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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

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

Check the Answer


But try the problem first…

Answer: is 330.

Source
Suggested Reading

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.

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