2000 AIME I Problems/Problem 8
Problem
A container in the shape of a right circular cone is 
inches tall and its base has a 
-inch radius. The liquid that is sealed inside is 
inches deep when the cone is held with its point down and its base horizontal. When the liquid is held with its point up and its base horizontal, the height of the liquid is ![$m - n\sqrt [3]{p},$](http://latex.artofproblemsolving.com/1/5/8/1586f2bca69882cdcb09b653822a61d6a5931db7.png)
from the base where 
and 
are positive integers and is not divisible by the cube of any prime number. Find 
.
Solution
Solution 1
The scale factor is uniform in all dimensions, so the volume of the liquid is 
of the container. The remaining section of the volume is 
of the volume, and therefore 
of the height when the vertex is at the top.
So, the liquid occupies 
of the height, or 
. Thus 
.
Solution 2
(Computational) The volume of a cone can be found by 
. In the second container, if we let 
represent the height, radius (respectively) of the air (so 
is the height of the liquid), then the volume of the liquid can be found by 
.
By similar triangles, we find that the dimensions of the liquid in the first cone to the entire cone is 
, and that 
; equating,
![\begin{align*}\frac{\pi}{3}\left(\frac{3}{4}r\right)^2 \left(\frac{3}{4}h\right) &= \frac{\pi}{3}\left(r^2h - \left(\frac{rh'}{h}\right)^2h'\right)\\ \frac{37}{64}r^2h &= \frac{r^2}{h^2}(h')^3 \\ h' &= \sqrt[3]{\frac{37}{64} \cdot 12^3} = 3\sqrt[3]{37}\end{align*}](http://latex.artofproblemsolving.com/1/4/a/14ab28bada1a2faa03eb79845b8e1cef3c8a0d39.png)
Thus the answer is ![$12 - h' = 12-3\sqrt[3]{37}$](http://latex.artofproblemsolving.com/9/c/e/9cefade80e75d0a63439935c0016c5ed094a047c.png)
, and 
.
See also
| 2000 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 7 |
Followed by Problem 9 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
