# 2004 AIME I Problems/Problem 11

## 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 $C$ and a frustum-shaped solid $F,$ in such a way that the ratio between the areas of the painted surfaces of $C$ and $F$ and the ratio between the volumes of $C$ and $F$ are both equal to $k.$ Given that $k=m/n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

## Solution

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

Solution provided by 1337h4x.