Difference between revisions of "2011 AIME I Problems/Problem 13"
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A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers, and <math>r+s+t<{1000}</math>. Find <math>r+s+t</math>. | A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labeled <math>A</math>. The three vertices adjacent to vertex <math>A</math> are at heights 10, 11, and 12 above the plane. The distance from vertex <math>A</math> to the plane can be expressed as <math> \frac{r-\sqrt{s}}{t}</math>, where <math>r</math>, <math>s</math>, and <math>t</math> are positive integers, and <math>r+s+t<{1000}</math>. Find <math>r+s+t</math>. | ||
− | ==Solution== | + | ==Solution 1== |
Set the cube at the origin with the three vertices along the axes and the plane equal to <math>ax+by+cz+d=0</math>, where <math>a^2+b^2+c^2=1</math>. Then the (directed) distance from any point (x,y,z) to the plane is <math>ax+by+cz+d</math>. So, by looking at the three vertices, we have <math>10a+d=10, 10b+d=11, 10c+d=12</math>, and by rearranging and summing, <math>(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100</math>. | Set the cube at the origin with the three vertices along the axes and the plane equal to <math>ax+by+cz+d=0</math>, where <math>a^2+b^2+c^2=1</math>. Then the (directed) distance from any point (x,y,z) to the plane is <math>ax+by+cz+d</math>. So, by looking at the three vertices, we have <math>10a+d=10, 10b+d=11, 10c+d=12</math>, and by rearranging and summing, <math>(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100</math>. | ||
− | Solving the equation is easier if we substitute <math>11-d=y</math>, to get <math>3y^2+2=100</math>, or <math>y=\sqrt {98/3}</math>. The distance from the origin to the plane is simply d, which is equal to <math>11-\sqrt{98/3} =(33-\sqrt{294})/3</math>, so <math>33+294+3=330</math> | + | Solving the equation is easier if we substitute <math>11-d=y</math>, to get <math>3y^2+2=100</math>, or <math>y=\sqrt {98/3}</math>. The distance from the origin to the plane is simply <math>d</math>, which is equal to <math>11-\sqrt{98/3} =(33-\sqrt{294})/3</math>, so <math>33+294+3=330</math> |
==Solution 2== | ==Solution 2== |
Revision as of 19:00, 5 November 2017
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 1
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 , which is equal to , so
Solution 2
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 Squaring each equation and then adding yields , and we can proceed as in the first solution.
Solution 3
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 .
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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