Difference between revisions of "2011 AIME I Problems/Problem 13"

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== See also ==
 
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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 14:37, 18 May 2011

Problem

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled $A$. The three vertices adjacent to vertex $A$ are at heights 10, 11, and 12 above the plane. The distance from vertex $A$ to the plane can be expressed as $\frac{r-\sqrt{s}}{t}$, where $r$, $s$, and $t$ are positive integers, and $r+s+t<{1000}$. Find $r+s+t$.

See also

2011 AIME I (ProblemsAnswer KeyResources)
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.