Solution to Problem #41

Let A with coordinates (a,u), B(b,v) and C(c,w) be the three points chosen
uniformly and at random from a unit square.

As the probability to have 2 or 3 coincident points is nil, we will
consider hereafter the strict inequalities between the abscissas/ordinates
of the 3 points.

There are 6 positions of the 3 abscissas having the same probability:

a<b<c a<c<b b<a<c b<c<a c<a<b c<b<a.

For each of them, there are 6 equally likely positions of the ordinates:

u<v<w u<w<v v<u<w v<w<u w<u<v w<v<u.

So there are globally 36 possible configurations having the same
probability 1/36.

These 36 configurations can be split into 2 subsets S1 and S2
representing respectively 12 and 24 possibilities:

- S1: one of the 3 points ( B for example) is within the square one of
the diagonals of which joins the 2 other points (A and C for example).In
other words, min(a,c)<b<max(a,c) and min(u,w)<v<max(u,w).

- S2: any other of the 24 remaining configurations.

It is well known that:

1) if a,b,c on one hand and u,v,w on the other hand are independent the
ones from the others,then min(a,b,c)=min(u,v,w)=1/4,
max(a,b,c)=max(u,v,w)=3/4 and med(a,b,c)=med(u,v,w)=1/2

2) if min(a,c)<b<max(a,c), min(u,w)<v<max(u,w) and b<v,
then the expected value of b is 1/2+1/3*1/4=7/12 and the expected value of
v is 1/2-1/3*1/4=5/12.

In these conditions,it is easy to check that the "average" position of
the triangle ABC with the subset S1 is the triangle T1 whose vertices are
(1/4,1/4)  (7/12,5/12)  (3/4,3/4) or one of the three triangles obtained
from T1 by successive rotations of angle 90°.The 4 triangles are the same
area equal to 1/24.

As far S2 is concerned, the average position of the triangle ABC is the
triangle T2 whose vertices are (1/4,1/2)  (1/2,3/4)  (1/4,3/4) or one of
the three triangles obtained from T2 by successive rotations of angle
90°.The 4 triangles are the same area equal to 3/32.

Therefore the expected area of the resulting triangle is equal to
12/36*1/24 + 24/36*3/32 = 11/144