2021 AMC 10A Problem 12

1. The liquid in each cone fills a smaller cone with the same shape as the container: a depth, and a round top.
2. A cone's volume is $\frac13 \pi r^2 h$. The narrow liquid has top radius $3$ and some depth $h_1$; the wide liquid has top radius $6$ and depth $h_2$. Equal volumes give
$$\tfrac13 \pi \cdot 9 \cdot h_1 = \tfrac13 \pi \cdot 36 \cdot h_2, \quad\text{so}\quad h_1 = 4h_2.$$
With real numbers: a wide depth of $1$ cm and a narrow depth of $4$ cm both give volume $12\pi$.
3. A useful fact about cones of the same shape: if you make the cone $t$ times as tall, the top radius also becomes $t$ times as big, so the volume becomes $t \times t \times t$ times as big.
4. Dropping in a marble raises the level as if the liquid had grown by the marble's volume $\frac43\pi$. Both cones start with the same volume $12\pi$ and both gain the same $\frac43\pi$, so both volumes are multiplied by the same number: $\frac{12\pi + \frac43\pi}{12\pi} = \frac{10}{9}$. (A different starting amount would give a different number, but it would still be the same for both cones, since they start equal and gain equal amounts.)
5. By step 3, both depths are then multiplied by the same number $t$ (the number with $t \times t \times t = \frac{10}{9}$, about $1.036$). We never need its exact value.
6. The narrow cone goes from depth $4$ to $4t$, a rise of $4t - 4 = 4(t - 1)$. The wide cone goes from $1$ to $t$, a rise of $t - 1$.
7. The ratio of the rises is $4(t - 1) : (t - 1) = 4 : 1$.

The answer is $\boxed{\textbf{(E)}\ 4:1}$.

Liquid in a point-down cone always makes a copy of the cone, so its depth depends only on its volume. Two situations that start with the same volume and gain the same volume get stretched by the same factor, so the rises keep the same ratio as the starting depths. That is why the marble's actual size drops out.