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

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==Solution 2==
 
==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)}{\sqrt{a^2 + b^2 + c^2}} = 10-d,</cmath> <cmath>\frac{b(10)}{\sqrt{a^2 + b^2 + c^2}} = 11-d,</cmath> and <cmath>\frac{c(10)}{\sqrt{a^2 + b^2 + c^2}} = 12-d.</cmath> Squaring each equation and then adding yields <math>100=(10-d)^2+(11-d)^2+(12-d)^2</math>, and we can proceed as in the first solution.
 
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)}{\sqrt{a^2 + b^2 + c^2}} = 10-d,</cmath> <cmath>\frac{b(10)}{\sqrt{a^2 + b^2 + c^2}} = 11-d,</cmath> and <cmath>\frac{c(10)}{\sqrt{a^2 + b^2 + c^2}} = 12-d.</cmath> Squaring each equation and then adding yields <math>100=(10-d)^2+(11-d)^2+(12-d)^2</math>, and we can proceed as in the first solution.
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==Solution 3==
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Let the vertices with distance <math>10,11,12</math> be <math>B,C,D</math>, respectively. An equilateral triangle <math>\triangle BCD</math> is formed with side length <math>10\sqrt{2}</math>. We care only about the <math>z</math> coordinate: <math>B=10,C=11,D=12</math>. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so <math>\text{centroid}=(10+11+12)/3=11</math>. Designate the midpoint of <math>BD</math> as <math>M</math>. Notice that median <math>CM</math> is parallel to the plane because the <math>\text{centroid}</math> and vertex <math>C</math> have the same <math>z</math> coordinate, <math>11</math>, and the median contains <math>C</math> and the <math>\text{centroid}</math>. We seek the angle of the line through the centroid perpendicular to the plane formed by <math>\triangle BCD</math> with the plane under the cube, <math>\theta</math>. Since the median is parallel to the plane, this orthogonal line is also perpendicular <math>in slope</math> to <math>AC</math>. Since <math>AC</math> makes a <math>2-14-10\sqrt{2}</math> right triangle, the orthogonal line makes the same right triangle rotated <math>90^\circ</math>. Therefore, <math>\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}</math>.
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It is also known that the centroid of <math>\triangle BCD</math> is a third of the way between vertex <math>A</math> and <math>H</math>, the vertex farthest from the plane. Since <math>AH</math> is a diagonal of the cube, <math>AH=10\sqrt{3}</math>. So the distance from the <math>\text{centroid}</math> to <math>A</math> is <math>10/\sqrt{3}</math>. So, the <math>\Delta z</math> from <math>A</math> to the centroid is <math>\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}</math>.
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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>.
  
 
== 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 18:07, 25 February 2016

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 $r+s+t$.

Solution

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 (directed) 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, $(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100$.

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 $33+294+3=330$

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 \[\frac{a(10)}{\sqrt{a^2 + b^2 + c^2}} = 10-d,\] \[\frac{b(10)}{\sqrt{a^2 + b^2 + c^2}} = 11-d,\] and \[\frac{c(10)}{\sqrt{a^2 + b^2 + c^2}} = 12-d.\] Squaring each equation and then adding yields $100=(10-d)^2+(11-d)^2+(12-d)^2$, and we can proceed as in the first solution.

Solution 3

Let the vertices with distance $10,11,12$ be $B,C,D$, respectively. An equilateral triangle $\triangle BCD$ is formed with side length $10\sqrt{2}$. We care only about the $z$ coordinate: $B=10,C=11,D=12$. It is well known that the centroid of a triangle is the average of the coordinates of its three vertices, so $\text{centroid}=(10+11+12)/3=11$. Designate the midpoint of $BD$ as $M$. Notice that median $CM$ is parallel to the plane because the $\text{centroid}$ and vertex $C$ have the same $z$ coordinate, $11$, and the median contains $C$ and the $\text{centroid}$. We seek the angle of the line through the centroid perpendicular to the plane formed by $\triangle BCD$ with the plane under the cube, $\theta$. Since the median is parallel to the plane, this orthogonal line is also perpendicular $in slope$ to $AC$. Since $AC$ makes a $2-14-10\sqrt{2}$ right triangle, the orthogonal line makes the same right triangle rotated $90^\circ$. Therefore, $\sin\theta=\frac{14}{10\sqrt{2}}=\frac{7\sqrt{2}}{10}$.

It is also known that the centroid of $\triangle BCD$ is a third of the way between vertex $A$ and $H$, the vertex farthest from the plane. Since $AH$ is a diagonal of the cube, $AH=10\sqrt{3}$. So the distance from the $\text{centroid}$ to $A$ is $10/\sqrt{3}$. So, the $\Delta z$ from $A$ to the centroid is $\frac{10}{\sqrt{3}}\sin\theta=\frac{10}{\sqrt{3}}\left(\frac{7\sqrt{2}}{10}\right)=\frac{7\sqrt{6}}{3}$.

Thus the distance from $A$ to the plane is $11-\frac{7\sqrt{6}}{3}=\frac{33-7\sqrt{6}}{3}=\frac{33-\sqrt{294}}{3}$, and $33+294+3=\boxed{330}$.

See also

2011 AIME I (Problems • Answer Key • Resources)
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