Probability that the triangle is acute

Question

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?

Answer 1

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 $1-\frac34=\frac14$.

Answer 2

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 $(-1,0)$?

Answer 3

If you're interested in an answer that doesn't directly make use of the facts used in the admittedly shorter approaches of JRen and Henning Makholm, consider the following answer.

Choose points $A, B$ and $C$ on the unit circle independently and uniformly distributed and let their coordinates be $\,(\cos(\theta_1), \sin(\theta_1)), (\cos(\theta_2), \sin(\theta_2)) $ and $(\cos(\theta_3), \sin(\theta_3))$ respectively where $0< \theta_1 < 2 \pi, \,\theta_1 < \theta_2 < 2 \pi$ and $\theta_2<\theta_3 < 2 \pi$

Our goal is to express angles $A, B$ and $C$ in terms of the $\theta_i.$ To do so, as in the picture above, draw lines from $A$ to $O$ and from $B$ to $O$ then $\angle AOB = \theta_2 - \theta_1.$ From the theorem in plane geometry that states that the angle subtended by an arc at the centre of a circle is twice the angle subtended by at any other point on the cirlce, we get that $\gamma = \angle ACB = \dfrac{\theta_2 - \theta_1}{2}.$

Repeating this process for vertices $A$ we can similarly see that $\alpha = \angle BAC = \dfrac{\theta_3 - \theta_2}{2}$ and then we can calculate $\beta = \angle CBA = \pi -\dfrac{\theta_3 - \theta_1}{2}.$

Thus if the triangle is to be acute-angled we need three conditions to be satisfied: $\alpha < \frac{\pi}{2}, \beta < \frac{\pi}{2}$ and $\gamma < \frac{\pi}{2}.$

Rewriting these conditions in terms of the $\theta_i$ we get:

$\pi + \theta_1 < \theta_3 < \pi + \theta_2$,

$\theta_1 < \theta_2 < \pi + \theta_1$ and

$0 < \theta_1 < 2 \pi.$

Thus we need to find the volume of the $3-$dimensional region described by the above relations which we do so using the following triple integral:

$\begin{align}\displaystyle \int_{0}^{2 \pi}\int_{\theta_1}^{\pi + \theta_1}\int_{\pi + \theta_1}^{\pi+\theta_2}\mathrm{d}\theta_3 \, \mathrm{d}\theta_2\, \mathrm{d}\theta_1 & = \displaystyle \int_{0}^{2 \pi}\int_{\theta_1}^{\pi + \theta_1}(\theta_2 - \theta_1) \mathrm{d}\theta_2\, \mathrm{d}\theta_1 \\&= \displaystyle \int_{0}^{2 \pi} \left(\dfrac{\theta_{2}^{2}}{2} - \theta_{1}\theta_{2}\right)_{\theta_2 = \theta_1}^{\theta_2 = \pi + \theta_1}\mathrm{d}\theta_1 \\&= \pi^{3} \end{align}.$

To find the volume of the permissible region from which points are selected we recall that from the way we drew the diagram, we have $0< \theta_1 < 2 \pi, \,\theta_1 < \theta_2 < 2 \pi$ and $\theta_2<\theta_3 < 2 \pi$ so we get the following triple integral

$\begin{align}\displaystyle \int_{0}^{2 \pi}\int_{\theta_1}^{2\pi}\int_{ \theta_2}^{2\pi}\mathrm{d}\theta_3 \, \mathrm{d}\theta_2\, \mathrm{d}\theta_1 & = \int_{0}^{2 \pi}\int_{\theta_1}^{2\pi}(2\pi - \theta_2) \, \mathrm{d}\theta_2\, \mathrm{d}\theta_1\\& = \displaystyle \int_{0}^{2 \pi} 2\pi\left(2 \pi - \theta_1\right) - \left(\dfrac{4\pi^{2}- \theta_{1}^{2}}{2}\right)\mathrm{d}\theta_1 \\&= 4\pi^{3} \end{align}$

Thus the probability of obtaining an acute-angled triangle is $$\dfrac{\displaystyle \int_{0}^{2 \pi}\int_{\theta_1}^{\pi + \theta_1}\int_{\pi + \theta_1}^{\pi+\theta_2}\mathrm{d}\theta_3 \, \mathrm{d}\theta_2\, \mathrm{d}\theta_1}{\displaystyle \int_{0}^{2 \pi}\int_{\theta_1}^{2\pi}\int_{ \theta_2}^{2\pi}\mathrm{d}\theta_3 \, \mathrm{d}\theta_2\, \mathrm{d}\theta_1} = \dfrac{\pi^{3}}{4\pi^{3}}= \dfrac{1}{4}.$$

Note: in the diagram shown, the circle has radius $3$. This does not matter to the solution.

Answer 4

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