# The probability that $3$ random points on the circumference form a right-angled triangle?

In my probability theory course, I dealt with a similar problem which asks for the probability that $3$ random points on the circumference of a circle lie on the same semi-circle. But it makes me think of the following:

Given a circle, we randomly mark $3$ distinct points on the circumference and name them $X_1, X_2$ and $X_3$ respectively. What is the probability that $\Delta X_1X_2X_3$ is a right-angled triangle?

Here is my work. Please tell me if there is any problem in my solution. Feel free to leave me your solution so that I can learn from you.

Solution:

Let $E$ be the event that "$\Delta ABC$ is a right-angled triangle" and let $E_i$, $i=1,2,3$, be the event that "Starting from $X_i$, traversing the circle counterclockwise, one can visit all the points by turning exactly $180$ degrees".

I then claim that $$P(E) = P(E_1)+P(E_2)+P(E_3).$$ We then consider $P(E_i)$ by conditioning on the position of $X_i$, $$P(E_i) = \int_0^{2\pi} P(E_i | X_i = x) \frac{1}{2\pi} dx.$$ Furthermore, we consider $P(E_i | X_i = x)$. For a fixed $X_i$, if we can visit all three points in exactly $180$ degrees, one of the remaining points, say $X_j$, should be exactly 'opposite' to $X_i$, which forms a diameter with $X_i$. Thus there is only one possible position of $X_j$. For the last point, we note that $X_iX_j$ divides the whole circle into two equal semi-circles. For $E_i$ to occur, we must have the last point lying in one of the semi-cirlces. Thus, I claim that $$P(E_i | X_i = x)=\frac{1}{2\pi} \frac{1}{2}$$ which quickly implies $$P(E)=\frac3{4\pi}.$$

It seems strange to me. Thanks in advance for leaving your ideas!

• I don't know probability but I think you could use Thales theorem and the problem collapses to finding the probability that for 3 random points, 2 are antipodal. Commented Oct 24, 2014 at 11:18
• To calculate this correctly is probably to go beyond what is expected of you. But in the real world, there is some margin of accuracy beyond which you cannot measure. If you specify what that is, in proportion to the radius of your circle, you can calculate the probability that your points fall within that range of some right angled triangle. Every right-angled triangle has the diameter as its hypoteneuse so you want the probability that any two points are within the specified range of being exactly opposite each other. You will need to make a call as to whether having all three within the Commented Apr 26, 2017 at 17:47
• ...specified range counts too. As you could call that a right angled triangle of zero width, or alternatively not a triangle. If your range of accuracy is $\frac{a}{r}$ where $r$ is the radius, meaning you can be $\frac{a}{2}$ either side of some point and that's close enough that you're considered on it, and you don't worry about whether two corners are within the same point, and the points are small relative to the circle then the probability is $\frac{a}{r}\cdot\frac{1}{\pi}$ Commented Apr 26, 2017 at 17:59
• In justifying this approach you may find it important to point out that although a point on a circle can be chosen randomly, it is impossible to select uniformly randomly. Commented Apr 26, 2017 at 18:01

The probability is $0$, assuming that each point is chosen with uniform distribution on the circle.

For a right-angled triangle you need two of the points to be the two endpoints of a diameter. The probability for that is obviously $0$, since there isn't any range of possibilities, and the distribution is continuous.

For anyone who is not satisfied so far, below is a more rigorous argument.

Consider the following question, which is pretty much the same, only simpler to answer in this format: When choosing two points on a circle, where the two choices are independent, and each point has a uniform distribution, what is the probability that the chosen points are the two endpoints of a diameter?

Having chosen one of the points, we need the other one to be exactly the first one's antipodal. As we know, uniform distribution yields that the probability that a point is between $\alpha$ and $\beta$ is just $$P(\alpha\leq p\leq\beta)=\frac{1}{2\pi}(\beta-\alpha).$$As $\alpha$ and $\beta$ get closer to one another, the probability gets smaller, and eventually, if $\alpha=\beta$, the probability vanishes.

• I don't get the point 'there isn't any range of possibilities'. What does that mean? Commented Oct 24, 2014 at 11:31
• @LeeKM I added something to the answer. Have a look. Commented Oct 24, 2014 at 16:30
• That's much better! Thank you!:) Commented Oct 25, 2014 at 9:46
• Great explanation! If the reader is familiar with probability density functions and their properties, that could also make for a brief explanation. Commented Mar 15, 2021 at 9:35