Difference between revisions of "2011 AIME I Problems/Problem 13"
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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 d, which is equal to <math>11-\sqrt{98/3} =(33-\sqrt{294})/3</math>, so <math>33+294+3=330</math> |
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+ | ==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 <cmath>\frac{a * (10-d)}{\sqrt{a^2 + b^2 + c^2}} = 10</cmath>, <cmath>\frac{b * (10-d)}{\sqrt{a^2 + b^2 + c^2}} = 11</cmath>, and <cmath>\frac{c * (10-d)}{\sqrt{a^2 + b^2 + c^2}} = 10</cmath>. An easy algebraic manipulation gives the equation in the first problem. | ||
== See also == | == See also == | ||
{{AIME box|year=2011|n=I|num-b=12|num-a=14}} | {{AIME box|year=2011|n=I|num-b=12|num-a=14}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 20:11, 22 March 2014
Contents
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
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
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
. An easy algebraic manipulation gives the equation in the first problem.
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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