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2004 AIME I Problems/Problem 11

Revision as of 21:08, 16 December 2015 by Jktkops (talk | contribs) (→‎Solution 2)

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 $k=\frac m n,$ where $m$ and $n$ are relatively prime positive integers, find $m+n.$

Solution

Solution 1

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

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

Next, we can calculate $V_c=\frac{1}{3}\pi r^2h=\frac{1}{3}\pi x^2 \left(\frac{4}{3}x\right)=\frac{4}{9}\pi x^3$. 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 can take reciprocals of both sides to simplify this equation to $\frac{72}{5x^2} - 1 = \frac{27}{x^3} - 1$ and so $x = \frac{15}{8}$. Then $k = \frac{\frac{5}{3}\pi x^2}{24\pi - \frac{5}{3}\pi x^2}= \frac{125}{387} = \frac mn$ so the answer is $m+n=125+387=\boxed{512}$.

Solution 2

Our original solid $V$ has surface area $A_v = \pi r^2 + \pi r \ell$, where $\ell$ is the slant height of the cone. Using the Pythagorean Theorem or Pythagorean Triple knowledge, we obtain $\ell = 5$ and lateral area $A_\ell = 15\pi$. The area of the base is $A_B = 3^2\pi = 9\pi$.

$V$ and $C$ are similar cones, because the plane that cut out $C$ was parallel to the base of $V$. Let $x$ be the scale factor between the original cone and the small cone $C$ in one dimension. Because the scale factor is uniform in all dimensions, $x^2$ relates corresponding areas of $C$ and $V$, and $x^3$ relates corresponding volumes. Then, the ratio of the painted areas $A_c$ to $A_f$ is $\frac{15\pi x^2}{9\pi + 15\pi - 15\pi x^2} = \frac{5 x^2}{8\pi - 5 x^2} = k$ and the ratio of the volumes $V_c$ to $V_f$ is $\frac{x^3}{1 - x^3} = k$. Since both ratios are equal to $k$, they are equal to each other. Therefore, $\frac{5 x^2}{8\pi - 5 x^2} = \frac{x^3}{1 - x^3}$.

Now we must merely solve for x and substitute back into either ratio. Cross multiplying gives $5 x^2(1 - x^3) = x^3(8 - 5 x^2)$. Dividing both sides by $x^2$ and distributing the $x$ on the right, we have $5 - 5 x^3 = 8 x - 5 x^3$, and so $8 x = 5$ and $x = \frac{5}{8}$. Substituting back into the easier ratio, we have $\frac{(\frac{5}{8})^3}{1 - (\frac{5}{8})^3} = \frac{\frac{125}{512}}{\frac{387}{512}} = \frac{125}{387}$. And so we have $m + n = 125 + 387 = \boxed{512}$.

See also

2004 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 10 		Followed by
Problem 12
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions

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