2014 AMC 10B Problem 23

Let the top radius be $1$, the bottom radius be $R$, and the inscribed sphere radius be $a$. In the cross-section, the sphere is tangent to the two bases, so the frustum height is $2a$. A slanted side joins $(1,a)$ to $(R,-a)$, if the sphere’s center is the origin. Its equation is $$2ax+(R-1)y-a(R+1)=0.$$ Because this line is tangent to the circle of radius $a$, its distance from the origin is $a$. Thus $$\frac{a(R+1)}{\sqrt{4a^2+(R-1)^2}}=a,$$ which simplifies to $R=a^2$. The frustum volume is $\frac13\pi(R^2+R+1)(2a)$ $=\frac{2a\pi}{3}(a^4+a^2+1)$. This is twice the sphere volume, $\frac{8a^3\pi}{3}$. Cancelling gives $a^4-3a^2+1=0$, so $R^2-3R+1=0$. Since the bottom radius is larger than the top radius, $R>1$. Thus $R=\frac{3+\sqrt5}{2}$, and the correct answer is E .