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**Find the average distance between 2 points on a square**

This is another excellent mathematical puzzle from the MindYourDecisions youtube channel. I like to try these without looking at the answer – and then to see how far I get. This one is pretty difficult (and the actual solution exceptionally difficult!) The problem is to take a square and randomly choose 2 points somewhere inside. If you calculate the distance between the 2 points, then do this trial approaching an infinite number of times what will the average distance be? Here is what I did.

**Simplify the situation: 1×1 square**

This is one of the most important strategies in tackling difficult maths problems. You simplify in order to gain an understanding of the underlying problem and possibly either develop strategies or notice patterns. So, I started with a unit square and only considered the vertices. We can then list all the possible lengths:

We can then find the average length by simply doing:

**2×2 square**

We can then follow the same method for a 2×2 square. This gives:

Which gives an average of:

**Back to a 1×1 square**

Now, we can imagine that we have a 1 x 1 square with dots at every 0.5. This is simply a scaled version of the 2×2 square, so we can divide our answer by 2 to give:

**3×3 square**

Following the same method we have:

This gives an average of:

and if we imagine a 1×1 square with dots at every 1/3. This is simply a scaled version of the 3×3 square, so we can divide our answer by 3 to give:

We can then investigate what happens as we consider more and more dots inside our 1×1 square. When we have considered an infinite number then we will have our average distance – so we are looking the limit to infinity. This suggests using a graph. First I calculated a few more terms in the sequence:

Then I plotted this on Desmos. The points looked like they fit either an exponential or a reciprocal function – both which have asymptotes, so I tried both. The reciprocal function fit with an R squared value of 1. This is a perfect fit so I will use that.

This was plotted using the regression line:

And we can find the equation of the horizontal asymptote by seeing what happens when x approaches infinity. This will give a/c. Using the values provided by Desmos’ regression I got 0.515004887. Because I have been using approximate answers throughout I’ll take this as 0.52 (2sf). **Therefore I predict that the average distance between 2 points in a 1×1 square will be approximately 0.52**. And more generally, the average distance in an n x n square will be 0.52(n). This is somewhat surprising as a result – it’s not obvious why it would be a little over half the distance from 0 to 1.

**Brute forcing using Python**

We can also write a quick code to approximate this answer using Python (This is a Monte Carlo method). I generate 4 random numbers to represent the 2 x-coordinates and 2-y coordinates of 2 random points. I then work out the distance between them and repeat this 10 million times, then calculate the average distance. This gives:

**Checking with the actual answer**

Now for the moment of truth – and we watch the video to find out how accurate this is. The correct answer is indeed 0.52 (2sf) – which is great – our method worked! The exact answer is given by:

Our graphical answer is not quite accurate enough to 3 sf – probably because we relied on rounded values to plot our regression line. Our Python method with 10 million trials was accurate to 4 sf. Just to keep my computer on its toes I also calculated this with 100 million trials. This gave 0.5214126210834646 (now accurate to 5 sf).

We can also find the percentage error when using our graphical method. This is only:

Overall this is a decent result! If you are feeling *extremely* brave you might want to look at the video to see how to do this using calculus.

**Extension: The average distance between 2 points in a unit circle**

I modified the Python code slightly to now calculate the average distance between 2 points in a unit circle. This code is:

which returns an answer of 0.9054134561871364. I then looked up what the exact answer is. For the unit circle it is 128/(45 pi). This is approximately 0.9054147874. We can see that our computer method was accurate to 5 sf here. Again, the actual mathematical proof is extremely difficult.

**Reflection**

This is a nice example of important skills and techniques useful in mathematics – simplification of a problem, noticing patterns, graphical methods, computational power and perseverance!

**Classical Geometry Puzzle: Finding the Radius **

This is another look at a puzzle from Mind Your Decisions. The problem is to find the radius of the following circle:

We are told that line AD and BC are perpendicular and the lengths of some parts of chords, but not much more! First I’ll look at my attempt to solve this. It’s not quite as “nice” as the solution in the video as it requires the use of a calculator, but it still does the job.

**Method 1, extra construction lines:**

These are the extra construction lines required to solve this problem. Here is the step by step thought process:

- Find the hypotenuse of triangle AGC.
- Use the circle theorem angles in the same segment are equal to show that angle CBD = angle CAG.
- Therefore triangle AGC and GBD are similar, so length BG = 4. We can now use Pythagoras to find length BD.
- We can find length CD by Pythagoras.
- Now we have 3 sides of a triangle, CDB. This allows use to find angle BDC using the cosine rule.
- Now we the circle theorem angles in the same segment are equal to show that angle BDC = angle BEC.
- Now we use the circle theorem angles in a semi circle are 90 degrees to show ECB = 90.
- Now we have a right angled triangle BCE where we know both an angle and a side, so can use trigonometry to find the length of BE.
- Therefore the radius is approximately 4.03.

**Method 2, creating a coordinate system**

This is a really beautiful solution – which does not require a calculator (and which is discussed in the video above). We start by creating a coordinate system based around point G at (0,0). Because we have perpendicular lines we can therefore create coordinates for A, B and C. We also mark the centre of the circle as (p,q).

First we start with the equation of a circle centre (p.q):

Next we create 3 equations by substituting in our coordinates:

Next we can do equation (3) – equation (1) to give:

Next we can substitute this value for p into equations (1) and (3) and equate to get:

Lastly we can substitute both values for p and q into equation (1) to find r:

We get the same answer as before – though this definitely feels like a “cleaner” solution. There are other ways to solve this – but some of these require the use of equations you may not already know (such as the law of sines in a circumcircle, or the equation for perpendicular chords and radius). Perhaps explore any other methods for solving this – what are the relative merits of each approach?