2003 AIME I Problems/Problem 5

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

2003 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 4	Followed by Problem 6
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2003 AIME I Problems/Problem 5

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