Breaking a stick into a triangle

Question

A stick of length 1 is broken at two points chosen independently and uniformly at random along its length, producing three pieces. What is the probability that the three pieces can form a triangle?

Accepted Answer

The answer is \boxed{	frac14}. The triangle condition. Three positive lengths a, b, c with a+b+c = 1 form a (non-degenerate) triangle iff each satisfies the triangle inequality, i.e. each length is less than the sum of the other two. Since the total is 1, "a < b + c" is the same as "a < 1 - a", i.e. a < 	frac12. So the condition is simply: 	ext{every piece has length } < 	frac12. Set up the sample space. Let the two break points be x, y, each uniform on [0,1] and independent, a uniform point in the unit square [0,1]^2 (area 1). The three piece lengths depend on which break is to the left.…

Breaking a stick into a triangle

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