1998 AHSME Problems/Problem 18

Problem

A right circular cone of volume $A$ , a right circular cylinder of volume $M$ , and a sphere of volume $C$ all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then

$\textbf{(A)}\ A-M+C = 0\qquad\textbf{(B)}\ A+M = C\qquad\textbf{(C)}\ 2A = M+C$

$\textbf{(D)}\ A^{2}-M^{2}+C^{2}= 0\qquad\textbf{(E)}\ 2A+2M = 3C$

Solution

Let $r$ be the radius of the cone, cynlinder, and sphere.

Then $2r$ will be the diameter of the sphere, and thus $2r$ is also the height of the cone and cylinder.

$A_{cone} = \frac{1}{3}(\pi r^2)(2r) = \frac{2\pi}{3} r^3$

$M_{cyl} = (\pi r^2)(2r) = 2\pi r^3$

$C_{sph} = \frac{4}{3}\pi r^3$

Notice that $A + C = M$ . With a slight rearrangement, we get answer $\boxed{A}$ .

Solution

Using the radius $r$ the three volumes can be computed as follows:

$A = \frac 13 (\pi r^2) \cdot 2r$

$M = (\pi r^2) \cdot 2r$

$C = \frac 43 \pi r^3$

Clearly, $M = A+C \Longrightarrow$ the correct answer is $\mathrm{(A)}$ .

The other linear combinations are obviously non-zero, and the left hand side of $\mathrm{(D)}$ evaluates to $(\pi r^3)^2 \cdot \left( \frac 49 - 4 + \frac {16}9 \right)$ which is negative. Thus $\mathrm{(A)}$ is indeed the only correct answer.

1998 AHSME (Problems • Answer Key • Resources)
Preceded by Problem 17	Followed by Problem 19
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1998 AHSME Problems/Problem 18

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Solution

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