We randomly select 3 points independently on the circumference of a circle. What is the probability that the triangle determined by them is acute?
I. solution. For the sake of simplicity, let's assume that the circumference of the circle is a unit length, and let $A$ be a fixed point on the circumference. By symmetry, the sought probability is the same as the probability that if we independently choose a point $B$ and a point $C$ on the circumference, the center $O$ of the circle will lie inside the triangle $ABC$. If the reflection of points $A, B$ over $O$ is denoted by $A', B'$, respectively, then this precisely means that

$B \neq A, A'$ and $C$ is an interior point of the (shorter) arc $A'B'$.

![](https://cdn.mathpix.com/cropped/2024_05_02_1932e0fb0d7aef91fc8fg-1.jpg?height=376&width=355&top_left_y=392&top_left_x=880)

By fixing a direction of traversal, we can establish a one-to-one correspondence between all pairs of points $(B, C)$ and the points $(b, c)$ in the unit square $[0,1) \times [0,1)$, where for a point $X$, $x$ denotes the length of the directed arc $AX$.

The set of points $(b, c)$ corresponding to the acute-angled triangle $ABC$ is illustrated by the shaded region in the diagram, ignoring the boundary points.

![](https://cdn.mathpix.com/cropped/2024_05_02_1932e0fb0d7aef91fc8fg-1.jpg?height=439&width=458&top_left_y=1033&top_left_x=823)

Since the probability that a point $P$ lies on a given arc $XY$ is equal to the length of the arc $XY$, due to the independent selection of points $B$ and $C$, we can argue that the sought probability is equal to the area of the shaded region (more precisely, the ratio of this area to the total area of the unit square), which is exactly $1/4$.

II. solution. We will continue to use the notation from the first solution. Consider the vertex $A$ as fixed, and suppose we have already chosen the vertex $B$; let the arc distance of $B$ from $A$ (the length of the shorter arc $AB$) be $\alpha (0 < \alpha < 1/2)$. We determine the (conditional) probability that the third point, $C$, when chosen, will form an acute-angled triangle $ABC$. As we have seen, this happens precisely when $C$ is an interior point of the shorter arc $A'B'$. The length of this arc, which is congruent to $AB$, is also $\alpha$, so the probability of a suitable choice of $C$ is $\alpha / 1 = \alpha$. The answer to the problem is the "average" of $\alpha$ (more precisely, the integral mean of the function $\alpha \rightarrow \alpha$ over the interval $(0, 1/2)$), which is $(1/8) / (1/2) = 1/4$.

Remark. We formulated the spatial variant of the problem in problem B. 4086. The solution provided there explains the suspicious "regularity" of the results: alongside the planar $1/4 = 1/2^2$, the corresponding probability for the tetrahedron is $1/8 = 1/2^3$. Any solution to the spatial variant can be easily adapted to the plane; the formulation of the resulting third solution (or solutions) is left to the Reader.