The Oxford Dictionary defines a paradox as “a seemingly absurd or contradictory statement or proposition which when investigated may prove to be well founded or true”. The one we are going to investigate today – named after its author Joseph Bertrand – in fact cannot be proven to be true nor false. But, before we get to this seemingly contradictory conclusion, let’s consider the problem setup:
Consider an equilateral triangle perfectly enclosed by (inscribed in) a circle. Suppose a straight line joining two points on the boundary of the circle (a chord) is chosen at random. What is the probability that the chord is longer than a side of the triangle?
To understand the question, lets first draw an equilateral triangle that fits perfectly inside a circle:
Now draw a couple of chords (line segments joining two points on the circle) at random:
The question asks what is the probability that one of these randomly drawn chords is longer than a side of this triangle? If you’re thinking the question seems pretty vague, then you’re right. There are indeed a few ways to think about this – Bertrand himself came up with 3 different solutions which is what we are going to explore.
In the first of the solutions we draw a chord using a corner of the triangle and then try to find when this chord will be longer than the side of the triangle. The argument begins with choosing randomly one point on the circle, A, and drawing an equilateral triangle starting from this point. Take a note that it does not matter which point we choose here – the points on the circle are “identical” to one another, so each one has equal probability of being chosen, meaning it won’t influence the probability later on. Since the triangle has to be equilateral and fit perfectly inside the circle, there is in fact only one triangle that we can draw from our chosen starting point as shown below:
Now, we choose a second random point, D, on the circle and draw a line joining it to A:
For the example we have shown the point D is between the points labelled B and C on the boundary of the circle – we can say that it is along the arc BC. Now, if we had instead selected D to be along the arc AB, how would this change our answer?
With a little bit of thought – and no doubt a few more drawings – you will hopefully eventually spot that for all points D along the arc AB or the arc CA the chord will be shorter than the side-length of the triangle, and for all points D along the arc BC it will be longer.
Due to the fact that the triangle is equilateral, each of the arcs are equal in length. Since there are 3 of them, the probability that a randomly chosen chord is longer than the side-length of the triangle is 1/3.
In this solution we will choose a chord in a different way to before. And I should warn you, it may seem quite a bizarre way to draw a chord, but it does give rise to some interesting maths.
Begin by drawing a straight line from the centre of the circle to a random point on the circumference (again all points are identical and equally likely to be chosen). This line is in fact a randomly chosen radius of the circle. Once we have our specified radius OA, we want to again draw an equilateral triangle, but this time we say that one of the sides of the triangle must cross our radius at exactly 90 degrees – in other words it must be perpendicular to the radius. Fixing this condition, in fact leaves us with only one possible equilateral triangle that can be drawn, as shown below:
Now we randomly choose a point B along the radius OA:
Next we draw a chord that is perpendicular to the radius and goes through point B:
We can see that our chord is longer than the side of the triangle. But what if we had chosen our point B’ to be on the segment CA – i.e. outside of the triangle?
You can hopefully now see that if the point lies between the side of the triangle and the centre of the circle (in other words, it lies on OC), then the chord will be longer than the side of the triangle and if it lies between the side of the triangle and the edge of the circle (along CA) then it will be shorter than the triangle side length. With a little more work (left as an exercise for the keen reader), we can in fact show that OC and CA have equal lengths, so when we choose the point B we are equally likely to place it on OC and on CA, and so the probability that the point lies on OC is 1/2. Hence, the probability that the chord is longer than the side of the triangle is 1/2. This is a different result than before!
We have two different solutions so far, but we promised you 3 – courtesy of Bertrand – so here is the third and final solution using yet another method to choose the chord. This time, instead of lines and points we will use areas. First, we choose a point at random inside the circle, let’s call it A, and use it as a midpoint (a point exactly half-way along a line) of a chord. Again, there is only one chord that will have this point as its midpoint.
Next, we draw our equilateral triangle fitting perfectly inside the circle. At this point it doesn’t matter where we draw the triangle (as mentioned in solution 1 they are all identical and therefore equally likely by just rotating the circle).
Now, let us draw a circle perfectly enclosed by the triangle, which we can show (again an exercise for the keen reader) will have a radius that is 1/2 of the radius of the larger circle.
By testing out a few different points, you will hopefully begin to realise that only when our point A is in the smaller circle will the chord be longer than the side of the triangle. See the diagram below for an example of a second point A’ which is outside of the circle and gives rise to a chord shorter than the side-length of the triangle.
Now, I mentioned this could be thought of in terms of areas, so let’s do some calculating. The area of the smaller circle is 1/4 of the area of the bigger circle – we have half the radius so its area is 𝜋(r/2)2 = 𝜋𝑟2/4 – and so the probability that a random point will end up in the smaller circle is 1/4. Therefore the probability that the random chord is longer than the side of the triangle is also 1/4, which is different to both of the previous results.
So there we have it: 3 different methods of solution that give rise to 3 completely different answers… so the burning question is, which one is correct? You are probably expecting me to point out mistakes that Bertrand has not spotted when he was formulating these arguments, so that only one result will be left. The problem is, there are no mistakes in these arguments! All of the answers are correct. 1
But hold on a minute, how is it possible that there are three different – and correct – probabilities? The real world does not work this way… right? Absolutely, but our models of the real world can (and often do). The problem is that the question did not specify how we are supposed to choose the chord. In each of the solutions we chose the chord in a different way, and so the answers are different.
This might feel like an unsatisfying result: one right answer does not exist because the question was badly formulated. But I look at it in a different way. This paradox is a reminder of an incredibly important concept in mathematics (also physics, engineering, chemistry…) – we create a model of a real world and we “model the model”, not the real world.
When creating such a model we are almost always making some assumptions – sometimes because we want to simplify, sometimes because we do not know everything about the system we are modelling – and this is perfectly okay and more often than not gives us good information and a lot of insight into the problem. However, we must remember the assumptions and take them into consideration when interpreting the result back in the real world. Otherwise, we can find ourselves in a very weird situation, such as having 3 different answers to a single question…
This is of course just one of many mathematical and logical paradoxes. From the Oxford Dictionary definition at the beginning of this article – a paradox is something that leads to a conclusion that seems contradictory with itself or completely contradicts our expectations. Sometimes it really is contradictory and you can’t solve it (such as the Unexpected hanging paradox), and sometimes it is perfectly valid (such as the the Monty Hall Problem). Paradoxes are not only an interesting weirdness, they are a great way to test your mathematical understanding, as they force you to think about small quirks and details that could otherwise be lost.
1A careful reader may argue that some of these solutions are counting the diameter multiple times, but it can be shown that excluding them does not change the probabilities.