# 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.

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