Acute Triangle from Random Points on a Circle
Three points are chosen uniformly at random on a circle. What is the probability that the triangle they form is acute?
Hints
- An inscribed triangle is acute if and only if the center of the circle lies inside the triangle -- what does this imply about the arc lengths?
- Fix one point and let the other two be uniform on $(0, 2\pi)$. Express the condition that all three arcs are less than $\pi$ as inequalities on the two angles.
- Rescale to the unit interval: if $u$ and $v$ are the ordered fractional positions, you need $u < 1/2$, $v - u < 1/2$, and $1 - v < 1/2$. Compute the area of this region relative to the triangle $0 < u < v < 1$.
Worked Solution
How to Think About It: An inscribed triangle is acute exactly when every inscribed angle is below $90^\circ$. By the inscribed-angle theorem, an inscribed angle is acute iff the arc it subtends is shorter than a semicircle, so the triangle is acute iff all three arcs between consecutive points are less than $\pi$ — equivalently, iff the center of the circle lies strictly inside the triangle. Translate that arc condition into inequalities on the points' positions and compute an area.
Quick Estimate: At most one arc can exceed $\pi$ (the three arcs sum to $2\pi$), so "obtuse" splits into three symmetric, mutually exclusive cases — one per vertex. If each has probability about $\frac14$, then $P(\text{obtuse}) \approx \frac34$ and $P(\text{acute}) \approx \frac14$. The exact count below confirms this.
Approach: Fix one point by rotational symmetry, place the other two by their ordered fractional positions, and find the area of the region where all three arcs are short.
Formal Solution:
Fix point $A$ at angle $0$. Let the other two points have fractional positions $u < v$ on $[0, 1]$, where position is measured as arc length divided by the full circumference. The pair $(u, v)$ is uniform on the triangle $0 < u < v < 1$, which has area $\frac12$.
The three arc fractions are $u$, $v - u$, and $1 - v$. The triangle is acute iff all three are less than $\frac12$: $u < \tfrac12, \qquad v - u < \tfrac12, \qquad 1 - v < \tfrac12.$ The last inequality gives $v > \frac12$; the middle gives $v < u + \frac12$. Combined with $u < \frac12$ and $u < v$, the favorable region is the triangle with vertices $\left(0, \tfrac12\right)$, $\left(\tfrac12, \tfrac12\right)$, $\left(\tfrac12, 1\right)$, whose area is $\frac12 \cdot \frac12 \cdot \frac12 = \frac18$.
Therefore $P(\text{acute}) = \frac{\text{favorable area}}{\text{sample area}} = \frac{1/8}{1/2} = \frac14.$
A quick Monte Carlo check (drawing three uniform points and testing that all arcs are under a half-circle) lands on $0.25$, confirming the result.
Answer: The probability that the inscribed triangle is acute is $\boxed{\dfrac14}$.
Intuition
The probability 1/4 is surprisingly low -- three-quarters of all inscribed triangles are obtuse. The reason is geometric: for the triangle to be acute, every arc must be less than a semicircle, meaning the three points must be "spread out" around the circle. But random points on a circle tend to cluster, leaving one large gap. In fact, the probability that some arc exceeds any threshold $\alpha > \pi$ increases as you allow more of the circle to be empty.
This problem illustrates a general pattern in geometric probability: uniform random configurations tend to be "degenerate" in some sense more often than intuition suggests. The same flavor appears in problems about random convex hulls, random simplices containing the origin, and coverage problems on the circle. The inscribed angle theorem converts a geometric question (acute triangle?) into a combinatorial one (all arcs short?), which is a powerful technique that shows up repeatedly in competition math and stochastic geometry.