# Forming a triangle out of 3 rods

A rod of the length 1 is randomly broken in two halves, so that the length of the right half is uniformly distributed on the unit interval. After that, the longer of the two halves is also broken into two pieces. How big is the probability that a triangle can be formed from the three rods that have been created?

My idea: Let $x,y,z$ be the length of the lower, middle and upper part, so that $x+y+z=1$. If we know $x,y$ we automatically also know $z$. With that, we have: $\Omega = \left\{ \, (x,y) \in \mathbb{R}^2 \, \left| \, x,y \geq 0 , x+y \leq 1 \, \right. \right\}$ as a sample space.

A triangle can be formed from the 3 rods if the triangular equations are fullfilled. With $z=1-x-y$, we receive the following equations:

$x+y \geq z \ \ \Leftrightarrow \ \ x+y \geq \frac{1}{2}$

$x+z \geq y \ \ \Leftrightarrow \ \ y \leq \frac{1}{2}$

$y+z \geq x \ \ \Leftrightarrow \ \ x \leq \frac{1}{2}$

The event "A triangle can be formed" is therefore described by:

$A = \left\{ \, (x,y) \in \Omega \, \left| \, x \leq \frac{1}{2} , y \leq \frac{1}{2} , x+y \geq \frac{1}{2} \, \right. \right\}$

Now thats the part, where I'm not really sure: If the lower part has the length $x$, than the restpiece has the lenth $1-x$, resulting in:

$P(A) \ = \ \int_{A}^{}~\frac{1}{1-x}~\mathrm{d}(x,y)$

$\to$ $P(A) \ = \ \int_{A}^{}~\frac{1}{1-x}~\mathrm{d}(x,y) \ = \ \int_0^{\frac{1}{2}}~\int_{\frac{1}{2} - x}^{\frac{1}{2}}~\frac{1}{1-x}~\mathrm{d}y~\mathrm{d}x \ = \ \ln{2} - \frac{1}{2} \ \approx \ 19{,}3 \ \mbox{%}$.

I would be really thankful if someone could help me here!

$$\frac{1-A}{2},\quad B\,\frac{1+A}{2},\quad (1-B)\,\frac{1+A}{2}$$ where $A,B$ are independent random variables, uniformly distributed over $[0,1]$.
These lengths are the side lengths of a triangle iff $\frac{1}{A+1}>B>\frac{A}{1+A}$.
It follows that the wanted probability is $$\int_{0}^{1}\int_{\frac{A}{A+1}}^{\frac{1}{A+1}}1\,dB\, dA=\int_{0}^{1}\frac{1-A}{1+A}\,dA=-1+2\log 2 \approx \color{red}{38.63\%}.$$