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

## Check the Answer

Answer: is 330.

AIME I, 2011, Question 13

Geometry Revisited by Coxeter

## Try with Hints

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

\(\frac{a10}{({a^{2}+b^{2}+c^{2}})^\frac{1}{2}}\)=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

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.

## Other useful links

- https://www.cheenta.com/rational-number-and-integer-prmo-2019-question-9/
- https://www.youtube.com/watch?v=lBPFR9xequ