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

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== Problem ==
 
== Problem ==
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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>.
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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. Find <math>r+s+t</math>.
  
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==Solution==
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==Solution 1==
  
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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>.
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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>.  The distance from a point <math>(X,Y,Z)</math> to a plane with equation <math>Ax+By+Cz+D=0</math> is
 +
<cmath>\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},</cmath>
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so the (directed) distance from any point <math>(x,y,z)</math> 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, <cmath>(10-d)^2+(11-d)^2+(12-d)^2= 100\cdot(a^2+b^2+c^2)=100.</cmath>
  
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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=\boxed{330}</math>.
  
 
==Solution 2==
 
==Solution 2==
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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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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 <math>\theta</math> of the line:<math>(1)</math> through the centroid <math>(2)</math> perpendicular to the plane formed by <math>\triangle BCD</math>, <math>(3)</math> with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular <math>\textit{in slope}</math> to <math>BD</math>. Since <math>BD</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>.
 +
 
 +
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>.
  
 
==Solution 3==
 
==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 <math>\theta</math> of the line:<math>(1)</math> through the centroid <math>(2)</math> perpendicular to the plane formed by <math>\triangle BCD</math>, <math>(3)</math> with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular <math>in\text{ }slope</math> to <math>BD</math>. Since <math>BD</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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[[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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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>.
 
  
−
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>.
+
==Video Solution==
 +
https://youtube.com/watch?v=Wi-aqv8Ron0
  
 
== 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}}
 +
[[Category:Intermediate Geometry Problems]]
 +
[[Category:3D Geometry Problems]]

Latest revision as of 12:20, 31 August 2024

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. Find $r+s+t$.

Solution 1

Set the cube at the origin with the three vertices along the axes and the plane equal to $ax+by+cz+d=0$, where $a^2+b^2+c^2=1$. The distance from a point $(X,Y,Z)$ to a plane with equation $Ax+By+Cz+D=0$ is \[\frac{AX+BY+CZ+D}{\sqrt{A^2+B^2+C^2}},\] so 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 $33+294+3=\boxed{330}$.

Solution 2

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 $\theta$ of the line:$(1)$ through the centroid $(2)$ perpendicular to the plane formed by $\triangle BCD$, $(3)$ with the plane under the cube. Since the median is parallel to the plane, this orthogonal line is also perpendicular $\textit{in slope}$ to $BD$. Since $BD$ 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 $33+294+3=\boxed{330}$.

Solution 3

2011 AIME I 13.png


Video Solution

2011 AIME I #13

MathProblemSolvingSkills.com


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
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All AIME Problems and Solutions

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