Acute from Three Random Lengths

Worked solution

The answer is 1 − π/4 ≈ 0.2146. Lengths X, Y, Z form an acute triangle iff each square is less than the sum of the other two squares; this implies the triangle inequality as well. At most one of the three conditions can fail, so P(acute) = 1 − 3·P(Z² ≥ X² + Y²). For Z² ≥ X² + Y² we need ρ = √(X² + Y²) ≤ 1 and Z ∈ [ρ, 1]. So the probability is the integral of (1 − ρ) over the quarter disk x, y ≥ 0, x² + y² ≤ 1: ∫₀^{π/2} ∫₀¹ (1 − ρ) ρ dρ dθ = (π/2)(1/6) = π/12. Hence P(acute) = 1 − 3·π/12 = 1 − π/4. A pleasant coincidence: this equals the area of the unit square outside the quarter disk.