315 words on Uni
Essentially you start with a circle and put n points on it. Then you join each point to all other points using straight lines. The question is how many areas this divides the circle’s interior into for any given n. And to give an argument for the answer you find, obviously.
That question can be rather hard to answer and we’ll add one more provision to it: You can assume that no three lines meet in a single point. E.g., for n=6, if the six points are arranged as the vertices of a regular hexagon inscribed in the circle, three of the hexagon’s ‘diagonals’ would meet right in the centre of the circle. That is a bit of a ‘singular’ situation and essentially removes a triangle – the one with parts of those three lines as its sides – from the count. Which messes things up. As we want a definite number as a result, we disallow such triple intersections and go for slightly less symmetric point arrangements.
It may be interesting to think about whether this additional provision actually makes sense. It’s obvious how we can move things just a bit in the case of six points. But will this still work when there are sixty, six hundred or six million points? Luckily this technique of moving things a little to avoid the undesired situations will work for arbitrarily large numbers of points thanks to Sard’s theorem. Which means that hiding in this seemingly harmless provision is a pointer to yet more maths.
As the exercise was just a fun and voluntary bonus one, people didn’t actually do it. Hence I rest my hopes on you to give the correct number and argument…