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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 <math>3</math> by <math>4</math> by <math>5</math> units. Given that the [[volume]] of this set is <math>\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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<center><asy>
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size(220);
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import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6));
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draw(box((0,-.1,0),(0.4,0.6,0.3)));
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draw(box((-.1,0,0),(0.5,0.5,0.3)));
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draw(box((0,0,-.1),(0.4,0.5,0.4)));
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draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1"));
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</asy></center>
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The set can be broken into several parts: the big <math>3\times 4 \times 5</math> parallelepiped, <math>6</math> external parallelepipeds that each share a face with the large parallelepiped and have a height of <math>1</math>, the <math>1/8</math> [[sphere]]s (one centered at each [[vertex]] of the large parallelepiped), and the <math>1/4</math> [[cylinder]]s connecting each adjacent pair of spheres.
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*The volume of the parallelepiped is <math>3 \times 4 \times 5 = 60 </math> cubic units.
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*The volume of the external parallelepipeds is <math>2(3 \times 4 \times 1)+2(3 \times 5 \times 1 )+2(4 \times 5 \times 1)=94 </math>.
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*There are <math>8</math> of the <math>1/8</math> spheres, each of radius <math>1</math>. Together, their volume is <math> \frac{4}{3}\pi </math>.
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*There are <math>12</math> of the <math>1/4</math> cylinders, so <math>3</math> 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 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 = \boxed{505} </math>.
  
 
== See also ==
 
== See also ==
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* [[2003 AIME I Problems]]
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{{AIME box|year=2003|n=I|num-b=4|num-a=6}}
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[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Latest revision as of 18:58, 4 July 2013

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 $\frac{m + n\pi}{p},$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$

Solution

[asy] size(220); import three; currentprojection = perspective(5,4,3); defaultpen(linetype("8 8")+linewidth(0.6)); draw(box((0,-.1,0),(0.4,0.6,0.3))); draw(box((-.1,0,0),(0.5,0.5,0.3))); draw(box((0,0,-.1),(0.4,0.5,0.4))); draw(box((0,0,0),(0.4,0.5,0.3)),linewidth(1.2)+linetype("1")); [/asy]

The set can be broken into several parts: the big $3\times 4 \times 5$ parallelepiped, $6$ external parallelepipeds that each share a face with the large parallelepiped and have a height of $1$, the $1/8$ spheres (one centered at each vertex of the large parallelepiped), and the $1/4$ cylinders connecting each adjacent pair of spheres.

  • The volume of the parallelepiped is $3 \times 4 \times 5 = 60$ cubic units.
  • The volume of the external parallelepipeds is $2(3 \times 4 \times 1)+2(3 \times 5 \times 1 )+2(4 \times 5 \times 1)=94$.
  • There are $8$ of the $1/8$ spheres, each of radius $1$. Together, their volume is $\frac{4}{3}\pi$.
  • There are $12$ of the $1/4$ 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 $m+n+p = 462+40+3 = \boxed{505}$.

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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