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?

  • $\begingroup$ 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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