2000 AIME I Problem 8

Attempt Problem 8 of the 2000 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2000 AIME I solutions, or check the answer key.

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8.

A container in the shape of a right circular cone is 1212 inches tall and its base has a 55-inch radius. The liquid that is sealed inside is 99 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 mnp3m - n\sqrt[3]{p} inches deep, where m,m, n,n, and pp are positive integers and pp is not divisible by the cube of any prime number. Find m+n+p.m + n + p.

Answer: 52
Concepts:conesimilaritypower scaling of length, area, and volumevolume
Difficulty rating: 2390
Solution:

Held point down, the liquid forms a cone similar to the container with ratio 912=34,\frac{9}{12} = \frac{3}{4}, so its volume is (34)3=2764\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 12764=37641 - \frac{27}{64} = \frac{37}{64} of the volume, so its height is 1237643=123734=337312 \sqrt[3]{\frac{37}{64}} = \frac{12 \sqrt[3]{37}}{4} = 3\sqrt[3]{37} inches. The liquid is therefore 12337312 - 3\sqrt[3]{37} inches deep.

Since 3737 is cube-free, m+n+p=12+3+37=52.m + n + p = 12 + 3 + 37 = 52.

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