2003 AIME I Problems/Problem 5

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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