# Difference between revisions of "2004 AIME I Problems/Problem 11"

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== Solution == | == Solution == | ||

− | {{solution}} | + | Let <math>x</math> denote the radius of the small cone. Let <math>A_c</math> and <math>A_f</math> denote the area of the painted surface on cone <math>C</math> and frustum <math>F</math>, respectively. Let <math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively. Using the Pythagorean Theorem, we find that the height and slant height of cone <math>C</math> are <math>\frac{4}{3}x</math> and <math>\frac{5}{3}x</math>, respectively. Using the formula for lateral surface area of a cone, we find that <math>A_c=\frac{1}{2}c\cdot \ell=\frac{1}{2}(2\pi x)\left(\frac{5}{3}x\right)=\frac{5}{3}\pi x^2</math>. By subtracting <math>A_c</math> from the lateral surface area of the original cone and adding to this the are of the base, we find that <math>A_f=12\pi - \frac{4}{9}\pi x^3</math>. Next, we find that <math>V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3.</math> Finally, we subtract <math>V_c</math> from the volume of the original cone to find that <math>V_f=12\pi - \frac{4}{9}\pi x^3</math>. We know that <math>\frac{A_c}{A_f}=\frac{V_c}{V_f}=k.</math> Plugging in our values for <math>A_c</math>, <math>A_f</math>, <math>V_c</math>, and <math>V_f</math>, we obtain the equation <math>\frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}=\frac{\frac{4}{9}\pi x^3}{12\pi - \frac{4}{9}\pi x^3}.</math> We solve for <math>x</math> to obtain <math>x=\frac{15}{8}</math> (Note that we have disregarded the trivial solution <math>x=0</math>). We plug this value of <math>x</math> into our expression for <math>k</math> and simplify to obtain <math>k=\frac{125}{387}</math>. Hence <math>m+n=125+387=512</math>. |

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+ | Solution provided by 177h4x. | ||

== See also == | == See also == |

## Revision as of 18:40, 30 November 2006

## Problem

A solid in the shape of a right circular cone is 4 inches tall and its base has a 3-inch radius. The entire surface of the cone, including its base, is painted. A plane parallel to the base of the cone divides the cone into two solids, a smaller cone-shaped solid and a frustum-shaped solid in such a way that the ratio between the areas of the painted surfaces of and and the ratio between the volumes of and are both equal to Given that where and are relatively prime positive integers, find

## Solution

Let denote the radius of the small cone. Let and denote the area of the painted surface on cone and frustum , respectively. Let and denote the volume of cone and frustum , respectively. Using the Pythagorean Theorem, we find that the height and slant height of cone are and , respectively. Using the formula for lateral surface area of a cone, we find that . By subtracting from the lateral surface area of the original cone and adding to this the are of the base, we find that . Next, we find that Finally, we subtract from the volume of the original cone to find that . We know that Plugging in our values for , , , and , we obtain the equation We solve for to obtain (Note that we have disregarded the trivial solution ). We plug this value of into our expression for and simplify to obtain . Hence .

Solution provided by 177h4x.