Wednesday, 14 November 2012

The Circle, Triangle and Random Line with an Answer

In A Circle, a Triangle and Some Random Lines, I posed the question “what is the probability, p, of a random chord in a circle being longer than the side of the largest equilateral triangle that can be drawn in that same circle?”  Although I did use somewhat different words in an attempt to not lead the question or give away the googling terms that would lead the reader to the answer in less than a minute.

I also asked the simpler question, “are there more shorter chords, more longer chords or more equally long chords?”

Here’s the largest equilateral triangle that can be drawn in a circle:

Or rather, this is one of an infinite number of largest equilateral triangles that can be drawn on a circle.  If we rotated the triangle around the common centre in one degree increments, there would be 360 equally large triangles and the 360 equally long chords (because at 180 degrees the chords start getting repeated).  But we are not restricted to one degree increments; we can use 0.0001 degree increments, leading to 360,000 triangles and 360,000 equally long chords.  Theoretically we can keep trying smaller and smaller increments until they are infinitesimally small leading to an infinite number of triangles and as many equally long chords.

So, you might think, you can’t get more than infinity so there can’t be more than an infinite number of shorter or longer chords, so … there must be either an even number of each (infinity, infinity and infinity) or more chords that are equally long.

Well, that doesn’t seem quite right.

Let’s just consider a single chord that is shorter than the sides of the triangle.
This chord has one end at the apex of the triangle, which for ease we can call 0 degrees on the circle.  The other end lies about 60 degrees anticlockwise from the apex of the triangle.  Now how many chords are there equal in length to this one?  We can do the same thing as we did with the triangle, rotating it around the centre of the circle and arrive at the conclusion that there are an infinite number of chords that are this long.  (Note that after rotating the chord 60 degrees clockwise, we arrive at a chord which has one end at the apex of the triangle and the other 60 degrees clockwise from that apex.)

However, we don’t have to start with the other end at 60 degrees from the apex.  All the chords with ends with less than 120 degrees apart will have sides shorter than the equilateral triangle.  Again, there are an infinite number of such chords because we are not limited to single degree divisions.

So, we have an infinite number of chord lengths which are shorter than the sides of the triangle, and an infinite number of each of them.  There’s only the single infinity of chords equal in length to the sides of the equilateral triangle … so at the moment we have more shorter chords than equal length chords, by a factor of infinity.

Now let’s consider chords that are longer than the sides of the triangle.

The chord shown is between the apex of the triangle and a point about 150 degrees anticlockwise from the apex.  Using the same logic as above, we have an infinite number of chords of this length.  And again, we don’t have to start at 150 degrees from the apex.  However, this time we are limited to a range between 120 and 180 degrees from the apex, a range of only 60 degrees, rather than 120 degrees.

Imagine rotating the chord illustrated above by 150 degrees clockwise.  The lower end will now be co-incident with the apex of the triangle and the lower end will now be 150 degrees clockwise from that apex.  While there are an infinite number of possible chords which are longer than the sides of the triangle, there are still only half as many as there are shorter chords.

So, the answer (apparently) is that there is a greater number of shorter chords than of longer chords by a factor of two and effectively zero chords that are equal in length to the sides of the triangle, meaning that if you take a random chord, then you have p=1/3 as the chance of getting a longer one.

Note that a variant of this was suggested by Daniel Antone in the comments.  Slightly edited:  “In order to be longer than the side of the triangle: if the line started intersecting the circle at any randomly chosen point it would need to intersect the circle again at a point greater than 120 degree around the circle, but less than 240 degrees.  That means 1/3 of the circle is open for that line to intersect with.”

However, the methods described above are not the only way to figure things out.  Another way is to think about a single radius, like this:

This is just one of an infinite number of possible radii, since we can rotate it around the centre of the circle like we did with the triangle.  The radius is bisected by the side of the equilateral triangle when it crosses at a right angle.  This means that half of all the chords which are perpendicular to the radius (of which there are an infinite number) will be longer than the sides of the equilateral triangle and half will be shorter.  A single chord will be the same length but it’s infinitely insignificant.

The probability, therefore, of picking a random chord which is longer than sides of the triangle is p=1/2.

Alternatively, you could just consider the mid-point on the chord.  Pick a random point in the circle and draw the chord on which that point is the mid-point.  This isn’t a particular good method for one very good reason.  Every single point that you select has a unique chord associated with it, except the locus of the circle.  The locus midpoint has an infinite number of different chords.  If you ignore this little embarrassment, this method gives you p=1/4 as the chance of picking a longer chord, or p=1/2, depending on how you handle your granularity problem.  All chords with a mid-point inside a new smaller circle with half the radius of the original circle will be longer than the side of the equilateral triangle.

The method that results in p=1/2 feels wrong, but is probably right for two reasons.

To illustrate the first of these two reasons, let’s revisit the first method and consider a slight variation.  These chords were created by taking a point on the circle and drawing lines that intersect again with the circle at another point.  This starting point, although there are admittedly an infinite number of variations of it, is pretty special.  We could generalise it by using point well outside the circle.  For example:

I’ve been fiddling with the mathematics to show that the angles shown by θ approach equality (they aren’t exactly equal until you get to infinity), but it’s a little bit too complicated to show here, so I’ll just have to be satisfied with the illustration (or, if you are a glutton for punishment, look here).  Here’s what it looks like when you take a reference point which is even further away:

 Any line originating from a distant point that lies in the middle two segments, on each side of the line which passes through the locus of the circle, will create a chord that is longer than the sides of the equilateral triangle.  In other words, half the possible chords will be longer, so you have p=1/2 as the chance of selecting a longer chord at random.  If you are in any doubt, look at this piece of graphical proof (you can draw it yourself if you don’t believe me – the points of origin are 120 degrees apart and about three and a half times the radius distant from the locus of the circle, each black tangential line is 13.33 degrees from the green pole and the red lines are equidistant from the pole and the tangential lines at 6.66 degrees from the pole).  It’s quite a bit wonkier than I would have liked, but such is the pleasure of working with inexpensive graphics programs.

Now what is possibly less intuitive is that you can select a point at random (any point anywhere) and then select another point at random from a reduced set (the set of points which combined with your first point defines a line which passes through the circle) and you will have exactly the same result – p=1/2.  When you consider this, then you might realise that the idea of selecting a point on the circle and then selecting another point on the circle is in fact a special subset of this much larger set of possible options.

The second reason is that the staff solution to what is known as the Bertrand paradox is indeed p=1/2.


So, what does this tell you?

The bottom line is that when you get infinity involved in a problem, you can start getting weird answers.  And the wrong answer to a problem can seem incredibly convincing if you start from the wrong position and are unwilling to consider another perspective – or to think a little further than the first answer that seems intuitive.

When an Apologist starts involving infinity in his attempts to solve a theological problem, you’d better stand clear.  Logically speaking, things are likely to get messy.


This issue is expanded on further in three articles: A Slightly More Mathematical Look, Response to Melchior and A Farewell to the Betrand Paradox.


  1. Hey, I think I have a simpler way to solve this. I came up with p=1/2. I apologize for having to link you to imgur, but that was the best way I could think to make the explanation clear:

    Do please let me know if I've overlooked something.

    1. This is functionally equivalent to the half a radius method. You're using the half a diameter method, which (unsurprisingly) ends up with the same answer of p=0.5 - don't take it as criticism, I think it's a more valid method than any method that ends up with any other value of p.


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