Moser circle problem: maximum case for N? Moser's circle problem sets the upper bound of regions the chords connecting n points can divide a circle into at ${n \choose 4} + {n \choose 2} + 1$. But how can we construct a set of points that gets there?
There must be infinitely many. For any set of n points, there are at most $n{n-1 \choose 4}$ points on the circle that might cause an intersection, e.g. you could pick any two intersecting chords and a point in the set, and the line between them would intersect the circle in one more place.
So this exists. But how to construct a set of points that provably works?
My guess is that if we have a list of the prime numbers $p_1=2, p_2=3, p_3=5$ etc. and a unit circle, we could have a set of points $x_n=(cos(\pi/p_n), sin(\pi/p_n))$. That seems like it would work, but I'm not sure how to prove it. And there might be something simpler, too. Any suggestions?
 A: The solution I found uses induction on N. Let's assume we have a circle with N points that maximize the number of regions. Then look at the arc between, say, $x_n$ and $x_n-1$, and pick a point $x_j$ where $0<j<n-1$. We will find a point on the arc that satisfies our condition.
There will be at most ${n \choose 4}$ rays from j through to an intersection of two points that go through the arc $x_n x_{n-1}$ but don't touch $x_n$ or $x_{n-1}$, since there are at most ${n \choose 4}$ intersections of chords. Let's say there are y such rays, and let's call their intersections $z_0$ to $z_{y-1}$, with $z_0$ being closest to $x_{n-1}$.
If y=0 it's easy. Any point on the arc works. If y=1 then it's also easy. We can just take the midpoint of the arc $x_n z_0$ or $x_n z_1$. In fact if y > 1 we can always take the midpoint of $x_{n-1} z_0$ or $x_n z_{y-1}$.
This is one construction. I suppose if we want the points spread out we could choose the midpoint of the arc between $z_{y-1/2}$ and $z_{y+1/2}$.
I also had an idea for having the points $j_n= (\frac{2\sqrt{p_n}}{p_n + 1}, \frac{p_n-1}{p_n+1})$, because each intersection of $j_a j_b$ and $j_c j_d$ will have $\sqrt{a}$, $\sqrt{b}$, $\sqrt{c}$, and $\sqrt{d}$ (or some product) terms in the x-coordinate + y-coordinate, and this sum is unique to any point pair, because if it were not, we would contradict this finding. Again, if we wanted to place things semi-evenly, we could let $k_n$ be $(\cos{n\phi},\sin{n\phi})$ where $j_n = (\cos\phi,\sin\phi)$. But this requires a lot of algebra.
