Breaking a Stick into a Triangle

Three lengths form a triangle exactly when each one is shorter than the other two added together. The pieces add up to $1$; what does that say about each piece?

Call the break points $X$ and $Y$. Draw the unit square of possible $(X, Y)$ and shade where every piece is shorter than $\tfrac12$.

Step 1: The triangle inequality. Three lengths $a$, $b$, $c$ can form a triangle exactly when each one is shorter than the other two added together:

(If one side is as long as the other two together, the triangle is squashed flat.)

Step 2: Use that the pieces add up to $1$. Since $a + b + c = 1$, the first condition $a < b + c$ is the same as $a < 1 - a$, i.e. $a < \tfrac12$. The same goes for $b$ and $c$. So:

Step 3: Write the pieces in terms of the break points. Let the break points be $X$ and $Y$, each uniform on $[0, 1]$ and independent. Every pair $(X, Y)$ is a point in the unit square, and probability is area. Take the case $X < Y$ first. The pieces are $X$, $Y - X$ and $1 - Y$, and Step 2 asks for

Step 4: Find the area. These three lines cut out the triangle with corners $(0, \tfrac12)$, $(\tfrac12, \tfrac12)$ and $(\tfrac12, 1)$. It has two sides of length $\tfrac12$ at a right angle, so its area is $\tfrac12 \cdot \tfrac12 \cdot \tfrac12 = \tfrac18$.

The case $Y < X$ is the mirror image and gives another $\tfrac18$.

Answer: