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Difference between revisions of "2003 AIME I Problems/Problem 5"

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== Solution ==
 
== Solution ==
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The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds, and the 1/8-th spheres (one centered at each vertex).
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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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There are 8 1/8-th spheres, each of radius 1. Together, their volume is <math> \frac43\pi </math>.
 
There are 8 1/8-th spheres, each of radius 1. Together, their volume is <math> \frac43\pi </math>.
  
−
The combined volume of these parts is <math> 60+94+\frac43\pi = \frac{464+4\pi}3 </math>.
+
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 = 464+4+3 = 471 </math>
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<math> m+n+p = 462+40+3 = 505 </math>
  
 
== See also ==
 
== See also ==

Revision as of 02:15, 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 $(m + n \pi)/p,$ where $m, n,$ and $p$ are positive integers, and $n$ and $p$ are relatively prime, find $m + n + p.$

Solution

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.


The volume of the parallelepiped is $60$ cubic units.

The volume of the external parallelepipeds is $2(12)+2(15)+2(20)=94$.

There are 8 1/8-th spheres, each of radius 1. Together, their volume is $\frac43\pi$.

There are 12 1/4-th cylinders, so 3 complete cylinders can be formed. Their volumes are $3\pi$, $4\pi$, and $5\pi$.

The combined volume of these parts is $60+94+\frac43\pi+3\pi+4\pi+5\pi = \frac{462+40\pi}3$.


$m+n+p = 462+40+3 = 505$

See also

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