I recently came upon this puzzle (inspired by this article), which asks how many triangles do you get when you mark N points (in this case 6) on a circle and draw lines joining every point to every other point?

This 6 sided-shape is symmetrical, so good way to count the triangles is to count how many there are in one orientation and multiply by the number of points. And that’s what I did.

This image shows 5 ‘rays’ to the top point in the shape and the horizontal lines which cross them.

You may notice that, as the counting stretches from right to left, it doesn’t go all the way to the left hand edge like this:

That’s because I don’t want to count the same triangle twice. But there’s another way to count triangles which span the whole row, which I’ll get to shortly.

Counting this way, it’s clear that there are 3+2+1 = 6 triangles in the top row (green circles), and 1 triangle in the next row (the red one).

We can carry on like this with the lines which cross the rays diagonally too, and similar triangular patterns emerge. I’ve only shown half the diagonals here, the other half slope the other way.

Counting this way we get this:

T(3)+T(1) = 7

2 x T(2) = 6

2 x T(1) = 2

For a total of 6 x 15 = 90, but that doesn’t include triangles where all 3 vertices are on the circle – the easy way to do that is to calculate 6 choose 3, but we’re on a triangular roll, so let’s carry on.

So, that’s T(4)+T(3)+T(2)+T(1) = 10+6+3+1 = 20, and adding in the 90 we counted before, we get a total of 110 triangles.

So, is this enough for a general solution? No. Because when we go over 6 points, we start to get triangles which have no vertices touching the circle.

Let’s do that now. This is a little tricky to explain, so I may have to edit-in a proper description later.

Essentially the way I did this was to, again, look at triangles in one orientation, but this time relative to a line rather than a point.

Any pair of lines which cross inside the circle and also pass through this ‘baseline’ will create a triangle, where the baseline forms one of the sides. As you can see here, the lines which cross the baseline create T(2)=3 triangles.

You might notice that these triangles will also be counted by neightbouring baselines, which is true, but every triangle is counted exactly 3 times, so all we have to do is divide the final result by 3.

So, when N=7, there are 7 x T(2)/3 = 7 triangles which don’t touch the circle.

While we’re here, lets do the rest of the arithmetic for N=7.

All points on the circle = T(5)+T(4)+T(3)+T(2)+T(1) = 15+10+6+3+1 = 35

Some points on the circle = 7 x ( T(4)+T(2) + 2 x(T(3)+T(1)+T(2)+T(1))) = 245

So, the total, for N=7, is 245 + 35 + 7 = 287.