2014 AMC 10B Problem 23

Call the bottom radius $R$, the top radius $r$, and the sphere's radius $s$. We want $\frac{R}{r}$.

Step 1: turn it into a flat picture. Slice the solid straight down through its middle. The cut-off cone becomes a trapezoid with parallel sides $2R$ (bottom) and $2r$ (top), and the sphere becomes a circle of radius $s$ touching all four sides. The circle touches the top and the bottom, so the height is $h = 2s$.

Step 2: find the slanted side. From any point outside a circle, the two touching segments to the circle have equal length. Use this at the two ends of a slanted side:
- From the top corner, the segment to the top touching point (the middle of the top side) is $r$, so the segment along the slanted side is also $r$.
- From the bottom corner, the segment to the bottom touching point is $R$, so the segment along the slanted side is also $R$.

So the slanted side is $R + r$.

Step 3: link $s$ to $R$ and $r$. Drop a height from a top corner to the bottom. It makes a right triangle with a horizontal leg of $R - r$, a vertical leg of $2s$, and the slanted side $R + r$ as hypotenuse:
$$(R + r)^2 = (R - r)^2 + (2s)^2.$$
Expanding, $R^2 + 2Rr + r^2 = R^2 - 2Rr + r^2 + 4s^2$, so $4Rr = 4s^2$ and
$$s^2 = Rr.$$

Step 4: use the volumes. The volume of a cone with its top cut off (height $h$, radii $R$ and $r$) is $\frac{\pi h}{3}\left(R^2 + Rr + r^2\right)$. With $h = 2s$ this is $\frac{2\pi s}{3}\left(R^2 + Rr + r^2\right)$. The sphere's volume is $\frac43 \pi s^3$. The solid is twice the sphere:
$$\frac{2\pi s}{3}\left(R^2 + Rr + r^2\right) = 2 \cdot \frac43 \pi s^3.$$
Divide both sides by $\frac{2\pi s}{3}$: $R^2 + Rr + r^2 = 4s^2 = 4Rr$. So
$$R^2 - 3Rr + r^2 = 0.$$

Step 5: solve for the ratio. Divide by $r^2$ and write $t = \frac{R}{r}$: $t^2 - 3t + 1 = 0$. The quadratic formula gives
$$t = \frac{3 \pm \sqrt5}{2}.$$
The bottom is the larger base, so $t > 1$, and $t = \frac{3 + \sqrt5}{2}$.

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

A ball inside a round solid with the same axis is really a flat problem: slice through the middle and you get a circle inside a trapezoid. The fact that the two touching segments from a point to a circle are equal gives the slanted side for free, and one Pythagorean step turns "the ball fits snugly" into $s^2 = Rr$. Since the question only asks for a ratio, dividing by $r^2$ leaves one equation in one unknown.

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