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