2011 AIME I Problems/Problem 13

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

2011 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
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2011 AIME I Problems/Problem 13

Contents

Problem

Solution

Solution 2

Solution 3

See also