Difference between revisions of "2003 AIME I Problems/Problem 5"
m |
I_like_pie (talk | contribs) (Added Solution) |
||
| Line 3: | Line 3: | ||
== Solution == | == Solution == | ||
| − | { | + | The set can be broken into several parts: the big parallelepiped, the 6 external parallelepipeds, and the 1/8-th spheres (one centered at each vertex). |
| + | |||
| + | |||
| + | The volume of the parallelepiped is <math> 60 </math> cubic units. | ||
| + | |||
| + | The volume of the external parallelepipeds is <math> 2(12)+2(15)+2(20)=94 </math>. | ||
| + | |||
| + | There are 8 1/8-th spheres, each of radius 1. Together, their volume is <math> \frac43\pi </math>. | ||
| + | |||
| + | The combined volume of these parts is <math> 60+94+\frac43\pi = \frac{464+4\pi}3 </math>. | ||
| + | |||
| + | |||
| + | <math> m+n+p = 464+4+3 = 471 </math> | ||
| + | |||
== See also == | == See also == | ||
* [[2003 AIME I Problems/Problem 4 | Previous problem]] | * [[2003 AIME I Problems/Problem 4 | Previous problem]] | ||
Revision as of 01:06, 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, and the 1/8-th spheres (one centered at each vertex).
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 
.
The combined volume of these parts is 
.
