Three points A , B , C are chosen at random on a circle of…

Three points $A$, $B$, $C$ are chosen at random on a circle of radius $R$. What is the probability that the triangle $ABC$ is acute-angled?

Step 1: Angles and arcs

The points $A,B,C$ cut the circle into three arcs $\overset{\frown}{BC}$, $\overset{\frown}{CA}$, $\overset{\frown}{AB}$ (not containing the third point). By the inscribed angle theorem,

\[ \angle A = \tfrac12\cdot(\text{central angle of arc } BC), \]

and similarly for $B$ and $C$. So $\angle A\lt90^\circ$ iff the arc $BC$ is less than a semicircle.

\[ \text{acute triangle} \iff \text{each of the three arcs is shorter than half the circle}. \]

Step 2: Reduce to breaking a segment

By rotational symmetry we may fix $A$. Cut the circle at $A$ and unroll it into a segment of length $2\pi R$. The points $B$ and $C$ become two independent uniform points on this segment, and the three arcs become the three pieces into which they cut it.

Step 3: Use Problem 81

By Problem 81, the probability that each of the three pieces is shorter than half of the segment is $\frac14$:

\[ \boxed{P(\text{acute}) = \frac14.} \]

The radius $R$ plays no role.

Step 4: The other cases

• A right triangle requires an arc of exactly a semicircle — probability $0$.
• So the triangle is obtuse with probability $\frac34$.

Intuition: the triangle is obtuse when all three points lie in some semicircle. For $n$ points on a circle, the probability that they all lie in some semicircle is $n/2^{n-1}$; for $n=3$ this is $\frac34$. ✓