2003 AIME I Problem 5

Attempt Problem 5 of the 2003 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2003 AIME I solutions, or check the answer key.

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5.

Consider the set of points that are inside or within one unit of a rectangular parallelepiped (box) that measures 33 by 44 by 55 units. Given that the volume of this set is m+nπp,\frac{m + n\pi}{p}, where m,m, n,n, and pp are positive integers, and nn and pp are relatively prime, find m+n+p.m + n + p.

Answer: 505
Concepts:volumerectangular prismcylindersphere
Difficulty rating: 2210
Solution:

The region consists of the box itself, six slabs of thickness 11 projecting outward from the faces, quarter-cylinders of radius 11 along the twelve edges, and eighth-spheres of radius 11 at the eight corners. The box has volume 345=60,3 \cdot 4 \cdot 5 = 60, and the slabs total 2(34+35+45)=94.2(3 \cdot 4 + 3 \cdot 5 + 4 \cdot 5) = 94.

The four quarter-cylinders along edges parallel to each dimension combine into a full cylinder, so the cylinders total π12(3+4+5)=12π.\pi \cdot 1^2 (3 + 4 + 5) = 12\pi. The eight octants combine into one unit sphere of volume 4π3.\frac{4\pi}{3}.

The total volume is 60+94+12π+4π3=154+40π3=462+40π3, \begin{aligned} &60 + 94 + 12\pi \\ &\quad {}+ \frac{4\pi}{3} = 154 + \frac{40\pi}{3} \\ &= \frac{462 + 40\pi}{3}, \end{aligned} so m+n+pm + n + p =462+40+3= 462 + 40 + 3 =505.= 505.

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