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Difference between revisions of "2011 AIME I Problems/Problem 13"
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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>
 
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>
  
−
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>.
+
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==

Revision as of 19:44, 6 July 2020

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

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 $\theta$ 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 $33+294+3=\boxed{330}$.

Video Solution

https://youtu.be/Wi-aqv8Ron0 ~Shreyas S

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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