Polygons inscribed in a circle and stars
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If on a circumference they are marked $n$ equally spaced points, those points
can be joined by line segments contiguous (without lifting the pencil). If you join the consecutive points, you get a polygon regular of n sides (that's not funny). But if you join non-contiguous points (skipping from one, or two or three, etc.), are obtained polygons crashed sometimes and others Sometimes they are not crashed. Which are the Where are star-studded polygons? The 5-pointed star (so famous) is a example of them.
I came to the conclusion that if $n$ is odd we can build a star without lifting the pencil by jumping from a vertex, what more interesting things can be said? What happens if I jump from two vertices? I will form a star with what characteristics about $n$?
algebra-precalculus geometry elementary-number-theory
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up vote
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If on a circumference they are marked $n$ equally spaced points, those points
can be joined by line segments contiguous (without lifting the pencil). If you join the consecutive points, you get a polygon regular of n sides (that's not funny). But if you join non-contiguous points (skipping from one, or two or three, etc.), are obtained polygons crashed sometimes and others Sometimes they are not crashed. Which are the Where are star-studded polygons? The 5-pointed star (so famous) is a example of them.
I came to the conclusion that if $n$ is odd we can build a star without lifting the pencil by jumping from a vertex, what more interesting things can be said? What happens if I jump from two vertices? I will form a star with what characteristics about $n$?
algebra-precalculus geometry elementary-number-theory
1
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
1
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If on a circumference they are marked $n$ equally spaced points, those points
can be joined by line segments contiguous (without lifting the pencil). If you join the consecutive points, you get a polygon regular of n sides (that's not funny). But if you join non-contiguous points (skipping from one, or two or three, etc.), are obtained polygons crashed sometimes and others Sometimes they are not crashed. Which are the Where are star-studded polygons? The 5-pointed star (so famous) is a example of them.
I came to the conclusion that if $n$ is odd we can build a star without lifting the pencil by jumping from a vertex, what more interesting things can be said? What happens if I jump from two vertices? I will form a star with what characteristics about $n$?
algebra-precalculus geometry elementary-number-theory
If on a circumference they are marked $n$ equally spaced points, those points
can be joined by line segments contiguous (without lifting the pencil). If you join the consecutive points, you get a polygon regular of n sides (that's not funny). But if you join non-contiguous points (skipping from one, or two or three, etc.), are obtained polygons crashed sometimes and others Sometimes they are not crashed. Which are the Where are star-studded polygons? The 5-pointed star (so famous) is a example of them.
I came to the conclusion that if $n$ is odd we can build a star without lifting the pencil by jumping from a vertex, what more interesting things can be said? What happens if I jump from two vertices? I will form a star with what characteristics about $n$?
algebra-precalculus geometry elementary-number-theory
edited Aug 7 at 20:25
Ethan Bolker
35.8k54299
35.8k54299
asked Aug 7 at 20:23
Nash
43129
43129
1
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
1
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31
add a comment |Â
1
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
1
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31
1
1
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
1
1
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31
add a comment |Â
2 Answers
2
active
oldest
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up vote
2
down vote
In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $gcd(n,k)$, which provides the count of separate circuits.
--- rk
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
add a comment |Â
up vote
1
down vote
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.
After you have done enough examples you should guess something about when the number of points on the circle and the number you count to the next vertex have a common factor.
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $gcd(n,k)$, which provides the count of separate circuits.
--- rk
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
add a comment |Â
up vote
2
down vote
In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $gcd(n,k)$, which provides the count of separate circuits.
--- rk
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
add a comment |Â
up vote
2
down vote
up vote
2
down vote
In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $gcd(n,k)$, which provides the count of separate circuits.
--- rk
In fact it is not the "number you skip", as @EthanBolker said, it is more the $k$-th vertex you'd visit next in a total sequance of $n$ vertices, which has or has not a common factor. That is, whenever $gcd(n,k)=1$, you could trace a complete polygon, which finally visits all of the $n$ vertices.
It is this very number $gcd(n,k)$, which provides the count of separate circuits.
--- rk
answered Aug 7 at 20:40
Dr. Richard Klitzing
7566
7566
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
add a comment |Â
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
Right of course. I edited my answer.
â Ethan Bolker
Aug 7 at 21:00
add a comment |Â
up vote
1
down vote
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.
After you have done enough examples you should guess something about when the number of points on the circle and the number you count to the next vertex have a common factor.
add a comment |Â
up vote
1
down vote
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.
After you have done enough examples you should guess something about when the number of points on the circle and the number you count to the next vertex have a common factor.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.
After you have done enough examples you should guess something about when the number of points on the circle and the number you count to the next vertex have a common factor.
Hint. $n$ being odd is not enough. Do some experiments with $15$ points.
After you have done enough examples you should guess something about when the number of points on the circle and the number you count to the next vertex have a common factor.
edited Aug 7 at 21:00
answered Aug 7 at 20:25
Ethan Bolker
35.8k54299
35.8k54299
add a comment |Â
add a comment |Â
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1
It has to do with the greatest common divisor of the number of points in the polygon and size of the jumps when you connect the vertexes.
â Doug M
Aug 7 at 20:25
@DougM How is the greatest common divisor and the points that are deleted related?
â Nash
Aug 7 at 20:28
1
Suppose you have 12 points. If you connect adjacent points you get a do-decagon. If connect every other point (jumps of 2) you get 2 hexagons. If you make jumps of 3, you get 3 squares. If you make jumps of 4 you get 4 triangles. and 5 and 12 are co-prime and you get a star.
â Doug M
Aug 7 at 20:31