折断木棍成三段能否构成三角形

经典“折断木棍”问题，答案为

$$\frac{1}{4}.$$

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Original Explanation

Consider this problem in the 2D-Plane, let $x$ be the first break and $y$ be the second break such that $x < y$. Therefore, the lengths of the three segments are $x$, $y-x$, and $1-y$. As you recall from geometry, in order for three side lengths to form a triangle, each side length must be less than the sum of the other two side lengths. We can rewrite this as:

$$x < (y-x) + (1-y) \Rightarrow x < \frac{1}{2}$$ $$y-x < x + (1-y) \Rightarrow y < x + \frac{1}{2}$$ $$1-y < x + (y-x) \Rightarrow y > \frac{1}{2}$$

Given these constraints (including $x<y$), they cover $\frac{1}{8}$ of the sample space of $x,y \in [0, 1]$, which can be seen visually, along with the fact that the $x<y$ constraint only accounts for half of the possibilities since $x$ is equally likely to be greater than or less than $y$. The final answer is $2 \times \frac{1}{8} = \frac{1}{4}$.