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"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.

David G. Stork
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kitsuneFox
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    It's always best to try to make problems self-contained. – Thomas Andrews Apr 16 '15 at 04:50
  • @ThomasAndrews, sorry about that, here is the full problem : "Three points are chosen uniformly and at random from a unit square. What is the expected value of the area of the resulting triangle?" – kitsuneFox Apr 16 '15 at 04:52
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    Put it in the question. Volunteers are coming here to help you, and they think the question is complete, and don't look in comments for clarification. Help them help you. – Thomas Andrews Apr 16 '15 at 04:58
  • A much earlier post of exactly the same question. The answers are consistent. – Lee David Chung Lin Mar 29 '18 at 13:29
  • This problem is directly related to Sylvester's Four-Point Problem, which is well-studied. The expected area $E$ of a triangle of $3$ u.i.r. chosen points is in one-to-one relation with the probability $P$, that four u.i.r. chosen points have a convex hull which is a quadrilateral. More precisely, for fixed shape of area $A$, it holds $P=1-4E/A$. Exact values of $E$ for the shapes triangle, square, pentagon, hexagon and disk are known, see the following Wolfram MathWorld URL for more information – Manfred Scheucher Nov 09 '21 at 11:44

1 Answers1

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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.

Christian Blatter
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  • It would be interesting to see how to get the expected area of a triangle that is formed by three randomly chosen points inside the unit circle. – user164118 Apr 19 '15 at 09:36
  • @user164118, in case you would be interested, I have asked a similar question: the expected area of a circle inside a triangle https://math.stackexchange.com/q/1711389/269624 – Yuriy S Sep 15 '17 at 14:54