Solution
Obtuse in a Square
Show the problem again
Three points are chosen independently and uniformly at random in the unit square. What is the probability that the triangle they form is obtuse? Give an exact closed form, or a decimal rounded to 4 places.
Worked solution
The answer is 97/150 + π/40 ≈ 0.7252. A triangle has at most one obtuse angle, so P(obtuse) = 3·P(the angle at C is obtuse) for a specified vertex C. By Thales' theorem, the angle at C is obtuse exactly when C lies strictly inside the circle with diameter AB. Hence P(obtuse) = 3·E[area of (disk on diameter AB) ∩ square]. Without the boundary, the disk has area π|AB|²/4, and E|AB|² = 1/3 gives π/12 per vertex. The real work is subtracting the parts of these disks that stick out of the square. Conditioning on A and B and integrating the clipped circular areas is a long but elementary computation (polynomial terms plus a π term), carried out by Langford, and it yields 97/150 + π/40 in total. A Monte Carlo simulation with 10⁷ triangles gives 0.7253, in agreement. Random triangles are obtuse most of the time: in a square, roughly 72.5% of the time.
Source: Result due to Eric Langford, 'The probability that a random triangle is obtuse', Biometrika (1969). Statement written for AxiomIQ.