Question

A container in the shape of a right circular cone is $12$ inches tall and its base has a $5$ -inch radius. The liquid that is sealed inside is $9$ inches deep when the cone is held with its point down and its base horizontal. When the cone is held with its point up and its base horizontal, the liquid is $m - n\sqrt[3]{p}$ inches deep, where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p.$

Answer

Solution:

Held point down, the liquid forms a cone similar to the container with ratio $\frac{9}{12} = \frac{3}{4},$ so its volume is $\left(\frac{3}{4}\right)^3 = \frac{27}{64}$ of the container's volume.

Held point up, the empty space is a similar cone at the apex with $1 - \frac{27}{64} = \frac{37}{64}$ of the volume, so its height is $12 \sqrt[3]{\frac{37}{64}} = \frac{12 \sqrt[3]{37}}{4} = 3\sqrt[3]{37}$ inches. The liquid is therefore $12 - 3\sqrt[3]{37}$ inches deep.

Since $37$ is cube-free, $m + n + p = 12 + 3 + 37 = 52.$

2000 AIME I Problem 8

Solution:

Problem 8 in Other Years