AMC 12 2014b Problem 19

A sphere is inscribed in a truncated right circular cone as shown. The volume of the truncated cone is twice that of the sphere. What is the ratio of the radius of the bottom base of the truncated cone to the radius of the top base of the truncated cone?

real r=(3+sqrt(5))/2;
real s=sqrt(r);
real Brad=r;
real brad=1;
real Fht = 2*s;
import graph3;
import solids;
currentprojection=orthographic(1,0,.2);
currentlight=(10,10,5);
revolution sph=sphere((0,0,Fht/2),Fht/2);
//draw(surface(sph),green+white+opacity(0.5));
//triple f(pair t) {return (t.x*cos(t.y),t.x*sin(t.y),t.x^(1/n)*sin(t.y/n));}
triple f(pair t) {
triple v0 = Brad*(cos(t.x),sin(t.x),0);
triple v1 = brad*(cos(t.x),sin(t.x),0)+(0,0,Fht);
return (v0 + t.y*(v1-v0));
}
triple g(pair t) {
return (t.y*cos(t.x),t.y*sin(t.x),0);
}
surface sback=surface(f,(3pi/4,0),(7pi/4,1),80,2);
surface sfront=surface(f,(7pi/4,0),(11pi/4,1),80,2);
surface base = surface(g,(0,0),(2pi,Brad),80,2);
draw(sback,gray(0.9));
draw(sfront,gray(0.5));
draw(base,gray(0.9));
draw(surface(sph),gray(0.4));

$\text{(A) } \dfrac32 \text{(B) } \dfrac{1+\sqrt5}2 \text{(C) } \sqrt3 \text{(D) } 2 \text{(E) } \dfrac{3+\sqrt5}2$