Difference between revisions of "2003 AMC 12B Problems/Problem 13"
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Let <math>r</math> be the common radius of the sphere and the cone, and <math>h</math> be the cone’s height. Then | Let <math>r</math> be the common radius of the sphere and the cone, and <math>h</math> be the cone’s height. Then | ||
<cmath>75\% \cdot \left(\frac 43 \pi r^3\right) = \frac 13 \pi r^2 h \Longrightarrow h = 3r</cmath> | <cmath>75\% \cdot \left(\frac 43 \pi r^3\right) = \frac 13 \pi r^2 h \Longrightarrow h = 3r</cmath> | ||
| − | Thus <math>h:r = 3:1</math>. | + | Thus <math>h:r = 3:1</math> and the answer is <math>\boxed{B}</math>. |
== See also == | == See also == | ||
Latest revision as of 15:20, 17 October 2016
Problem
An ice cream cone consists of a sphere of vanilla ice cream and a right circular cone that has the same diameter as the sphere. If the ice cream melts, it will exactly fill the cone. Assume that the melted ice cream occupies 
of the volume of the frozen ice cream. What is the ratio of the cone’s height to its radius?
Solution
Let 
be the common radius of the sphere and the cone, and 
be the cone’s height. Then
![\[75\% \cdot \left(\frac 43 \pi r^3\right) = \frac 13 \pi r^2 h \Longrightarrow h = 3r\]](http://latex.artofproblemsolving.com/a/b/e/abe59fc13b9d7b3ab5893c9a287df3ce7ff21e74.png)
Thus 
and the answer is 
.
See also
| 2003 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 12 |
Followed by Problem 14 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
