1998 AMC 12 Problem 18

Attempt Problem 18 of the 1998 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1998 AMC 12 solutions, or check the answer key.

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18.

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

A-M+C=0

A+M=C

2A=M+C

A^2-M^2+C^2=0

2A+2M=3C

Let the common radius be

r

, so the cone and cylinder height is

2r

Write all three volumes as multiples of

\pi r^3

The volumes are

A=\frac23\pi r^3,

M=2\pi r^3,

and

C=\frac43\pi r^3.

Therefore

A-M+C

=(\frac{2}{3}-2+\frac{4}{3})\pi r^3=0,

so A is correct.

Problem 18 in other years