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

Key concept: This is one of those circle probability problems where symmetry does most of the work. The event is really about where the three random points land on the circumference, not about the triangle itself as a fixed shape.

Step by step

Let the circle have radius $R$. For any chord subtending a central angle $\theta$, its length is

$\ell=2R\sin\left(\frac{\theta}{2}\right)$

We want every side of the triangle to satisfy $\ell\le R$. So

$2R\sin\left(\frac{\theta}{2}\right)\le R$

which gives

$\sin\left(\frac{\theta}{2}\right)\le \frac12$

Since $0\le \theta\le \pi$ for the smaller central angle, this means

$\frac{\theta}{2}\le \frac{\pi}{6}\quad\Rightarrow\quad \theta\le \frac{\pi}{3}$

So a side is longer than the radius exactly when the corresponding arc between two points is larger than $120^\circ$.

Now label the three random points around the circle in order. The triangle will have a side longer than the radius if and only if the three points all lie inside some semicircle of length $2\pi/3$? A cleaner standard result is this: the triangle has all sides at most $R$ exactly when the three points are not contained in any arc of length $2\pi/3$ or greater.

Using the well-known circle-arc probability for three random points, the probability that all three lie in some arc of length at least $2\pi/3$ is

$3\left(\frac{2}{3}\right)^2-2\left(\frac{2}{3}\right)^3=\frac{4}{9}$

so the complementary probability is

$1-\frac49=\frac59$

Answer

$\boxed{\frac59}$

Pitfall alert

A common mistake is mixing up chord length with arc length. The condition is about side length of the triangle, so you need the chord formula first. Another trap is assuming the longest side is controlled by the largest angle without checking the threshold carefully. The probability step is not a brute-force counting problem; it comes from a rotational symmetry argument on arcs.

Try different conditions

If the question asked for the probability that no side is longer than a different threshold $kR$, the angle cutoff would change to

$2R\sin\left(\frac{\theta}{2}\right)\le kR$

and the arc-based probability would have to be recomputed with that new angular window. If the threshold were very small, the event would become rare; if it were at least $2R$, then the probability would be $1$ because every chord in a circle has length at most $2R$.

Further reading

inscribed triangle probability, chord length formula, circle arc symmetry