Mixing
Moser circle problem: maximum case for N?
$begingroup$
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
2571515
$endgroup$
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
2571515
$endgroup$
$begingroup$
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
$begingroup$
@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 |
$begingroup$
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
2571515
$endgroup$
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
2571515
asked Jan 11 at 3:57
aschultzaschultz
2571515
edited Jan 11 at 4:04
aschultz
asked Jan 11 at 3:57
aschultzaschultz
2571515
asked Jan 11 at 3:57
aschultzaschultz
2571515
asked Jan 11 at 3:57
aschultzaschultz
2571515
2571515
$begingroup$
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 |
1 Answer
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oldest
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$begingroup$
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 $j_n= (frac{2sqrt{p_n}}{p_n + 1}, frac{p_n-1}{p_n+1})$, 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 $(cos{nphi},sin{nphi})$ where $j_n = (cosphi,sinphi)$. But this requires a lot of algebra.
answered Jan 29 at 3:26
aschultzaschultz
2571515
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add a comment |
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$begingroup$
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 $j_n= (frac{2sqrt{p_n}}{p_n + 1}, frac{p_n-1}{p_n+1})$, 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 $(cos{nphi},sin{nphi})$ where $j_n = (cosphi,sinphi)$. But this requires a lot of algebra.
answered Jan 29 at 3:26
aschultzaschultz
2571515
$endgroup$
add a comment |
$begingroup$
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 $j_n= (frac{2sqrt{p_n}}{p_n + 1}, frac{p_n-1}{p_n+1})$, 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 $(cos{nphi},sin{nphi})$ where $j_n = (cosphi,sinphi)$. But this requires a lot of algebra.
answered Jan 29 at 3:26
aschultzaschultz
2571515
$endgroup$
add a comment |
$begingroup$
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 $j_n= (frac{2sqrt{p_n}}{p_n + 1}, frac{p_n-1}{p_n+1})$, 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 $(cos{nphi},sin{nphi})$ where $j_n = (cosphi,sinphi)$. But this requires a lot of algebra.
answered Jan 29 at 3:26
aschultzaschultz
2571515
$endgroup$
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 $j_n= (frac{2sqrt{p_n}}{p_n + 1}, frac{p_n-1}{p_n+1})$, 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 $(cos{nphi},sin{nphi})$ where $j_n = (cosphi,sinphi)$. But this requires a lot of algebra.
answered Jan 29 at 3:26
aschultzaschultz
2571515
answered Jan 29 at 3:26
aschultzaschultz
2571515
answered Jan 29 at 3:26
aschultzaschultz
2571515
answered Jan 29 at 3:26
aschultzaschultz
2571515
2571515
add a comment |
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