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# Planes and distance | AIME I, 2011 | Question 13

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

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

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.