Proving the count of symmetric configurations of pentagon












2












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In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confuse of how to prove that it is just really 5. Can anyone enlighten me?










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  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    1 hour ago
















2












$begingroup$


In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confuse of how to prove that it is just really 5. Can anyone enlighten me?










share|improve this question









$endgroup$












  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    1 hour ago














2












2








2





$begingroup$


In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confuse of how to prove that it is just really 5. Can anyone enlighten me?










share|improve this question









$endgroup$




In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confuse of how to prove that it is just really 5. Can anyone enlighten me?







combinatorics






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asked 2 hours ago









Sierra SorongonSierra Sorongon

365




365












  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    1 hour ago


















  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    1 hour ago
















$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
1 hour ago




$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
1 hour ago










2 Answers
2






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3












$begingroup$

Here are 5 symmetric pentagons on a $3times3$ grid:




symmetric pentagons


It can be proved by examining the $binom95=126$ cases. After reduction by symmetry and rotation, and removing obvious cases, such as 3 in a row, there are only 5 left.







share|improve this answer











$endgroup$













  • $begingroup$
    I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
    $endgroup$
    – Hugh
    1 hour ago












  • $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    56 mins ago



















1












$begingroup$

@JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:




  • The axis of symmetry must go through one of the 5 vertices.

  • The other 4 vertices must be symmetric to each other about the axis of symmetry.


Now note that there are only 3 vertices to choose from for the first vertex: The middle, the edge, and the corner.



The middle can have a horizontal axis of symmetry or a diagonal one.



The edge and corner will be symmetric about the line through that vertex and the center vertex.



Putting this together, there are only four cases which leads to the 5 cases already identified.






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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    active

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    3












    $begingroup$

    Here are 5 symmetric pentagons on a $3times3$ grid:




    symmetric pentagons


    It can be proved by examining the $binom95=126$ cases. After reduction by symmetry and rotation, and removing obvious cases, such as 3 in a row, there are only 5 left.







    share|improve this answer











    $endgroup$













    • $begingroup$
      I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
      $endgroup$
      – Hugh
      1 hour ago












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      56 mins ago
















    3












    $begingroup$

    Here are 5 symmetric pentagons on a $3times3$ grid:




    symmetric pentagons


    It can be proved by examining the $binom95=126$ cases. After reduction by symmetry and rotation, and removing obvious cases, such as 3 in a row, there are only 5 left.







    share|improve this answer











    $endgroup$













    • $begingroup$
      I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
      $endgroup$
      – Hugh
      1 hour ago












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      56 mins ago














    3












    3








    3





    $begingroup$

    Here are 5 symmetric pentagons on a $3times3$ grid:




    symmetric pentagons


    It can be proved by examining the $binom95=126$ cases. After reduction by symmetry and rotation, and removing obvious cases, such as 3 in a row, there are only 5 left.







    share|improve this answer











    $endgroup$



    Here are 5 symmetric pentagons on a $3times3$ grid:




    symmetric pentagons


    It can be proved by examining the $binom95=126$ cases. After reduction by symmetry and rotation, and removing obvious cases, such as 3 in a row, there are only 5 left.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 1 hour ago

























    answered 1 hour ago









    JonMark PerryJonMark Perry

    18.9k63891




    18.9k63891












    • $begingroup$
      I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
      $endgroup$
      – Hugh
      1 hour ago












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      56 mins ago


















    • $begingroup$
      I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
      $endgroup$
      – Hugh
      1 hour ago












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      56 mins ago
















    $begingroup$
    I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
    $endgroup$
    – Hugh
    1 hour ago






    $begingroup$
    I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
    $endgroup$
    – Hugh
    1 hour ago














    $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    56 mins ago




    $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    56 mins ago











    1












    $begingroup$

    @JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:




    • The axis of symmetry must go through one of the 5 vertices.

    • The other 4 vertices must be symmetric to each other about the axis of symmetry.


    Now note that there are only 3 vertices to choose from for the first vertex: The middle, the edge, and the corner.



    The middle can have a horizontal axis of symmetry or a diagonal one.



    The edge and corner will be symmetric about the line through that vertex and the center vertex.



    Putting this together, there are only four cases which leads to the 5 cases already identified.






    share|improve this answer









    $endgroup$


















      1












      $begingroup$

      @JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:




      • The axis of symmetry must go through one of the 5 vertices.

      • The other 4 vertices must be symmetric to each other about the axis of symmetry.


      Now note that there are only 3 vertices to choose from for the first vertex: The middle, the edge, and the corner.



      The middle can have a horizontal axis of symmetry or a diagonal one.



      The edge and corner will be symmetric about the line through that vertex and the center vertex.



      Putting this together, there are only four cases which leads to the 5 cases already identified.






      share|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        @JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:




        • The axis of symmetry must go through one of the 5 vertices.

        • The other 4 vertices must be symmetric to each other about the axis of symmetry.


        Now note that there are only 3 vertices to choose from for the first vertex: The middle, the edge, and the corner.



        The middle can have a horizontal axis of symmetry or a diagonal one.



        The edge and corner will be symmetric about the line through that vertex and the center vertex.



        Putting this together, there are only four cases which leads to the 5 cases already identified.






        share|improve this answer









        $endgroup$



        @JonMarkPerry got it and indicated that he'd looked through all the possibilities. But to outline the proof, you can note that:




        • The axis of symmetry must go through one of the 5 vertices.

        • The other 4 vertices must be symmetric to each other about the axis of symmetry.


        Now note that there are only 3 vertices to choose from for the first vertex: The middle, the edge, and the corner.



        The middle can have a horizontal axis of symmetry or a diagonal one.



        The edge and corner will be symmetric about the line through that vertex and the center vertex.



        Putting this together, there are only four cases which leads to the 5 cases already identified.







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 37 mins ago









        Dr XorileDr Xorile

        11.8k22566




        11.8k22566






























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