An Acute Break

Worked solution

The answer is 3 ln 2 − 2 ≈ 0.0794. The pieces (a, b, c) are uniform on the simplex a + b + c = 1. They form an acute triangle iff each square is less than the sum of the other two squares; this already implies the triangle inequalities. At most one piece can violate it, so P(acute) = 1 − 3·P(a² ≥ b² + c²). Use (b, c) as coordinates: uniform on the triangle b, c > 0, b + c < 1, with density 2. With a = 1 − b − c, the inequality (1 − b − c)² > b² + c² simplifies to 1 − 2b − 2c + 2bc > 0, i.e. c < (1 − 2b)/(2(1 − b)) for b < 1/2. Its area is ∫₀^{1/2} [1 − 1/(2(1 − b))] db = 1/2 − (ln 2)/2, so P(a² > b² + c²) = 2·(1/2 − (ln 2)/2) = 1 − ln 2. Therefore P(acute) = 1 − 3(1 − ln 2) = 3 ln 2 − 2: only about 8%, against 25% for any triangle at all.