Difference between revisions of "2003 AMC 10B Problems/Problem 17"
(→Solution) |
|||
Line 7: | Line 7: | ||
==Solution== | ==Solution== | ||
− | Let <math>r</math> be the radius of both the cone and the sphere of ice cream. The cone can hold a maximum of <math>\pi r^2h/3</math> melted ice cream. The volume of the frozen ice cream is <math>4\pi r^3/3.</math> The volume of the melted ice cream is <math>75 \%</math> of that, which is <math>\pi r^3.</math> Since the melted ice cream fills the cone exactly, | + | Let <math>r</math> be the radius of both the cone and the sphere of ice cream, and let <math>h</math> be the height of the cone. The cone can hold a maximum of <math>\pi r^2h/3</math> melted ice cream. The volume of the frozen ice cream is <math>4\pi r^3/3.</math> The volume of the melted ice cream is <math>75 \%</math> of that, which is <math>\pi r^3.</math> Since the melted ice cream fills the cone exactly, |
<cmath>\begin{align*}\pi r^3&=\pi r^2h/3\\ | <cmath>\begin{align*}\pi r^3&=\pi r^2h/3\\ |
Revision as of 10:34, 20 July 2019
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? (Note: a cone with radius and height has volume and a sphere with radius has volume .)
Solution
Let be the radius of both the cone and the sphere of ice cream, and let be the height of the cone. The cone can hold a maximum of melted ice cream. The volume of the frozen ice cream is The volume of the melted ice cream is of that, which is Since the melted ice cream fills the cone exactly,
Then the ratio of the cone's height to its radius is
See Also
2003 AMC 10B (Problems • Answer Key • Resources) | ||
Preceded by Problem 16 |
Followed by Problem 18 | |
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 10 Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.