Three sticks forming a triangle, Geometric Probability

Problem 1.1

Let $a, b, c$ be independent uniform random variables in $[0,1]$ . What is the probability that sticks of lengths $a, b, c$ form a triangle?

Solution. Three positive lengths form a triangle iff each is less than the sum of the other two. The parameter space is the unit cube $[0,1]^3$ , with total volume $1$ .

The triangle inequality $a < b + c$ fails in the region $\{a \geq b + c\}$ , which is the corner tetrahedron with vertices $(0,0,0), (1,0,0), (1,1,0), (1,0,1)$ . Its volume is $\tfrac16$ (one-sixth of the unit cube). By symmetry, the analogous tetrahedra for $b \geq a + c$ and $c \geq a + b$ also have volume $\tfrac16$ each. These three bad regions are disjoint: at most one side can exceed the sum of the other two.

Hence the bad volume is $3 \cdot \tfrac16 = \tfrac12$ , and the good volume — the triangle region — is $1 - \tfrac12 = \tfrac12$ . The probability of forming a triangle is $\boxed{\mathbb{P} = \tfrac12.} \qedhere$

The unit cube [0,1]^3 of stick lengths (a,b,c) . The three shaded corner tetrahedra (copper) are the configurations where one stick is at least as long as the other two combined; each has volume \tfrac16 . The remaining volume \tfrac12 is the triangle region. — The unit cube $[0,1]^3$ of stick lengths $(a,b,c)$ . The three shaded corner tetrahedra (copper) are the configurations where one stick is at least as long as the other two combined; each has volume $\tfrac16$ . The remaining volume $\tfrac12$ is the triangle region.

import numpy as np
a, b, c = np.random.rand(3, 10**7)
p = ((a+b>c) & (b+c>a) & (c+a>b)).mean()
print(f"sim: {p:.5f}   exact: 0.5")
# sim: 0.50005   exact: 0.5