Difference between revisions of "2011 AIME I Problems/Problem 13"
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{{AIME box|year=2011|n=I|num-b=12|num-a=14}} | {{AIME box|year=2011|n=I|num-b=12|num-a=14}} | ||
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| + | Set the cube at the origin with the three vertices along the axes and the plane equal to ax+by+cz+d=0, where a^2+b^2+c^2=1. Then the distance from any point (x,y,z) to the plane is ax+by+cz+d. So, by looking at the three vertices, we have 10a+d=10, 10b+d=11, 10c+d=12, and by rearranging and summing, (10-d)^2+(11-d)^2+(12-d)^2= 100*(a^2+b^2+c^2)=100. Solving the equation is easier if we substitute 11-d=y, to get 3y^2+2=100, or y=sqrt )(98/3). The distance from the origin to the plane is simply d, which is equal to 11-sqrt(98/3) =(33-sqrt(294))/3, so 33+294+3=330. | ||
Revision as of 13:37, 18 May 2011
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 
.
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 | ||
Set the cube at the origin with the three vertices along the axes and the plane equal to ax+by+cz+d=0, where a^2+b^2+c^2=1. Then the distance from any point (x,y,z) to the plane is ax+by+cz+d. So, by looking at the three vertices, we have 10a+d=10, 10b+d=11, 10c+d=12, and by rearranging and summing, (10-d)^2+(11-d)^2+(12-d)^2= 100*(a^2+b^2+c^2)=100. Solving the equation is easier if we substitute 11-d=y, to get 3y^2+2=100, or y=sqrt )(98/3). The distance from the origin to the plane is simply d, which is equal to 11-sqrt(98/3) =(33-sqrt(294))/3, so 33+294+3=330.
