2011 AIME I Problems/Problem 13
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 .
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
Set the cube at the origin and the adjacent vertices as (10, 0, 0), (0, 10, 0) and (0, 0, 10). Then consider the plane ax + by + cz = 0. Because A has distance 0 to it (and distance d to the original, parallel plane), the distance from the other vertices to the plane is 10-d, 11-d, and 12-d respectively. The distance formula gives and An easy algebraic manipulation gives the equation in the first solution.
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