Probability that the triangle is acute

A triangle is formed by randomly choosing three distinct points on the circumference of a circle and joining them.

What is the probability that the formed triangle is an acute triangle?

Hint: Consider the unit circle, centered at $$(0,0)$$ and fix the first point as $$A=(-1,0)$$. Consider the point $$B$$ in quadrant II. There must be a region of the circle where we can put our third point, $$C$$, such that $$\triangle ABC$$ is acute.

Label the points on the circle opposite $$A$$ and $$B$$ as $$A'$$ and $$B'$$, respectively. Thales' Theorem and the Inscribed Angle Theorem tell us how the angles of the triangle change when $$C$$ is moved around the circle. This shows us that the region (shown in green) where $$C$$ makes $$\triangle ABC$$ acute is the arc between $$A'$$ and $$B'$$. The region where $$C$$ forms an obtuse triangle is shown in red. This holds true if $$B$$ is in quadrant I. Now, constrain $$B$$ to quadrant II again but label its reflection in the $$y$$-axis as $$D$$. Label the point opposite $$D$$ on the circle as $$D'$$ and draw the region where $$ADC$$ forms an acute triangle in dashed blue. Hint (a): What is the average size of the arcs $$A'B'$$ and $$A'D'$$ as a proportion of the circle?

Hint (b): What does this tell us about the average size of the region where the triangle is acute, if we randomly select a point on the upper semicircle?

Hint (c): Does this generalise to when $$B$$ is in the bottom semicircle? And to when $$A$$ is not $$(-1,0)$$?

Choose points $$A$$, $$B$$, $$C$$ independently and uniformly distributed on the circle.

Now we can see on geometric grounds that

The angle at $$C$$ is obtuse if and only if $$C$$ lies on the same side of a diameter through $$A$$ as $$B$$ does, and $$C$$ is closer to $$A$$ than $$B$$ is.

The two conditions here are independent, and each of them has probability $$\frac12$$. So the probability of $$C$$ being obtuse is $$\frac14$$.

With the same reasoning the probability that each of $$A$$ and $$B$$ is obtuse is $$\frac14$$ too.

Since a triangle has at most one obtuse angle, we can add the probabilities, so the probability that some angle in the triangle is obtuse is $$\frac34$$.

A right (or degenerate) triangle occurs with probability $$0$$, so the probability of an acute triangle is $$1-\frac34=\frac14$$.

• Nice answer - I think it's slightly more direct than mine since only one of your angles can be obtuse. – Jam Oct 29 '18 at 16:25

A triangle with its vertices lying on a circle is acute if and only if it contains the center of the circle. Then the answer is clear to be $$0.25$$. Refer to https://www.youtube.com/watch?v=OkmNXy7er84

• That is not clear to me (with that reasoning) at all. – Henning Makholm Oct 29 '18 at 16:20
• @Henning Makholm Please check the video. – JRen Oct 29 '18 at 18:44