Modified broken stick problem The following question is taken from Mark Joshi's Quant Job Interview.

Question: Let $x,y$ be uniformly distributed on $[0,1]$ and separate the unit interval $[0,1]$ into $3$ pieces. What is the probability of forming a triangle using the $3$ segments if the second point is chosen to break the longer segment left after the first point has been chosen?

This is a modified version of the well-known broken stick problem. The original problem does not have restriction on point to break the longer segment. For that problem, the answer is $\frac{1}{4}$ by merely checking triangle inequality.
However, in this modified question, I have no idea how to start at all.
Any hint is appreciated.
 A: Let's go in the geometric way. 

case a) - classic - breakings independent
The two breaking points ,  $x$ and $y$, are each uniformly and independently distributed on $[0,1]$.
That it is the same as saying that the point $(x,y)$ is uniformly distributed on the unit square $[0,1]^2$.
We can partition the outcomes into two different sets: $0 \le x \le y \le 1$ and the symmetric $0 \le y < x \le 1$ 
("above" and "below" the diagonal).
For a triangle to be formed, each piece shall be smaller that the sum of the others.
In the set "above", the three pieces have lengths $x, y-x, 1-y$.
Therefore we shall have
$$
\eqalign{
  & \left\{ \matrix{
  0 \le x \le y \le 1 \hfill \cr 
  \left( {1 - y < y} \right)\; \wedge \left( {x < 1 - y + y - x} \right)\; \wedge \;\left( {y - x < x + 1 - y} \right) \hfill \cr}  \right.\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad \left\{ \matrix{
  0 \le x \le y \le 1 \hfill \cr 
  \left( {1/2 < y} \right)\; \wedge \left( {x < 1/2} \right)\; \wedge \;\left( {y - x < 1/2} \right) \hfill \cr}  \right. \cr} 
$$
and it is easy to see that the conditions define a right triangle (that in yellow in the sketch) with sides $1/2$, and thus  with an area which is $1/4$
of the "above" diagonal triangle, and the two symmetric possible areas total as well $1/4$ of the unit square.
case b) -present - second break dependent on the first
The first cutting point $x$ is uniformly distributed on $[0,1]$.
Once $x$ has been selected,  $y$
is uniformly chosen on the longer segment, which is either $x$ or $1-x$.
That means that either $x \le 1/2$ and $x \le y$ or $1/2 < x$ and $y < x$.
The possible cases are represented by the unit square, deducting the two shaded triangles.
The positive results remain as above, but now their ratio is $1/3$.
A: Let $L_1,L_2,L_3$ denote the length of the $3$ segments.
A triangle cannot be made if and only if one of the following conditions is satisfied:


*

*$L_1+L_2\leq L_3$

*$L_1+L_3\leq L_2$

*$L_2+L_3\leq L_1$
Taking into account that $L_1+L_2+L_3=1$ the conditions can be rewritten as:


*

*$L_3\geq\frac12$

*$L_2\geq\frac12$

*$L_1\geq\frac12$
So we conclude that a triangle can be made iff $L_i<\frac12$ for $i=1,2,3$.
Projecting this on $x,y$ we find that a triangle can be made iff:
$$\left[x<\frac12\text{ and }\frac12<y<x+\frac12\right]\text{ or }\left[y<\frac12\text{ and }\frac12<x<y+\frac12\right]$$
Denoting this condition as $P(x,y)$ it now remains to find the area of: $$\left\{(x,y)\in[0,1]^2\mid P(x,y)\right\}$$
I leave that to you.
