Expected area of random triangle

Question

I want to check the following theorem by using Mathematica:

$\textbf{Theorem} $. If three point $P,Q,R$ are picked independently at random in the triangle $ABC$, than expected value of the area of triangle $PQR$ is

$S_{PQR}=\frac{1}{12}S_{ABC}$

I tried:

first I define triangle $ABC$ with area $S_{ABC}$

triangle = Triangle[{{0, 0}, {1, 1}, {2, 0}}];

area = Area[triangle]

but I don't know how to define new random triangle (random three point) inside the $ABC$ triangle.

I want to do something like this

t1=Graphics[{Red,triangle}]
t2=Graphics[{Black, rundomtriangle}]
Show[t1,t2]

than

S1=Area[triangle]
S2=Area[randomtriangle]

$\frac{S1}{S2}$ is a random number.

than plot this random numbers

ListPlot[S1/S2, S1/S2, S1/S2,...}]

but.. how to define randomtriangle ?

Answer 1

Edit I did not know about RandomPoint! New in v10.2 So just use

triangle = Triangle[{{0, 0}, {1, 1}, {2, 0}}];
randomtriangle = Triangle@RandomPoint[triangle, 3];

For the ListPlot:

areas = Area /@ Triangle /@ RandomPoint[triangle, {n, 3}];
ListPlot[Area[triangle]/area]

Unchanged:

For stuff like this (sampling from a region) the RegionDistribution function as defined in this post can be very useful. Copying the function unchanged:

RegionDistribution /: 
 Random`DistributionVector[RegionDistribution[reg_MeshRegion], 
  n_Integer, prec_?Positive] := 
 Module[{d = RegionDimension@reg, cells, measures, s, m}, 
  cells = Developer`ToPackedArray@MeshPrimitives[reg, d][[All, 1]];
  s = RandomVariate[DirichletDistribution@ConstantArray[1, d + 1], 
    n];
  measures = PropertyValue[{reg, d}, MeshCellMeasure];
  m = RandomVariate[#, n] &@
    EmpiricalDistribution[measures -> Range@Length@cells];
  #[[All, 1]] (1 - Total[s, {2}]) + Total[#[[All, 2 ;;]] s, {2}] &@
   cells[[m]]]

Then we can use it as follows:

triangle = Triangle[{{0, 0}, {1, 1}, {2, 0}}];

region = DiscretizeRegion@triangle;
pts = RandomVariate[RegionDistribution[region], 3];

randomtriangle = Triangle@pts;

t1 = Graphics[{
   {Red, triangle},
   {Black, randomtriangle}
 }]

Answer 2

The brute force approach:

        p0 = RandomReal[ {-1, 1}, {3, 2}]

{{-0.639468, -0.471533}, {-0.0621599, -0.952355}, {-0.253727, -0.133992}}

triangle = Triangle[p0];
a0 = Area[triangle];
p = Select[RandomReal[ {-1, 1}, {30000, 2}], 
   RegionMember[triangle, #] &];
Table[Area@Triangle[RandomSample[p, 3]]/a0, {1000}] // Mean

0.0826126 (* ~1/12 *)

I posted this mostly to point out a seemingly good approach that does not work:

p2 = {x, y} /. 
   FindInstance[ Element[ {x, y} , triangle], {x, y}, 1500];
Table[Check[Area@Triangle[RandomSample[p2, 3]]/a0, 0], {1000}] // Mean

0.151361

Why? Well the FindInstance distribution is not even close to uniform..

GraphicsRow[{Graphics[{Line[Append[p0, First@p0]], Point@p}], 
  Graphics[{Line[Append[p0, First@p0]], Point@p2}]}]

FindInstance give a lot of points exactly on the edges (So that many point triples are colinear, hence the need for Check )