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

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== Problem == | == 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 <math> C </math> and a frustum-shaped solid <math> F, </math> in such a way that the ratio between the | + | 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 <math> C </math> and a [[frustum]]-shaped solid <math> F, </math> in such a way that the [[ratio]] between the [[area]]s of the painted surfaces of <math> C </math> and <math> F </math> and the ratio between the [[volume]]s of <math> C </math> and <math> F </math> are both equal to <math> k</math>. Given that <math> k=\frac m n, </math> where <math> m </math> and <math> n </math> are [[relatively prime]] [[positive integer]]s, find <math> m+n. </math> |

== 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 | + | Our original solid has volume equal to <math>V = \frac13 \pi r^2 h = \frac13 \pi 3^2\cdot 4 = 12 \pi</math> and has [[surface area]] <math>A = \pi r^2 + \pi r \ell</math>, where <math>\ell</math> is the [[slant height]] of the cone. Using the [[Pythagorean Theorem]], we get <math>\ell = 5</math> and <math>A = 24\pi</math>. |

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+ | 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, and let <math>V_c</math> and <math>V_f</math> denote the volume of cone <math>C</math> and frustum <math>F</math>, respectively. Because the plane cut is parallel to the base of our solid, <math>C</math> is [[similar]] to the uncut solid and so 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 surface area of the original solid, we find that <math>A_f=24\pi - \frac{5}{3}\pi x^2</math>. | ||

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+ | Next, we can calculate <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 can take [[reciprocal]]s of both sides to simplify this [[equation]] to <math>\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1</math> and so <math>x = \frac{15}{8}</math>. Then <math>k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn</math> so the answer is <math>m+n=125+387=512</math>. | ||

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

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* [[2004 AIME I Problems]] | * [[2004 AIME I Problems]] | ||

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+ | [[Category:Intermediate Geometry Problems]] |

## Revision as of 19:45, 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

Our original solid has volume equal to and has surface area , where is the slant height of the cone. Using the Pythagorean Theorem, we get and .

Let denote the radius of the small cone. Let and denote the area of the painted surface on cone and frustum , respectively, and let and denote the volume of cone and frustum , respectively. Because the plane cut is parallel to the base of our solid, is similar to the uncut solid and so 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 surface area of the original solid, we find that .

Next, we can calculate . 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 can take reciprocals of both sides to simplify this equation to and so . Then so the answer is .