The acute-triangle probability, Geometric Probability

Problem 26.1

Three points are chosen independently and uniformly on a circle. What is the probability that they form an acute triangle?

Solution. A classical theorem of elementary geometry says: a triangle inscribed in a circle is acute iff the circumcentre lies strictly inside the triangle. (The circumcentre of an inscribed triangle is the centre of the circle.) So the inscribed triangle is acute iff it contains the centre of the circle.

In Chapter 25 we showed that the probability that three random points on a circle form a triangle containing the centre is $\tfrac14$ . Hence the probability that they form an acute triangle is also $\boxed{\mathbb{P}(\text{acute}) = \tfrac14.} \qedhere$

An inscribed triangle in a circle is acute iff it contains the circumcentre. Left: an acute triangle, circumcentre inside. Right: a right triangle (centre on hypotenuse) or obtuse triangle (centre outside). Hence “acute” and “contains centre” are the same event.

import numpy as np
theta = np.random.rand(3, 10**7) * 2*np.pi
x, y = np.cos(theta), np.sin(theta)
sq = lambda i, j: (x[i]-x[j])**2 + (y[i]-y[j])**2
a2, b2, c2 = sq(1,2), sq(0,2), sq(0,1)
acute = (a2+b2 > c2) & (b2+c2 > a2) & (c2+a2 > b2)
print(f"sim: {acute.mean():.5f}   exact: 0.25")
# sim: 0.24993   exact: 0.25