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1998 AHSME Problems/Problem 18

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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 ...

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)
Preceded by
Problem 17
Followed by
Problem 19
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