A stick of length $a$ is broken in three parts. Find the probability that the length of each part is less than $b$, where $b>a/3$. A stick of length $a$ is broken in three parts. Find the probability that the length of each part is less than $b$, where $b>a/3$.
A sample space, $\Omega$, is defined as:
$$\Omega=\{(x,y): x>0,y>0,a-x-y>0\}$$
$$=\{(x,y): x>0,y>0,x+y<a\}$$
where $x,y,a-x-y$ are lenghts of broken parts.
Event $A$: "The length of every part is less than $b,b>a/3$."
$$A=\{(x,y)\in\Omega\::0<x<b,0<y<b,0<a-x-y<b\}$$
$$=\{(x,y)\in\Omega\::0<x<b,0<y<b,a-b<x+y<a\}$$
Now, what I don't understand is the following:
In my book's solution it says that we consider two cases:
First case  $$0<\frac{a}{3}<b\le \frac{a}{2}$$
In this case, $\Omega$ is a right angled triangle with sides $a$.
The problem in this case is how to determine event $A$ (I am given the geometric approach).
In my book's solution it says that event $A$ is also a right angled triangle.
How?
Shouldn't it be quadrilateral surface?
We have that
$$m(A)=\int_{a-2b}^b(b-(-x+a-b))dx$$
How?
Second case $$b > \frac{a}{2}$$
Here, $\Omega$ is defined as a square with side $b$, and event $A$ as a hexagonal surface.
How?
We have that $$m(A)=b^2-\frac{1}{2}(a-b)^2-\frac{1}{2}(2b-c)^2$$
Why do we choose $\frac{a}{2}$ as a bound in both cases?
 A: The most natural sample space is the square $\Omega:=[0,1]^2$ with area as probability measure. A point $(x,y)\in\Omega$ signifies that  the stick $[0,1]$ is partitioned into three pieces by making two cuts  independently at $x\in[0,1]$ and at $y\in[0,1]$. Assume that a number $\beta\in\bigl[{1\over3},1\bigr]$ is given. We want to know the probability $p_\beta$ that all three pieces have a length $\leq\beta$.
In the following figures only the case $x\leq y$ is represented in detail. In this case the lengths of the pieces are $x$, $y-x$, and $1-y$. It follows that a point $(x,y)$ is admissible iff
$$0\leq x\leq \beta\quad\wedge\quad  x\leq y\leq \beta+x\quad\wedge\quad  y\geq1-\beta\ .$$It turns out that we have to distinguish the cases ${1\over3}\leq\beta\leq{1\over2}$ and ${1\over2}\leq\beta\leq 1$. In the figures the admissible area is shaded, and $p_\beta$ is twice this area.

From the figures we can  read off
$$\eqalign{p_\beta&=2\cdot{1\over2}\bigl(\beta-(1-2\beta)\bigr)^2=(3\beta-1)^2\qquad\left({1\over3}\leq\beta\leq{1\over2}\right)\>,\cr p_\beta&=2\left(\beta^2-{1\over2}(1-\beta)^2-{1\over2}(2\beta-1)^2\right)=-3\beta^2+6\beta-2\qquad\left({1\over2}\leq\beta\leq1\right)\ .\cr}$$
