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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
  
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<math> \textbf{(A)}\ A-M+C = 0\qquad\textbf{(B)}\ A+M = C\qquad\textbf{(C)}\ 2A = M+C </math>
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<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> \textbf{(D)}\ A^{2}-M^{2}+C^{2}= 0\qquad\textbf{(E)}\ 2A+2M = 3C </math>
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<math>\qquad \mathrm{(D) \ }A^2 - M^2 + C^2 = 0 \qquad \mathrm{(E) \ } 2A + 2M = 3C </math>
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==Solution==
 
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Let <math>r</math> be the radius of the cone, cynlinder, and sphere.
 
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Then <math>2r</math> will be the diameter of the sphere, and thus <math>2r</math> is also the height of the cone and cylinder.
 
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<math>A_{cone} = \frac{1}{3}(\pi r^2)(2r) = \frac{2\pi}{3} r^3</math>
 
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<math>M_{cyl} = (\pi r^2)(2r) = 2\pi r^3</math>
 
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<math>C_{sph} = \frac{4}{3}\pi r^3</math>
 
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Notice that <math>A + C = M</math>.  With a slight rearrangement, we get answer <math>\boxed{A}</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}}
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{{MAA Notice}}

Latest revision as of 13:29, 5 July 2013

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.

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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