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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?

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  • $\begingroup$ Wouldn't the chords of every set of $n$ points on a circle divide the interior into $\binom{n}4+\binom{n}2+1$ regions? Can you give a single example of $n$ points which product fewer regions? $\endgroup$ – Mike Earnest Jan 11 at 18:28
  • $\begingroup$ @Mike Earnest If you have 6 points a1-a6 that form, in order, a regular hexagon, then you would not have 31 regions, because at the very least, a1a4 and a2a5 and a3a6 would all intersect in one point. $\endgroup$ – aschultz Jan 12 at 5:08
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    $\begingroup$ I see now, that was a brain fart on my part. $\endgroup$ – Mike Earnest Jan 12 at 5:11
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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.

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