Probability that a random inscribed triangle has no side longer than the radius

Key takeaway: This is one of those geometry-probability questions where the first move is to check whether the event is even possible. If it cannot happen, the probability is zero and the proof is short.

Let the circle have radius $R$.

For a chord of a circle, the chord length is at most the radius only when the corresponding minor central angle is at most $60^\circ$, since

\quad\Longrightarrow\quad \sin\left(\frac{\theta}{2}\right)\le \frac12 \quad\Longrightarrow\quad \theta\le 60^\circ.$$ Now place the three random points around the circle. The three side lengths of the inscribed triangle correspond to the three minor arcs between the points. Those three arc measures add up to $360^\circ$. If every side were at most the radius, then each of the three corresponding arc measures would have to be at most $60^\circ$. But then their sum would be at most $$60^\circ+60^\circ+60^\circ=180^\circ,$$ which is impossible because the arc measures must total $360^\circ$. So the event cannot happen at all. ### Answer $$\boxed{0}$$ --- **Pitfalls the pros know** 👇 A subtle trap is to think the three pairwise chords are independent. They are not. Once the points are fixed on the circle, the three arc gaps must add to the full circumference, and that kills the event immediately. **What if the problem changes?** If the question were changed to ask for the probability that **at least one** side is longer than the radius, the answer would be $1$, because the complement event has probability $0$. If the threshold were larger than the radius, then the probability would no longer be trivial. `Tags`: inscribed triangle, chord length, central angle