2014 AMC 10B Problem 23

Let the bottom and top radii be $R$ and $r$, and the sphere's radius $s$. Slice through the axis: the frustum becomes an isosceles trapezoid with bases $2R$ and $2r$, and the sphere becomes its incircle, so the height is $h = 2s$.

A quadrilateral with an incircle has equal sums of opposite sides, so $2R + 2r = 2\ell$ where $\ell$ is the slant side: $\ell = R + r$. Dropping a perpendicular from the top base gives a right triangle with legs $R - r$ and $2s$ and hypotenuse $R + r$:
$$(R + r)^2 = (R - r)^2 + 4s^2 \quad\Longrightarrow\quad s^2 = Rr.$$

Now the volumes. Frustum: $\frac{\pi h}{3}(R^2 + Rr + r^2) = \frac{2\pi s}{3}(R^2 + Rr + r^2)$. Sphere: $\frac{4}{3}\pi s^3$. The condition "frustum is twice the sphere" gives
$$\frac{2\pi s}{3}(R^2 + Rr + r^2) = \frac{8\pi s^3}{3} \quad\Longrightarrow\quad R^2 + Rr + r^2 = 4s^2 = 4Rr,$$
so $R^2 - 3Rr + r^2 = 0$. Divide by $r^2$ and set $t = \frac{R}{r}$:
$$t^2 - 3t + 1 = 0, \qquad t = \frac{3 \pm \sqrt5}{2}.$$
Since the bottom base is larger, $t > 1$, so $t = \frac{3 + \sqrt5}{2}$.

The answer is $\boxed{\textbf{(E)}\ \frac{3+\sqrt5}{2}}$.

Solids of revolution with a common axis are really plane problems: sphere in frustum is incircle in isosceles trapezoid. The tangential-quadrilateral property (equal sums of opposite sides) plus one Pythagorean step converts "inscribed" into the algebraic relation $s^2 = Rr$, a geometric mean. From there the problem is only about the ratio, so dividing by $r^2$ is the natural move.

• Using the sphere's diameter as the slant height, or forgetting the middle term Rr in the frustum volume formula.
• Choosing the root $\frac{3 - \sqrt5}{2} < 1$, or reporting $\frac{1 + \sqrt5}{2}$ (which is $\sqrt{t}$, the golden ratio) instead of $t$ itself.