# 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

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

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

## 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.

## See also

 1998 AHSME (Problems • Answer Key • Resources) Preceded byProblem 17 Followed byProblem 19 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 All AHSME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

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