Amiya Kumar 
🧮 Probability of Forming a Triangle When a Rope is Cut Twice
A rope (or stick) of unit length is cut at two random points (uniformly). These two cuts divide the rope into three segments.
What is the probability that these three segments can form a triangle?
🔺 Triangle Inequality Condition
To form a triangle from three lengths \( a, b, c \), the triangle inequality must be satisfied:
- \( a + b > c \)
- \( b + c > a \)
- \( c + a > b \)
✂️ Cut Point Representation
Let the two cut points on the unit stick be \( x \) and \( y \), with \( 0 < x < y < 1 \). The three segment lengths will be:
- \( a = x \)
- \( b = y - x \)
- \( c = 1 - y \)
📊 Graphical Probability Area Approach
The valid region of all \((x, y)\) where \( 0 < x < y < 1 \) forms a triangle of area \( \frac{1}{2} \) on the coordinate plane. Within this, the region where triangle inequality holds has area \( \frac{1}{8} + \frac{1}{8} = \frac{1}{4} \).
Illustration of valid region (shaded) inside the triangle where triangle formation is possible.
Let’s verify each triangle condition:
- \( a + b = y \Rightarrow y > 1 - y \Rightarrow y > \frac{1}{2} \)
- \( b + c = 1 - x \Rightarrow 1 - x > x \Rightarrow x < \frac{1}{2} \)
- \( c + a = 1 - x \Rightarrow 1 - x > y - x \Rightarrow 1 > y \) (Always True)
Therefore, the valid region is bounded by:
- \( x < \frac{1}{2} \)
- \( y > \frac{1}{2} \)
- \( x < y \)
The overlap area (probability region) is \( \frac{1}{4} \) of the triangle defined by \( x < y \) in unit square.
🎯 Final Answer
Probability = \( \boxed{\frac{1}{4}} \)
