Difference between revisions of "2003 AIME I Problems/Problem 5"
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== Solution == | == Solution == | ||
− | The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds, | + | 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. |
Line 12: | Line 12: | ||
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>. | ||
− | + | 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>. | |
+ | The combined volume of these parts is <math> 60+94+\frac43\pi+3\pi+4\pi+5\pi = \frac{462+40\pi}3 </math>. | ||
− | <math> m+n+p = | + | |
+ | <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 where
and
are positive integers, and
and
are relatively prime, find
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 cubic units.
The volume of the external parallelepipeds is .
There are 8 1/8-th spheres, each of radius 1. Together, their volume is .
There are 12 1/4-th cylinders, so 3 complete cylinders can be formed. Their volumes are ,
, and
.
The combined volume of these parts is .