Triangles Containing the Origin

You may have found this site by searching for information regarding Project Euler Problem 184. This is not a solution to that problem. In fact, it really bothers me when I find people who post their solutions to these problems online, especially the higher level ones. In this blog post, I talk about a simple problem from probability that was motivated from this Project Euler problem, but the solution to this problem is not likely to help you solve that one.

Pick three points uniformly at random along the circumference of a unit circle centered at the origin. What is the probability that the triangle connecting these three points contains the origin?

Like I said, this problem is not as difficult as the problems that I usually write about, but I decided to write a blog post about it for two main reasons:

I wanted to create the animation/simulation for this problem
I eventually want to extend this problem to explore all convex polygons instead of just triangles (which is a more difficult problem) and I will be able to easily reference this problem if needed
As promised, here is the animation that describes this process. I created this animation with Racket: a functional programming language that makes it extremely easy to work with images. In a future blog post, I will probably talk more about Racket and what I like to use it for, but for now here is the animation.

Solution

If you would like to attempt this problem for yourself, stop reading here and refer back to this section once you have a solution. Otherwise, read on!
The first thing to notice is that the location of the first point is not important. The only thing we must consider is the position of the second point relative to the first, and the position of the third point relative to the first 2. Using these three pieces of information, it is possible to determine whether the triangle contains the origin. Assume we've already picked the first two points on the circumference of the circle. For example:

Given these two starting points, where can the third point be in order to form a triangle that contains the origin? What is the probability of a point chosen at random to land in that location? Well to answer this question consider this figure:

Any point on the circumference of the semi-circle given by the green region is a valid point that will result in the triangle containing the origin. The probability of a point lying in that region is just $\frac{L}{2 \pi}$ where $L$ is just the arc length of the green semi-circle. Now, we must go back a step and figure out how to deal with the second point. Recall that we are only interested in where the second point is relative to the first, so it can end up being anywhere between a distance of $0$ from the original point to a distance of ππ from the original point. Let $x$ denote the distance, then using the result we just came up with, the probability of forming a triangle with point 2 $x$ units from point 1 is simply $\frac{x}{2 \pi}$. In order to average this probability over all the possible values for $x$ we must use an integral.
$$ \begin{align*}
P &= \frac{1}{2 \pi} \int_{x=0}^{\pi} \frac{x}{2 \pi} \\
P &= \frac{1}{2 \pi} \frac{x^2}{2 \pi} \Big|_0^{\pi} \\
P &= \frac{1}{4}
\end{align*} $$

Alternatively, you could use a geometrical argument by noticing that the probability can be as small as $0$ and as large as $\frac{1}{2}$ (and this is uniform), so the average probability is going to be $\frac{0+\frac{1}{2}}{2} = \frac{1}{4}$.

Conclusion

An interesting thing to notice is that the the answer (and reasoning) would be the same even if we were allowed to choose points anywhere on the interior of the unit circle (instead of on the circumference). It is left to the reader to figure out why this works.

Stay posted for a future blog posts! I'm hoping to generalize this problem to all convex polygons that can be formed by connecting $n \geq 3$ points on the circumference of the circle.