Calculating probability for forming a triangle I am having trouble coming up with a solution for this problem:

There is a stick of unit length. We break it into two parts.
  Now, we pick the bigger one and break it into two parts.
  I want to calculate the probability that the three pieces form a triangle.

The problem is from "Introduction to Probability, Charles M. Grinstead", Chapter 2.2, Exercise 13
 A: Let $\lambda$ be the length of the bigger piece and we split it into two smaller  pieces $\lambda\mu$ and $\lambda(1-\mu)$. It is clear $\lambda$ and $\mu$ are
uniform random variables $\sim \mathcal{U}(\frac12,1)$ and $\mathcal{U}(0,1)$ respectively. 
In order for the three pieces with lengths $\;1-\lambda, \lambda\mu, \lambda(1-\mu)\;$ to form a triangle, the necessary and sufficient conditions are the fulfillment of following three triangular inequalities:
$$\begin{cases}
\lambda \mu + \lambda (1-\mu) &\ge 1-\lambda\\
\lambda \mu +  (1 - \lambda)  &\ge \lambda (1-\mu)\\
\lambda (1-\mu) + (1-\lambda) &\ge \lambda \mu
\end{cases}
\quad\iff\quad
\begin{cases}
\lambda \ge \frac12\\
\mu     \ge 1 - \frac{1}{2\lambda}\\
\frac{1}{2\lambda} \ge  \mu
\end{cases}$$
The first inequality is trivially satisfied because we are told to break the 
bigger piece.
The probability we seek is given by:
$$2\int_{1/2}^1 \int_{1-\frac{1}{2\lambda}}^{\frac{1}{2\lambda}} d\mu d\lambda
= 2\int_{1/2}^1 \left(\frac{1}{\lambda} - 1 \right) d\lambda
= 2\log 2 - 1
\approx 38.6294\%$$
A: Let $A$, $B$, and $C$ be the lengths of the three resulting pieces. Let $A$ be the length of the shorter of the two segments from the (presumably uniformly-distributed) initial cut. Clearly $A$ can be written as $A = u_1/2$, where $u_1 \sim \mathcal{U}(0,1)$. Now let $u_2 \sim \mathcal{U}(0,1)$ be independent of $u_1$. We can write $B = (1-A)u_2 = (1-u_1/2)u_2$ and $C = (1 - A)(1 - u_2) = (1-u_1/2)(1 - u_2)$.
In order for these to form a triangle, we need $A\le B+C$, $B\le C+A$, and $C\le A+B$. Substituting the above expressions for $A$,$B$, and $C$ in terms of $u_1$ and $u_2$, and using the fact that $A+B+C = 1$, these three conditions reduce to
$$
u_2 \le \frac{1}{2 - u_1}
$$
and
$$
u_2 \ge \frac{1-u_1}{2 - u_1}\, .
$$
The probability that this occurs is given by the area between the two curves defined by making these into equalities:
\begin{align}
\mathrm{P(triangle)} &= 
\int_0^1 du_1\, \left(\frac{1}{2 - u_1} - \frac{1 - u_1}{2 - u_1}\right) \\
&=  \log(4) - 1\\
&= 0.386294...
\end{align}
For what it's worth, I've checked this numerically with 100,000 trials of random "line-segment making", and got 0.38589.
A: Following is my approach which gives the conditional and joint probability densities explicitly, then shows the feasible region in which it is possible to form a triangle, and finally computes the probability of forming a triangle by integrating the joint PDF over the feasible region.
Explicit expression for the Joint PDF $f_{X,Y}(x,y)$:
Let $(X, Y)$ be the bivariate r.v. in which $X$ represents the position of the first break and $Y$ represents the position of the second break. Note that $X$ and $Y$ are not independent.
Since the stick is of of unit length, the distribution of $X$ is given by $X \sim \mathcal{U}(0,1)$. Hence the density function of $X$ is given by $f_X (x) = 1$ when $x \in (0,1)$, and $0$ elsewhere.
Now, let the first break occur at $X = x$. (Note that we are dealing with continuous r.v. So we will be using the formula described here.)
If $x < \frac{1}{2}$, the bigger half of the stick is the right half. Hence the second break $Y$ will be distributed uniformly between $x$ and $1$. i.e. $(Y | X = x) \sim \mathcal{U}(x,1)$ when $x < \frac{1}{2}$. Hence the conditional density function is given by $f_{Y | X} (y | X = x) = \frac{1}{1 - x}$ for all $y \in (x, 1)$, and $0$ everywhere else in the region $0 < x < \frac{1}{2}$. (This is the region $AEFGA$ shown in the following picture.)
Similarly, if $x > \frac{1}{2}$, the bigger half of the stick is the left half. Hence the second break $Y$ will be distributed uniformly between $0$ and $x$. i.e. $(Y | X = x) \sim \mathcal{U}(0,x)$ when $x > \frac{1}{2}$. Hence the conditional density function is given by $f_{Y | X} (y | X = x) = \frac{1}{x}$ for all $y \in (0, x)$, and $0$ everywhere else in the region $\frac{1}{2} < x < 1$. (This is the region $ABCDA$ shown in the following picture.)
Now, using the formula $f_{X,Y}(x,y) = f_{Y|X}(y | X = x) \cdot f_X (x)$, we get the following joint density function:
\begin{equation*}
f_{X,Y}(x,y) = 
\begin{cases}
\frac{1}{1-x} & \text{when $(x,y)$ is in region $AEFGA$} \\
\frac{1}{x} & \text{when $(x,y)$ is in region $ABCDA$} \\
0 & \text{elsewhere}
\end{cases}
\end{equation*}
(You can check that this is a legitimate PDF which satisfies $\iint_{\mathbb{R}^2} f_{X,Y}(x,y) dx dy = 1$).
Feasible Region

In region $AEFGA$, we have $0 < x < y < 1$, and the triangle will be formed if and only if the following three triangle inequalities are satisfied:
\begin{eqnarray*}
(x - 0) + (y - x) > 1 - y &\iff& y > \frac{1}{2} \\
(y - x) + (1 - y) > (x - 0) &\iff& x < \frac{1}{2} \\
(1 - y) + (x - 0) > y - x &\iff& y < x + \frac{1}{2}
\end{eqnarray*}
These three inequalities form the region inside $\Delta AIG$.
Similarly, in the region $ABCDA$, we can form a triangle if and only if $y < \frac{1}{2}$, $x > \frac{1}{2}$, and $y > x - \frac{1}{2}$. This is the region inside $\Delta AHD$.
Probability of forming a triangle
The probability of forming a triangle is given by:
\begin{eqnarray*}
P(\text{forming a triangle}) &=& 
\iint \limits_{\Delta AIG} f_{X,Y}(x,y) dx dy
+ \iint \limits_{\Delta AHD} f_{X,Y}(x,y) dx dy \\
&=& \int \limits_{x \;=\; 0}^{1/2} \;\;\; \int \limits_{y \;=\; 1/2}^{x + 1/2}
        \frac{1}{1-x} \;dx \;dy + 
    \int \limits_{x \;=\; 1/2}^{1} \;\;\; \int \limits_{y \;=\; x - 1/2}^{1/2}
        \frac{1}{x} \;dx \;dy \\
&=& \big( \ln(2) - \frac{1}{2} \big) + \big( \ln(2) - \frac{1}{2} \big) \\
&=& 2\ln(2) - 1 \\
&\approx& 38.6294.. \%
\end{eqnarray*}
