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 $\frac{m + n\pi}{p},$ 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 $m+n+p = 462+40+3 = \boxed{505}$ .

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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