Finding the average distance in an equilateral triangle

In the previous post I looked at the average distance between 2 points in a rectangle.  In this post I will investigate the average distance between 2 randomly chosen points in an equilateral triangle.

Drawing a sketch.

The first step is to start with an equilateral triangle with sides 1.  This is shown above.  I sketched this using Geogebra – and used some basic Pythagoras to work out the coordinates of point C.

I can then draw a square of sides 1 around this triangle as shown above.  I’m then going to run a Python program to randomly select points and then work out the distance between them – but I need to make sure that the 2 points chosen are both inside this triangle.  For this I need to work out the equation of the line AC and CB.

Using basic coordinate geometry we can see that the line AC has equation y = √3x.  We want the inequality y < √3x so that we are on the correct side of this line.

The line BC has equation y = -√3x + √3.  Therefore the triangle must also satisfy the inequality y < -√3x + √3.

I can then run the following code on Python, with finds the average distance between points (a,c) and (b,d) both within the unit square but also subject to the 2 inequality constraints above.

When this is run it performs 999,999 trials and then finds the average distance.  This returns the following value:

So we can see that the average distance is just over a third of a unit.

Finding the average distance of an equilateral triangle of length n.

We can then draw the sketch above to find the equation of lines AC and CB for an equilateral triangle with lengths n.  This leads to the following inequalities:

y < √3x

y < -√3x + √3n

So we can then modify the code as follows:

This then returns the average distances for equilateral triangles of sizes 1 to 10.

And when we plot this on Desmos we can see that there is a linear relationship:

The regression line has gradient 0.36 (2sf) so we can hypothesise that for an equilateral triangle of size n, the average distance between 2 points is approximately 0.36n.

Checking the maths

I then checked the actual equation for the average distance between 2 points in an equilateral triangle of sides n:

This gives us:

So we can see that we were accurate to 2 significant figures.  So this is a nice mixture of geometry, graphing and computational power to provide a result which would be otherwise extremely difficult to calculate.