Solution
Chord Through a Random Point
Show the problem again
A point P is chosen uniformly at random inside a disk of radius 1. Then a line through P is drawn in a direction chosen uniformly at random, independent of P. The line meets the disk in a chord. What is the probability that this chord is longer than √3 (the side of the inscribed equilateral triangle)? Give an exact closed form, or a decimal rounded to 4 places.
Worked solution
The answer is 1/3 + √3/(2π) ≈ 0.6090. The chord is longer than √3 exactly when the line passes within distance 1/2 of the centre. If P is at distance r from the centre and the line makes angle ψ with OP (ψ uniform), the line's distance from the centre is r|sin ψ|. If r ≤ 1/2 this is always below 1/2; that has probability 1/4. If r > 1/2, we need |sin ψ| < 1/(2r), which has probability (2/π) arcsin(1/(2r)). With r having density 2r, P = 1/4 + (2/π) ∫_{1/2}^1 2r arcsin(1/(2r)) dr. An antiderivative of 2r arcsin(1/(2r)) is r² arcsin(1/(2r)) + (1/4)√(4r² − 1), so the integral is (π/6 + √3/4) − π/8 = π/24 + √3/4. Then P = 1/4 + 1/12 + √3/(2π) = 1/3 + √3/(2π). Bertrand's three classic schemes give 1/3, 1/2 and 1/4; this one gives yet another value, which underlines that "a random chord" is meaningless until the sampling procedure is specified.
Source: Original problem written for AxiomIQ.