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 $n$?







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    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.
    – Doug M
    Aug 7 at 20:25











  • @DougM How is the greatest common divisor and the points that are deleted related?
    – 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.
    – Doug M
    Aug 7 at 20:31














up vote
1
down vote

favorite
1












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 $n$?







share|cite|improve this question

















  • 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.
    – Doug M
    Aug 7 at 20:25











  • @DougM How is the greatest common divisor and the points that are deleted related?
    – 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.
    – Doug M
    Aug 7 at 20:31












up vote
1
down vote

favorite
1









up vote
1
down vote

favorite
1






1





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 $n$?







share|cite|improve this question













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 $n$?









share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








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.
    – Doug M
    Aug 7 at 20:25











  • @DougM How is the greatest common divisor and the points that are deleted related?
    – 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.
    – Doug M
    Aug 7 at 20:31












  • 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.
    – Doug M
    Aug 7 at 20:25











  • @DougM How is the greatest common divisor and the points that are deleted related?
    – 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.
    – 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.
– 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.
– Doug M
Aug 7 at 20:25













@DougM How is the greatest common divisor and the points that are deleted related?
– Nash
Aug 7 at 20:28




@DougM How is the greatest common divisor and the points that are deleted related?
– 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.
– 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.
– Doug M
Aug 7 at 20:31










2 Answers
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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






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  • Right of course. I edited my answer.
    – Ethan Bolker
    Aug 7 at 21:00

















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.






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    2 Answers
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    2 Answers
    2






    active

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    active

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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






    share|cite|improve this answer





















    • Right of course. I edited my answer.
      – Ethan Bolker
      Aug 7 at 21:00














    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






    share|cite|improve this answer





















    • Right of course. I edited my answer.
      – Ethan Bolker
      Aug 7 at 21:00












    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






    share|cite|improve this answer













    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







    share|cite|improve this answer













    share|cite|improve this answer



    share|cite|improve this answer











    answered Aug 7 at 20:40









    Dr. Richard Klitzing

    7566




    7566











    • Right of course. I edited my answer.
      – Ethan Bolker
      Aug 7 at 21:00
















    • Right of course. I edited my answer.
      – Ethan Bolker
      Aug 7 at 21:00















    Right of course. I edited my answer.
    – Ethan Bolker
    Aug 7 at 21:00




    Right of course. I edited my answer.
    – Ethan Bolker
    Aug 7 at 21:00










    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.






    share|cite|improve this answer



























      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.






      share|cite|improve this answer

























        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.






        share|cite|improve this answer















        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.







        share|cite|improve this answer















        share|cite|improve this answer



        share|cite|improve this answer








        edited Aug 7 at 21:00


























        answered Aug 7 at 20:25









        Ethan Bolker

        35.8k54299




        35.8k54299






















             

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