TITLE: Probability of the Center of a Square Being Contained in A Triangle With Vertices on its Boundary
QUESTION [9 upvotes]: Background : I happen to love solving tough problems. Problem is, I simply cannot answer some! It happened again today, as I attempted to solve the questions in this site: http://www.skytopia.com/project/imath/imath.html
This site seems to have really, really difficult questions, and in fact, I'm struggling to answer the second question! That struggle is not without reason, as you shall understand after reading the below question...

The Question is: 3 points are chosen at random along the square's outline. They then combine to form a triangle. What is the probability that the centre of the square will be 'engulfed' by this triangle?

My effort: I divided the square as below (I tried my best people,- assume the gaps between the line segments are points; I've been only ably to name the vertices of the square), -
                                 a1 a2 a3 a4
                               A __ __ __ __ B
                            d1  |           |  b1
                            d2  |           |  b2
                            d3  |           |  b3
                            d4  |__ __ __ __|  b4
                               D             C
                                 c1 c2 c3 c4

After having made these divisions and naming them accordingly...

Considering point to be random, evaluated the probability of a point lying on any one of the sixteen segments to be 1/16.
Realised centre of the square is engulfed in some cases of 3 points lying on (some examples):-        
_(a1,b3,c3);(a2,b4,c4);(a3,c1,d1);(a4,c2,d2)_
_(b1,c3,d3);(b2,c4,d4);(b3,d1,a1);(b4,d2,a2)_
_(c1,d3,a3);(c2,d4,a4);(c3,a1,b1);(c4,a2,b2)_
_(d1,a3,b3);(d2,a4,b4);(d3,b1,c1);(d4,b2,c2)_

Realised that I'm not accounting for so many more favourable cases (Not able to aptly apply concept of permutation and combination in this question).
Realised that I'm not able to calculate the sample space.
Realised that the answer is much tougher to compute because I'm not accounting for special cases of collinearity, as well as cases where the vertices and points of separation in my diagram (Not able to assign those points any segment; even if one for each, square's vertices cause problems when evaluating cases, preventing successful generalisation)


A convincing, ingenious, enlightening and accurate solution to question bolded above will be immensely appreciated.

REPLY [4 votes]: To extend achille hui's answer even further, we can prove the following statement:

Let $\mu$ be a probability measure on the plane which is centrally symmetric around some point $O$ (i.e. such that if $S'$ is the reflection of $S$ through $O$, then $\mu(S)=\mu(S')$) and such that the measure of any line through $O$ is $0$. Then, the probability of three points chosen according that measure containing $O$ is $\frac{1}4$.

This eliminates the "convex" requirement of achille hui's answer (and is even slightly more general than that). Intuitively this means:

If we choose three points according to the same distribution, such that $A$ and its reflection through $O$ are equally likely to show up, then the probability that $O$ will be in the triangle formed by those points is $\frac{1}4$.

And it of course applies here, as choosing random points on the boundary of a square satisfies the condition.
Consider that once we have chosen two points $a$ and $b$, we can easily calculate the probability of the third forming a triangle as the measure of a certain "wedge" of the space:

where the shaded wedge is where the third point must lie. We will say that $p$ is the measure of that wedge (and the one opposite it), which is the probability of choosing an appropriate $c$ given $a$ and $b$.
As symmetry will allow us to extend this argument to either case, assume $b$ is to the right of the line $aO$, as is shown in the diagram above. If we fix $a$, then we can calculate that the measure of the set of $b$ such that the shaded region has measure of at least $p$ - in particular, this means that $b$ lies in region IV as noted in the diagram, which happens with probability $2p$ (the $2$ appears since we assumed $b$ was in region II or IV) - so the CDF of $p$ upon choosing $b$ at random is $P(p<k)=2k$. This probability is unconditional as the symmetries of the problem and the fact that $a$ was chosen arbitrarily suffice to eliminate the fact that we fixed $a$ and the side of the line $aO$ that $b$ was on. 
Knowing this, we can reduce the original problem to choosing some $a$, which doesn't matter, then picking $p$ out of a uniform distribution on $\left[0,\frac{1}2\right]$ (which represents choosing $b$), and finally picking a $c$ in the desired wedge with probability $p$. The overall probability of $c$ being in that wedge (and hence forming a triangle containing the origin) is thus expressed as $$\int_{0}^{\frac{1}2}2p\,dp=\frac{1}4.$$