AIME 2011 I · 第 13 题

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

Solution 4 (Intuition, not very detailed)

First, shift the cube down by 11 so the vertices adjacent to $A$ are $-1$, $0$, and $1$ above the plane.

Consider the two points 1 above and 1 below the plane:

Now we rotate the cube so that all 3 vertices adjacent to $A$ are on the plane. We can calculate $A$ to be $\frac{10\sqrt3}{3}$ below the plane.

Notice that when $A$ is rotated, it gets slightly closer to the plane.

The triangle shown in the diagram has legs of ration $1:7$, because it is similar to the triangle in the first diagram, since we are rotating. (kind of bad explanation)

So, we can compute $h$, where $h^2 + \left(\dfrac{h}{7}\right)^2 = \left(\dfrac{10\sqrt3}{3}\right)^2$.

Solving yields $h=\frac{7\sqrt{6}}{3}$.

Remember, this is the distance that $A$ is below the plane, so when we shift up by 11, $A$ is $11-\frac{7\sqrt{6}}{3}$ above the plane. Answer extraction yields $\boxed{330}$.

Video Solution

2011 AIME I #13

MathProblemSolvingSkills.com

Video Solution

https://youtube.com/watch?v=Wi-aqv8Ron0