A segment of length l is divided at random into three parts.…

A segment of length $l$ is divided at random into three parts. Find the probability that a triangle can be constructed from these three parts.

Step 1: The model

Choose two points $x,y$ independently and uniformly on $[0,l]$. The point $(x,y)$ is uniform in the square $[0,l]^2$ of area $l^2$.

Step 2: The triangle condition

Three lengths $p,q,s$ with $p+q+s=l$ form a triangle iff $p\lt q+s$, $q\lt p+s$, $s\lt p+q$. Since $q+s=l-p$, the first inequality is $p\lt l-p$, i.e. $p\lt\frac l2$. So:

\[ \text{triangle} \iff \text{each of the three parts is shorter than } \tfrac l2. \]

Step 3: The case $x\lt y$

The parts are $x$, $y-x$, $l-y$. The conditions $x\lt\frac l2$, $y-x\lt\frac l2$, $l-y\lt\frac l2$ give

\[ x\lt\tfrac l2,\qquad y\gt\tfrac l2,\qquad y\lt x+\tfrac l2, \]

a triangle with vertices $(0,\frac l2)$, $(\frac l2,\frac l2)$, $(\frac l2,l)$ and legs $\frac l2$: area $\frac{l^2}{8}$.

Step 4: The case $x\gt y$

By symmetry (swap $x$ and $y$) we get another triangle of area $\frac{l^2}8$.

Step 5: The probability

\[ \boxed{P = \frac{2\cdot\frac{l^2}{8}}{l^2} = \frac14.} \]

The answer does not depend on $l$.

Step 6: A quick intuition

A triangle fails exactly when one part is at least half of the segment. Each of the three parts is the long one with probability $\frac14$, and at most one part can be longer than half, so $P(\text{no triangle}) = 3\cdot\frac14 = \frac34$ and $P = \frac14$.