The expected area of a triangle formed by three points randomly chosen from the unit square

"Three points are chosen uniformly and at random from a unit square. What is the expected value of the area of the resulting triangle?"

I need to do a research about that problem and i found this suggested solution: here.

Now, I understand almost everything except anecdote (2) when he computes the expected value of $b$ and $v$. I can't understand how he reaches those calculations.

If someone can explain to me this that would be great.

Thanks.

Solution 1:

Here is an approach which is different from the one in the link, does not involve so many cases, and is self-contained.

Let $A=(a_1,a_2)$, $B=(b_1,b_2)$, $C=(c_1,c_2)$ be the three vertices of the random triangle $T$. It is sufficient to consider the case $a_2<b_2<c_2$, which takes ${1\over6}$ of the total "volume". Fix $a_2$, $b_2$, $c_2$ for the moment, and write $$b_2=(1-t)a_2+t c_2,\qquad0\leq t\leq 1\ .$$ The side $AC$ of $T$ intersects the horizontal level $y=b_2$ at the point $S=(s ,b_2)$ with $$s=s(a_1,c_1,t)=(1-t)a_1+t c_1\ .\tag{1}$$ The area $X$ of $T$ is then given by $$X={1\over2}|b_1-s|(c_2-a_2)\ .$$

We now start integrating with respect to our six variables. The innermost integral is with respect to $b_1$ and gives $$\eqalign{X_1&:=\int_0^1 X\>db_1={1\over2}(c_2-a_2)\left(\int_0^s (s-b_1)\>db_1+\int_s^1(b_1-s)\>db_1\right)\cr &={1\over4}(c_2-a_2)(1-2s+2s^2)\ .\cr}$$ Next we integrate over $b_2$: $$X_2:=\int_{a_2}^{c_2} X_1\>db_2={1\over4}(c_2-a_2)^2\int_0^1(1-2s+2s^2)\>dt \ ,$$ whereby $s$ is given by $(1)$, and does not depend on $a_2$ and $c_2$. It follows that from now on the integration factorizes into $$X_3:={1\over4}\ \int_0^1\int_{a_2}^1(c_2-a_2)^2\>dc_2\> da_2\ \times\ \int_0^1\int_0^1\int_0^1 (1-2s+2s^2)\>dt\> dc_1\>da_1\ .$$ The easy computation gives $$X_3={1\over4}\cdot{1\over12}\cdot{11\over18}={11\over 6\cdot 144}\ .$$ From this the end result is obtained after multiplying by $6$, in order to take care of the assumption $a_2<b_2<c_2$ made at the beginning. The probability in question is therefore given by ${11\over144}$, as obtained in the source quoted by the OP.