# 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
• At your last paragraph, how to show that a right triangle occurs with probability $0?$ – Idonknow Mar 8 at 7:58

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. – hmakholm left over Monica Oct 29 '18 at 16:20
• @Henning Makholm Please check the video. – JRen Oct 29 '18 at 18:44

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.