Moser circle problem: maximum case for N?
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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?
combinatorics geometry
asked Jan 11 at 3:57
aschultzaschultz
1921415
$endgroup$
add a comment |
$begingroup$
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?
combinatorics geometry
asked Jan 11 at 3:57
aschultzaschultz
1921415
$endgroup$
$begingroup$
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
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@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
1
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11
add a comment |
$begingroup$
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?
combinatorics geometry
asked Jan 11 at 3:57
aschultzaschultz
1921415
$endgroup$
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?
combinatorics geometry
combinatorics geometry
asked Jan 11 at 3:57
aschultzaschultz
1921415
asked Jan 11 at 3:57
aschultzaschultz
1921415
edited Jan 11 at 4:04
aschultz
asked Jan 11 at 3:57
aschultzaschultz
1921415
asked Jan 11 at 3:57
aschultzaschultz
1921415
asked Jan 11 at 3:57
aschultzaschultz
1921415
1921415
$begingroup$
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
1
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11
add a comment |
$begingroup$
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
1
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11
$begingroup$
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
$begingroup$
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
$begingroup$
@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
1
1
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11
add a comment |
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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
1
$begingroup$
I see now, that was a brain fart on my part.
$endgroup$
– Mike Earnest
Jan 12 at 5:11