expected area of a triangle determined by randomly placed points

Question

Three points are placed at independently and at random in a unit square. What is the expected value of the area of the triangle formed by the three points?

Answer 1

See here.

Answer 2

This isn't a full solution, but it goes most of the way there.

The area of a triangle $(x_1, y_1),(x_2,y_2),(x_3,y_3)$ is given by the formula

$A(x_1,x_2,x_3,y_1,y_2,y_3)= | \frac{x_1(y_2-y_3) + x_2 (y_3-y_1) + x_3 (y_1-y_2)}{2}|$

If $x_1, x_2, x_3, y_1, y_2, y_3$ are all independently identically uniformly distributed over $[0,1]$, then the average area is just given by:

$\int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 \int_0^1 A(x_1,x_2,x_3,y_1,y_2,y_3) d x_1 d x_2 d x_3 d y_1 d y_2 d y_3$

At this point, it's a fairly simple, though tedious, calculation. I recommend using Mathematica or some other computational software if you have access to it. There are also ways to simplify the computation based on inherent symmetries in the problem.

I can't post the final answer because I want to avoid giving a wrong answer, which is totally possible (I don't claim to be able to do the above integral by hand without errors). I can check the answer in Mathematica if you don't have access to it, though it will have to wait until at least Monday.

Answer 3

Here's a perl script that confirms the answer Shai linked to via a Monte Carlo approach.

#!/usr/bin/perl -w

$numTrials = 1000000 ;

sub distance {
   my $point1 = $_[0] ;
   my $point2 = $_[1] ;
   return sqrt(($x[$point1]-$x[$point2])**2 + ($y[$point1]-$y[$point2])**2) ;
}

sub heron {
   my $a = $legLength[$_[0]] ;
   my $b = $legLength[$_[1]] ;
   my $c = $legLength[$_[2]] ;
   my $s = ( $a + $b + $c ) / 2 ;
   return sqrt( $s * ( $s - $a ) * ( $s - $b ) * ( $s - $c ) ) ;
}

sub doAtriangle() {
   for ( my $j = 0; $j <= 2 ; $j++ ) {
      $x[$j] = rand(1) ;
      $y[$j] = rand(1) ;
   }   
   $legLength[0] = distance(0,1) ;
   $legLength[1] = distance(1,2) ;
   $legLength[2] = distance(2,0) ;
   return heron(0,1,2) ;   
}

for ( $i = 0 ; $i < $numTrials ; $i++ ) {
   $sum += doAtriangle() ;
}

print $sum/$numTrials . "\n" ;