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Difference between revisions of "2003 AIME I Problems/Problem 5"
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== Problem ==
 
== Problem ==
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Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is <math> (m + n \pi)/p, </math> where <math> m, n, </math> and <math> p </math> are positive integers, and <math> n </math> and <math> p </math> are relatively prime, find <math> m + n + p. </math>
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Consider the [[set]] of [[point]]s that are inside or within one unit of a [[rectangular prism|rectangular parallelepiped]] (box) that measures 3 by 4 by 5 units. Given that the [[volume]] of this set is <math>\displaystyle \frac{m + n\pi}{p}, </math> where <math> m, n, </math> and <math> p </math> are [[positive]] [[integer]]s, and <math> n </math> and <math> p </math> are [[relatively prime]], find <math> m + n + p. </math>
  
 
== Solution ==
 
== Solution ==
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The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds, the 1/8-th spheres (one centered at each vertex), and the 1/4-th cylinders connecting each adjacent pair of spheres.
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{{image}}
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The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds that each share a [[face]] with the large parallelepiped, the <math>\frac{1}{8}</math>th [[sphere]]s (one centered at each [[vertex]] of the large parallelepiped), and the <math>\frac{1}{4}</math>th [[cylinder]]s connecting each adjacent pair of spheres.
  
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*The volume of the parallelepiped is <math>3 \cdot 4 \cdot 5 = 60 </math> cubic units.
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*The volume of the external parallelepipeds is <math>\displaystyle 2(3 \cdot 4)+2(3 \cdot 5)+2(4 \cdot 5)=94 </math>.
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*There are 8 <math>\frac{1}{8}</math>th spheres, each of radius <math>1</math>. Together, their volume is <math> \frac{4}{3}\pi </math>.
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*There are 12 <math>\frac{1}{4}</math>th cylinders, so 3 complete cylinders can be formed. Their volumes are <math> 3\pi </math>, <math> 4\pi </math>, and <math> 5\pi </math>, adding up to <math>12\pi</math>.
  
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The volume of the parallelepiped is <math> 60 </math> cubic units.
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The combined volume of these parts is <math> 60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3} </math>. Thus, the answer is <math> m+n+p = 462+40+3 = 505 </math>.
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The volume of the external parallelepipeds is <math> 2(12)+2(15)+2(20)=94 </math>.
 
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There are 8 1/8-th spheres, each of radius 1. Together, their volume is <math> \frac43\pi </math>.
 
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There are 12 1/4-th cylinders, so 3 complete cylinders can be formed. Their volumes are <math> 3\pi </math>, <math> 4\pi </math>, and <math> 5\pi </math>.
 
−
 
 
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The combined volume of these parts is <math> 60+94+\frac43\pi+3\pi+4\pi+5\pi = \frac{462+40\pi}3 </math>.
 
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<math> m+n+p = 462+40+3 = 505 </math>
 
  
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems/Problem 4 | Previous problem]]
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{{AIME box|year=2003|n=I|num-b=4|num-a=6}}
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* [[2003 AIME I Problems/Problem 6 | Next problem]]
 
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* [[2003 AIME I Problems]]
 
  
 
[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]

Revision as of 15:57, 8 March 2007

Problem

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 3 by 4 by 5 units. Given that the volume of this set is $\displaystyle \frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$

Solution

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The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds that each share a face with the large parallelepiped, the $\frac{1}{8}$th spheres (one centered at each vertex of the large parallelepiped), and the $\frac{1}{4}$th cylinders connecting each adjacent pair of spheres.

  • The volume of the parallelepiped is $3 \cdot 4 \cdot 5 = 60$ cubic units.
  • The volume of the external parallelepipeds is $\displaystyle 2(3 \cdot 4)+2(3 \cdot 5)+2(4 \cdot 5)=94$.
  • There are 8 $\frac{1}{8}$th spheres, each of radius $1$. Together, their volume is $\frac{4}{3}\pi$.
  • There are 12 $\frac{1}{4}$th cylinders, so 3 complete cylinders can be formed. Their volumes are $3\pi$, $4\pi$, and $5\pi$, adding up to $12\pi$.

The combined volume of these parts is $60+94+\frac{4}{3}\pi+12\pi = \frac{462+40\pi}{3}$. Thus, the answer is $m+n+p = 462+40+3 = 505$.

See also

2003 AIME I (Problems • Answer Key • Resources)
Preceded by
Problem 4 		Followed by
Problem 6
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
All AIME Problems and Solutions
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