Difference between revisions of "1998 AHSME Problems/Problem 18"
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A right circular cone of volume <math>A</math>, a right circular cylinder of volume <math>M</math>, and a sphere of volume <math>C</math> 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 right circular cone of volume <math>A</math>, a right circular cylinder of volume <math>M</math>, and a sphere of volume <math>C</math> all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then | ||
| − | <math> \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 </math> | + | <math> \mathrm{(A) \ } A-M+C = 0 \qquad \mathrm{(B) \ } A+M=C \qquad \mathrm{(C) \ } 2A = M+C </math> |
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| + | <math>\qquad \mathrm{(D) \ }A^2 - M^2 + C^2 = 0 \qquad \mathrm{(E) \ } 2A + 2M = 3C </math> | ||
== Solution == | == Solution == | ||
Revision as of 16:09, 8 August 2011
Problem
A right circular cone of volume 
, a right circular cylinder of volume 
, and a sphere of volume 
all have the same radius, and the common height of the cone and the cylinder is equal to the diameter of the sphere. Then
Solution
Using the radius 
the three volumes can be computed as follows:
Clearly, 
the correct answer is 
.
The other linear combinations are obviously non-zero, and the left hand side of 
evaluates to 
which is negative. Thus is indeed the only correct answer.
See also
| 1998 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 17 |
Followed by Problem 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 | ||
