The Probability of Forming a Triangle from Three Random Pieces of a Broken Stick

Imagine breaking a stick into three pieces at random points. The question arises: what is the probability that the three pieces can form a triangle? This problem involves a blend of geometry and probability, and its solution is a fascinating exploration of the interplay between these two areas.

Breaking the Stick and the Triangle Inequality

To form a triangle, the lengths of the three pieces must satisfy the triangle inequality. This states that the sum of any two sides of a triangle must be greater than the third side. In mathematical terms, if the lengths of the three pieces are (a), (b), and (c), then:

(a b > c) (a c > b) (b c > a)

For a stick of length 1, we break it at two random points, say (x_1) and (x_2), where (0 le x_1 le x_2 le 1). The lengths of the three pieces will then be:

Piece 1: (x_1) Piece 2: (x_2 - x_1) Piece 3: (1 - x_2)

Geometric Interpretation and Valid Combinations

The problem can be visualized using a unit square, where the coordinates ((x_1, x_2)) represent the positions of the breaks. The triangle inequality conditions translate into geometric constraints on these coordinates:

(x_2 ge x_1) (x_2 ge 1 - x_1) (1 - x_2 ge x_1)

These inequalities define the region in the unit square where the lengths of the three pieces can form a triangle. This region can be analyzed through integration or geometric methods, revealing the area where the conditions are satisfied.

Calculating the Probability

The area of the entire unit square is 1. The area of the region satisfying the triangle inequality can be found to be (frac{1}{4}). This is derived by considering the valid combinations of ((x_1, x_2)) within the constraints mentioned above.

The probability (P) that the three pieces can form a triangle is given by the ratio of the area satisfying the triangle inequalities to the total area of the unit square:

[ P frac{text{Area satisfying inequalities}}{text{Total area}} frac{frac{1}{4}}{1} frac{1}{4} ]

Real-World Considerations

Theoretical probabilities sometimes differ from practical outcomes. In reality, breaking a stick at two random points often results in lengths that are closer to (frac{1}{3}) and (frac{2}{3}) of the original length, which naturally form a triangle. This is because breaking the stick at points significantly closer to the ends or middle can be constrained by the geometry of the break.

Using a vice or specific tools would allow for more precise breaking, but most unaided breaks tend to be in the middle range, making a triangle a common outcome.

Mathematical Derivation

Mathematically, if we break the stick at (x_1) and (x_2) in ([0,1]), the lengths of the three pieces are (y x_1), (z x_2 - x_1), and (1 - x_2). For the three pieces to form a triangle, no piece can be greater than half the length of the stick:

[1 - x_2 le frac{1}{2} implies x_2 ge frac{1}{2}]

Similarly, the other inequalities lead to:

[x_1 ge frac{1}{2}, quad x_2 ge 1 - x_1]

These inequalities chop off the three corners of the original unit square, leaving a central triangular region with an area of (frac{1}{4}). Therefore, the probability of forming a triangle is (frac{1}{4}).

Conclusion

In conclusion, the probability of breaking a stick into three pieces that can form a triangle is (frac{1}{4}). This analytical result, while simple, highlights the importance of understanding the underlying principles of geometry and probability. Such idealized problems can indeed be misleading in practical scenarios, as real-world constraints often influence outcomes.