General test to prove a set of points creates a cyclic polygon

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For $n>4$, is there a way to show that a set of points creates an $n$-sided, cyclic polygon? Or simply, that the set of points are concyclic?




All triangles are cyclic. For quadrilaterals, we can use Ptolemy's theorem about the product of diagonals, among others. But, can these tests be extended to arbitrary sided polygons?



The Japanese theorem for cyclic polygons is one good answer, but I am wondering if there are other simpler ways.




euclidean-geometry




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  • 4




    Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
    – Lord Shark the Unknown
    Jul 15 at 8:37










  • It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
    – Theo Bendit
    Jul 15 at 8:41










  • @LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
    – John Glenn
    Jul 15 at 8:46














up vote
2
down vote

favorite













For $n>4$, is there a way to show that a set of points creates an $n$-sided, cyclic polygon? Or simply, that the set of points are concyclic?




All triangles are cyclic. For quadrilaterals, we can use Ptolemy's theorem about the product of diagonals, among others. But, can these tests be extended to arbitrary sided polygons?



The Japanese theorem for cyclic polygons is one good answer, but I am wondering if there are other simpler ways.




euclidean-geometry




share|cite|improve this question

















  • 4




    Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
    – Lord Shark the Unknown
    Jul 15 at 8:37










  • It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
    – Theo Bendit
    Jul 15 at 8:41










  • @LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
    – John Glenn
    Jul 15 at 8:46












up vote
2
down vote

favorite









up vote
2
down vote

favorite












For $n>4$, is there a way to show that a set of points creates an $n$-sided, cyclic polygon? Or simply, that the set of points are concyclic?




All triangles are cyclic. For quadrilaterals, we can use Ptolemy's theorem about the product of diagonals, among others. But, can these tests be extended to arbitrary sided polygons?



The Japanese theorem for cyclic polygons is one good answer, but I am wondering if there are other simpler ways.




euclidean-geometry




share|cite|improve this question














For $n>4$, is there a way to show that a set of points creates an $n$-sided, cyclic polygon? Or simply, that the set of points are concyclic?




All triangles are cyclic. For quadrilaterals, we can use Ptolemy's theorem about the product of diagonals, among others. But, can these tests be extended to arbitrary sided polygons?



The Japanese theorem for cyclic polygons is one good answer, but I am wondering if there are other simpler ways.





euclidean-geometry





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 15 at 8:44









Bernard

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asked Jul 15 at 8:33









John Glenn

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  • 4




    Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
    – Lord Shark the Unknown
    Jul 15 at 8:37










  • It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
    – Theo Bendit
    Jul 15 at 8:41










  • @LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
    – John Glenn
    Jul 15 at 8:46












  • 4




    Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
    – Lord Shark the Unknown
    Jul 15 at 8:37










  • It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
    – Theo Bendit
    Jul 15 at 8:41










  • @LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
    – John Glenn
    Jul 15 at 8:46







4




4




Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
– Lord Shark the Unknown
Jul 15 at 8:37




Just check that for each $k$, $P_1P_2P_3P_k$ is concyclic, by say Ptolemy.
– Lord Shark the Unknown
Jul 15 at 8:37












It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
– Theo Bendit
Jul 15 at 8:41




It's probably good to note that Lord Shark's idea works because of Caratheodory's theorem.
– Theo Bendit
Jul 15 at 8:41












@LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
– John Glenn
Jul 15 at 8:46




@LordSharktheUnknown I was expecting to use all the points simultaneously, but that's a very good point too.
– John Glenn
Jul 15 at 8:46










1 Answer
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An alternative is the following.
Take any three points $p_1, p_2, p_3$ and compute the circle $C$ they determine:



         
Circ3

Then check each of the subsequent points, $p_4, p_5, ldots$
to lie on $C$. You may need to take care with numerical issues.

For how to compute a $3$-point circle, see, e.g., Ed Pegg's
Mathematica Demo.






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    1 Answer
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    active

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    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes








    up vote
    1
    down vote













    An alternative is the following.
    Take any three points $p_1, p_2, p_3$ and compute the circle $C$ they determine:



             
    Circ3

    Then check each of the subsequent points, $p_4, p_5, ldots$
    to lie on $C$. You may need to take care with numerical issues.

    For how to compute a $3$-point circle, see, e.g., Ed Pegg's
    Mathematica Demo.






    share|cite|improve this answer

























      up vote
      1
      down vote













      An alternative is the following.
      Take any three points $p_1, p_2, p_3$ and compute the circle $C$ they determine:



               
      Circ3

      Then check each of the subsequent points, $p_4, p_5, ldots$
      to lie on $C$. You may need to take care with numerical issues.

      For how to compute a $3$-point circle, see, e.g., Ed Pegg's
      Mathematica Demo.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        An alternative is the following.
        Take any three points $p_1, p_2, p_3$ and compute the circle $C$ they determine:



                 
        Circ3

        Then check each of the subsequent points, $p_4, p_5, ldots$
        to lie on $C$. You may need to take care with numerical issues.

        For how to compute a $3$-point circle, see, e.g., Ed Pegg's
        Mathematica Demo.






        share|cite|improve this answer













        An alternative is the following.
        Take any three points $p_1, p_2, p_3$ and compute the circle $C$ they determine:



                 
        Circ3

        Then check each of the subsequent points, $p_4, p_5, ldots$
        to lie on $C$. You may need to take care with numerical issues.

        For how to compute a $3$-point circle, see, e.g., Ed Pegg's
        Mathematica Demo.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 15 at 11:23









        Joseph O'Rourke

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