A Triangle on the Circle

Worked solution

The answer is 3/(2π) ≈ 0.4775. Let α, β, γ be the central angles of the three arcs between consecutive points, so α + β + γ = 2π. Splitting the triangle into three triangles with apex at the centre gives area = (1/2)(sin α + sin β + sin γ), which remains valid when the centre lies outside the triangle (the arc larger than π then contributes a negative sine, subtracting the extra piece). By linearity, E[area] = (3/2) E[sin α]. Each arc, as a fraction s = α/(2π) of the circle, has density 2(1 − s) on [0, 1] (the law of one spacing of three points on a circle). So E[sin α] = ∫₀¹ 2(1 − s) sin(2πs) ds = −2∫₀¹ s sin(2πs) ds = −2·(−1/(2π)) = 1/π. Therefore E[area] = 3/(2π), about 37% of the inscribed equilateral triangle.