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AIME I Algebra Arithmetic Geometry Math Olympiad USA Math Olympiad

Number of points and planes | AIME I, 1999 | Question 10

Try this beautiful problem from the American Invitational Mathematics Examination I, AIME I, 1999 based on Number of points and planes.

Number of points and planes – AIME I, 1999


Ten points in the plane are given with no three collinear. Four distinct segments joining pairs of three points are chosen at random, all such segments being equally likely.The probability that some three of the segments form a triangle whose vertices are among the ten given points is \(\frac{m}{n}\) where m and n are relatively prime positive integers, find m+n.

  • is 107
  • is 489
  • is 840
  • cannot be determined from the given information

Key Concepts


Number of points

Plane

Probability

Check the Answer


Answer: is 489.

AIME I, 1999, Question 10

Geometry Vol I to IV by Hall and Stevens

Try with Hints


\(10 \choose 3\) sets of 3 points which form triangles,

fourth distinct segment excluding 3 segments of triangles=45-3=42

Required probability=\(\frac{{10 \choose 3} \times 42}{45 \choose 4}\)

where \({45 \choose 4}\) is choosing 4 segments from 45 segments

=\(\frac{16}{473}\) then m+n=16+473=489.

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Categories
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.

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.

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