Difference between revisions of "2011 AIME I Problems/Problem 13"
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Thus the distance from <math>A</math> to the plane is <math>11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}</math>, and <math>33+294+3=\boxed{330}</math>. | Thus the distance from <math>A</math> to the plane is <math>11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}</math>, and <math>33+294+3=\boxed{330}</math>. | ||
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+ | ==Solution 3== | ||
+ | [[File:2011 AIME I 13.png|500px]] | ||
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+ | ==Video Solution== | ||
+ | [https://youtu.be/35Gdasd1dfo?si=MM-Ipg8AT8_-d3K3 2011 AIME I #13] | ||
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+ | [https://mathproblemsolvingskills.wordpress.com/ MathProblemSolvingSkills.com] | ||
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==Video Solution== | ==Video Solution== |
Latest revision as of 12:20, 31 August 2024
Contents
[hide]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. Find .
Solution 1
Set the cube at the origin with the three vertices along the axes and the plane equal to , where . The distance from a point to a plane with equation is so the (directed) distance from any point 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 , which is equal to , so .
Solution 2
Let the vertices with distance be , respectively. An equilateral triangle is formed with side length . We care only about the coordinate: . It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so . Designate the midpoint of as . Notice that median is parallel to the plane because the and vertex have the same coordinate, , and the median contains and the . We seek the angle of the line: through the centroid perpendicular to the plane formed by , with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular to . Since makes a right triangle, the orthogonal line makes the same right triangle rotated . Therefore, .
It is also known that the centroid of is a third of the way between vertex and , the vertex farthest from the plane. Since is a diagonal of the cube, . So the distance from the to is . So, the from to the centroid is .
Thus the distance from to the plane is , and .
Solution 3
Video Solution
Video Solution
https://youtube.com/watch?v=Wi-aqv8Ron0
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 |
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