Breaking a stick into a triangle

Idea. The triangle condition becomes “no piece is longer than half the stick”, and the two break points form a uniform point in the unit square, so the answer is an area.

Step 1: the condition. Lengths $a,b,c$ form a triangle when each is less than the sum of the other two. With $a+b+c=1$, $a<b+c$ means $a<\frac{1}{2}$. So all three pieces must be shorter than $\frac{1}{2}$.

Step 2: the region. Let $x<y$ be the two break points. The pieces are $x$, $y-x$ and $1-y$, so we need $$x<\frac{1}{2},y>\frac{1}{2},y-x<\frac{1}{2}.$$ In the triangle $0<x<y<1$, of area $\frac{1}{2}$, this is a smaller triangle with corners $(0,\frac{1}{2})$, $(\frac{1}{2},\frac{1}{2})$ and $(\frac{1}{2},1)$, of area $\frac{1}{8}$.

Step 3. $$\mathbb{P}(\text{triangle})=\frac{1/8}{1/2}=\frac{1}{4}.$$

Compare. Drawing three lengths independently and uniformly on $[0,1]$ gives a triangle with probability $\frac{1}{2}$. Here the lengths are tied together by their sum, which makes a triangle less likely.