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 $\frac{a(10-d)}{\sqrt{a^2 + b^2 + c^2}} = 10,$ $\frac{b(10-d)}{\sqrt{a^2 + b^2 + c^2}} = 11,$ and $\frac{c(10-d)}{\sqrt{a^2 + b^2 + c^2}} = 10.$ An easy algebraic manipulation gives the equation in the first problem.

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
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2011 AIME I Problems/Problem 13

Contents

Problem

Solution

Solution 2

See also