.$ Its distance from $(0,a)$ is $a,$ so $(r+1)^2=4a^2+(r-1)^2,$ giving $r=a^2.$ The frustum volume is $\tfrac13 \pi (r^2 + r + 1)(2a).$ Setting it equal to twice the sphere volume $\tfrac43 \pi a^3$ and using $r = a^2$ yields $$a^4 - 3a^2 + 1 = 0,$$ that is $r^2 - 3r + 1 = 0.$ The positive root is $r = \dfrac{3+\sqrt5}{2}.$ Thus, the correct answer is E.