2004 AIME I Problem 11

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 = \frac{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

Small hint:

If the cut is at fraction

t

of the way up,

\mathcal{C}

has painted area

15\pi t^2

of the total painted

24\pi,

and volume fraction

t^3

Big hint:

Equating

\frac{15t^2}{24 - 15t^2} = \frac{t^3}{1 - t^3}

simplifies to

15 = 24t

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}.