Question

A line segment is divided into 3 parts. What is the probability that they form sides of a possible triangle.

Answer 1

The probability is $\frac{1}{4}$.

Answer 2

Let the line segment have length $L$. We make two random cuts, dividing it into three parts of lengths $x, y, z$. The total length is $x+y+z = L$. For these three parts to form a triangle, they must satisfy the triangle inequality: $x+y>z$, $x+z>y$, and $y+z>x$. Since $x+y+z=L$, these inequalities simplify to $z<L/2$, $y<L/2$, and $x<L/2$. Thus, each part must be less than half the total length.

We can model this problem by considering two independent random variables $X$ and $Y$ representing the positions of the two cuts on a line segment of length 1, uniformly distributed in $[0, 1]$. The sample space is the unit square $[0,1]\times[0,1]$ with area 1. The lengths of the three parts depend on the order of $X$ and $Y$. Case 1: $X < Y$. The lengths are $X$, $Y-X$, and $1-Y$. The triangle inequalities are $X < 1/2$, $Y-X < 1/2$, and $1-Y < 1/2$ (which implies $Y > 1/2$). These translate to $X < 1/2$, $Y > 1/2$, and $Y < X+1/2$. The region satisfying these conditions within the $X<Y$ part of the unit square (the upper triangle) is a triangle with vertices $(0, 1/2)$, $(1/2, 1/2)$, and $(1/2, 1)$. Its area is $1/8$. Case 2: $Y < X$. The lengths are $Y$, $X-Y$, and $1-X$. The triangle inequalities are $Y < 1/2$, $X-Y < 1/2$, and $1-X < 1/2$ (which implies $X > 1/2$). These translate to $Y < 1/2$, $X > 1/2$, and $X < Y+1/2$. The region satisfying these conditions within the $Y<X$ part of the unit square (the lower triangle) is a triangle with vertices $(1/2, 0)$, $(1/2, 1/2)$, and $(1, 1/2)$. Its area is $1/8$.

The total favorable area is the sum of the areas from both cases: $1/8 + 1/8 = 1/4$. Since the total area of the sample space is 1, the probability is $1/4$.