Stick of unit length is broken into three random pieces, what is the expected length of the longest piece? In regards to this question: Average length of the longest segment
Can anybody explain why the cumulative distribution function is in the form that is given?

Take a one-foot stick,lay it down, and you simultaneously  cut it at two points in order to get three pieces. It is equivalent to choose two points $X$ and $Y$ on the stick.
From left to right, the first piece $A$ will be of size the minimum between $X$ and $Y$ minus $0$.
The piece on the right  will be of size $1$ minus the maximum between $X$ and $Y$, or $1-B$ where $B=max(X,Y)$
The one in the middle is of size  $X-Y$ fs $X>Y$, or $Y-X$ if $X \leq Y$, in other words , $|X-Y|=B-A$
Finally, $C$ is the longest stick, that is $max(A,1-B,B-A)$
The cdf is then defined as follows 
$$F_C(a)=P(C \leq a)= P(max(A,1-B,B-A)\leq a)$$
If the maximum between $A,1-B$ and $B-A$ is smaller than $a$, therefore each of them are smaller than $a$, and reciprocally.
We can rewrite the cdf as 
$$F_C(a)=P(C \leq a)= P(A\leq a,1-B\leq a,B-A\leq a)$$
$X$ and Y are totally symmetrical, you can write 
$$P(A\leq a,1-B\leq a,B-A\leq a)=P(A\leq a,1-B\leq a,B-A\leq a,X<Y)+P(A\leq a,1-B\leq a,B-A\leq a,X \geq Y)$$
Thus,
$$P(A\leq a,1-B\leq a,B-A\leq a)=2P(X\leq a,1-Y\leq a,Y-X\leq a,X<Y)$$
A bit of geometry will give you the result : draw a square $[0,1]*[0,1]$, the vertical line $\{x=a\}$, the horizontal line $\{y=1-a\}$, the straight lines $y=a+x$ and $y=x$.
Locate the intersection of the surfaces defined by $P(A\leq a,1-B\leq a,B-A\leq a)$ and you will get the results.
