Question

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

Answer

Answer: 512

Concepts:cone surface area power scaling of length, area, and volume

Difficulty rating: 2710

Solution:

The cone has radius $3,$ height $4,$ and slant height $5,$ so its painted surface consists of lateral area $\pi \cdot 3 \cdot 5 = 15\pi$ and base area $9\pi,$ totaling $24\pi.$ Suppose the cut is at similarity ratio $t,$ so $\mathcal{C}$ is a cone with radius $3t$ and slant height $5t.$ Then $\mathcal{C}$ 's painted surface is only its lateral area $15\pi t^2,$ and $\mathcal{F}$ 's painted surface is the rest, $24\pi - 15\pi t^2.$ The volumes are in ratio $t^3$ to $1 - t^3.$

Setting the two ratios equal, $\frac{15 t^2}{24 - 15 t^2} = \frac{t^3}{1 - t^3},$ so $15 t^2 (1 - t^3) = t^3 (24 - 15 t^2),$ which simplifies to $15 t^2 = 24 t^3,$ giving $t = \frac{5}{8}.$

Then $k = \frac{t^3}{1 - t^3} = \frac{125/512}{387/512} = \frac{125}{387},$ which is in lowest terms since $387 = 3^2 \cdot 43,$ so $m + n = 125 + 387 = 512.$

2004 AIME I Problem 11

Answer: 512

Difficulty rating: 2710

Solution:

Problem 11 in Other Years