Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "1998 AHSME Problems/Problem 18"

(New page: == Problem == 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 co...)
 
(→‎Problem)
Line 2: Line 2:
 
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> \textbf{(A)}\ A-M+C = 0\qquad\textbf{(B)}\ A+M = C\qquad\textbf{(C)}\ 2A = M+C </math>
 +
 
 +
<math> \textbf{(D)}\ A^{2}-M^{2}+C^{2}= 0\qquad\textbf{(E)}\ 2A+2M = 3C </math>
 +
 
 +
 
 +
==Solution==
 +
 
 +
Let <math>r</math> be the radius of the cone, cynlinder, and sphere.
 +
 
 +
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.
 +
 
 +
<math>A_{cone} = \frac{1}{3}(\pi r^2)(2r) = \frac{2\pi}{3} r^3</math>
 +
 
 +
<math>M_{cyl} = (\pi r^2)(2r) = 2\pi r^3</math>
 +
 
 +
<math>C_{sph} = \frac{4}{3}\pi r^3</math>
 +
 
 +
Notice that <math>A + C = M</math>.  With a slight rearrangement, we get answer <math>\boxed{A}</math>.
  
 
== Solution ==
 
== Solution ==

Revision as of 23:33, 7 August 2011

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

$\textbf{(A)}\ A-M+C = 0\qquad\textbf{(B)}\ A+M = C\qquad\textbf{(C)}\ 2A = M+C$

$\textbf{(D)}\ A^{2}-M^{2}+C^{2}= 0\qquad\textbf{(E)}\ 2A+2M = 3C$


Solution

Let $r$ be the radius of the cone, cynlinder, and sphere.

Then $2r$ will be the diameter of the sphere, and thus $2r$ is also the height of the cone and cylinder.

$A_{cone} = \frac{1}{3}(\pi r^2)(2r) = \frac{2\pi}{3} r^3$

$M_{cyl} = (\pi r^2)(2r) = 2\pi r^3$

$C_{sph} = \frac{4}{3}\pi r^3$

Notice that $A + C = M$. With a slight rearrangement, we get answer $\boxed{A}$.

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
Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=1998_AHSME_Problems/Problem_18&oldid=41246"