 Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 2011 based on Planes and Distance.

## Planes and distance- AIME I, 2011

A cube with side length 10 is suspended above a plane. The vertex closest to the plane is labelled 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 $\frac{r-s^\frac{1}{2}}{t}$, where r and s and t are positive integers and $r+s+t \lt 1000$, find r+s+t.

• is 107
• is 330
• is 840
• cannot be determined from the given information

### Key Concepts

Plane

Distance

Algebra

But try the problem first…

Source

AIME I, 2011, Question 13

Geometry Revisited by Coxeter

## Try with Hints

First hint

Cube at origin and adjacent vertices (10,0,0), (0,10,0) and (0,0,10) here plane ax+by+cz=0 A is distance 0 to this and distance d to given parallel plane and distance from other vertices to plane is 10-d,11-d,12-d

Second Hint

$\frac{a10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=10-d and $\frac{b10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=11-d and $\frac{c10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}$=12-d

Final Step

squaring and adding $100=(10-d)^{2}+(11-d)^{2}+(12-d)^{2}$ then having 11-d=y, 100=3$y^{2}$+2then y=$\frac{98}{3}^\frac{1}{2}$ then d=11-$\frac{98}{3}^\frac{1}{2}$=$\frac{33-294^\frac{1}{2}}{3}$ then 33+294+3=330.