Difference between revisions of "2011 AIME I Problems/Problem 13"
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Revision as of 19:25, 4 July 2013
Problem
A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled 
. The three vertices adjacent to vertex are at heights 10, 11, and 12 above the plane. The distance from vertex to the plane can be expressed as 
, where 
, 
, and 
are positive integers, and 
. Find 
.
Solution
Set the cube at the origin with the three vertices along the axes and the plane equal to 
, where 
. Then the (directed) distance from any point (x,y,z) to the plane is 
. So, by looking at the three vertices, we have 
, and by rearranging and summing, 
.
Solving the equation is easier if we substitute 
, to get 
, or 
. The distance from the origin to the plane is simply d, which is equal to 
, so 
.
See also
| 2011 AIME I (Problems • Answer Key • Resources) | ||
| Preceded by Problem 12 |
Followed by Problem 14 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
| All AIME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
