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 == | ||
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== See also == | == See also == | ||
{{AHSME box|year=1998|num-b=17|num-a=19}} | {{AHSME box|year=1998|num-b=17|num-a=19}} | ||
+ | {{MAA Notice}} |
Latest revision as of 13:29, 5 July 2013
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 |
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