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## 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

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.

Categories

## 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

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.