Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 AIME I Problems/Problem 8"

(sol2)
m (+img)
Line 4: Line 4:
 
__TOC__
 
__TOC__
 
== Solution ==
 
== Solution ==
−
{{image}}
+
[[Image:2000_I_AIME-8.png]]
 
=== Solution 1 ===
 
=== Solution 1 ===
 
The scale factor is uniform in all dimensions, so the volume of the liquid is <math>\left(\frac{3}{4}\right)^{3}</math> of the container. The remaining section of the volume is <math>\frac{1-\left(\frac{3}{4}\right)^{3}}{1}</math> of the volume, and therefore <math>\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height when the vertex is at the top.
 
The scale factor is uniform in all dimensions, so the volume of the liquid is <math>\left(\frac{3}{4}\right)^{3}</math> of the container. The remaining section of the volume is <math>\frac{1-\left(\frac{3}{4}\right)^{3}}{1}</math> of the volume, and therefore <math>\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}</math> of the height when the vertex is at the top.

Revision as of 18:04, 31 December 2007

Problem

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 liquid is held with its point up and its base horizontal, the height of the liquid is $m - n\sqrt [3]{p},$ from the base where $m,$ $n,$ and $p$ are positive integers and $p$ is not divisible by the cube of any prime number. Find $m + n + p$.

Solution

2000 I AIME-8.png

Solution 1

The scale factor is uniform in all dimensions, so the volume of the liquid is $\left(\frac{3}{4}\right)^{3}$ of the container. The remaining section of the volume is $\frac{1-\left(\frac{3}{4}\right)^{3}}{1}$ of the volume, and therefore $\frac{\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height when the vertex is at the top.

So, the liquid occupies $\frac{1-\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}}{1}$ of the height, or $12-12\left(1-\left(\frac{3}{4}\right)^{3}\right)^{1/3}=12-3\left(37^{1/3}\right)$. Thus $m+n+p=\boxed{052}$.

Solution 2

(Computational) The volume of a cone can be found by $V = \frac{\pi}{3}r^2h$. In the second container, if we let $h',r'$ represent the height, radius (respectively) of the air (so $12 -h'$ is the height of the liquid), then the volume of the liquid can be found by $\frac{\pi}{3}r^2h - \frac{\pi}{3}(r')^2h'$.

By similar triangles, we find that the dimensions of the liquid in the first cone to the entire cone is $\frac{3}{4}$, and that $r' = \frac{rh'}{h}$; 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*}

Thus the answer is $12 - h' = 12-3\sqrt[3]{37}$, and $m+n+p=52$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_AIME_I_Problems/Problem_8&oldid=21718"