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 confused about how to prove that it is really just 5. Can anyone enlighten me?






combinatorics







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
















2












$begingroup$


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






combinatorics







share|improve this question











$endgroup$












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














2












2








2





$begingroup$


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






combinatorics







share|improve this question











$endgroup$




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






combinatorics




combinatorics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 41 mins ago









JonMark Perry

18.9k63891




18.9k63891










asked 3 hours ago









Sierra SorongonSierra Sorongon

365




365












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


















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
















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




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










2 Answers
2






active

oldest

votes


















3












$begingroup$

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




symmetric pentagons


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


A shorter proof is to notice that there are only two lines of symmetry, vertical (or horizontal by rotation) and diagonal. The line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible, and there are only 3 variations here.







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
    2 hours ago












  • $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    2 hours 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 (call it $A$)

  • 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 vertex $A$: The center, the edge, and the corner.



The center can have an orthogonal 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:




  1. Orthogonal axis of symmetry through the center vertex: 1 possibility.

  2. Diagonal axis of symmetry through the center vertex: 1 possibility

  3. Orthogonal axis of symmetry through the edge vertex: 1 possibility

  4. Diagonal axis of symmetry through the corner vertex: 2 possibilities






share|improve this answer











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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

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    votes






    active

    oldest

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    3












    $begingroup$

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




    symmetric pentagons


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


    A shorter proof is to notice that there are only two lines of symmetry, vertical (or horizontal by rotation) and diagonal. The line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible, and there are only 3 variations here.







    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
      2 hours ago












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
















    3












    $begingroup$

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




    symmetric pentagons


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


    A shorter proof is to notice that there are only two lines of symmetry, vertical (or horizontal by rotation) and diagonal. The line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible, and there are only 3 variations here.







    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
      2 hours ago












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














    3












    3








    3





    $begingroup$

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




    symmetric pentagons


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


    A shorter proof is to notice that there are only two lines of symmetry, vertical (or horizontal by rotation) and diagonal. The line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible, and there are only 3 variations here.







    share|improve this answer











    $endgroup$



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




    symmetric pentagons


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


    A shorter proof is to notice that there are only two lines of symmetry, vertical (or horizontal by rotation) and diagonal. The line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible, and there are only 3 variations here.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 19 mins ago

























    answered 2 hours 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
      2 hours ago












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












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














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




    $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    2 hours 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 (call it $A$)

    • 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 vertex $A$: The center, the edge, and the corner.



    The center can have an orthogonal 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:




    1. Orthogonal axis of symmetry through the center vertex: 1 possibility.

    2. Diagonal axis of symmetry through the center vertex: 1 possibility

    3. Orthogonal axis of symmetry through the edge vertex: 1 possibility

    4. Diagonal axis of symmetry through the corner vertex: 2 possibilities






    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 (call it $A$)

      • 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 vertex $A$: The center, the edge, and the corner.



      The center can have an orthogonal 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:




      1. Orthogonal axis of symmetry through the center vertex: 1 possibility.

      2. Diagonal axis of symmetry through the center vertex: 1 possibility

      3. Orthogonal axis of symmetry through the edge vertex: 1 possibility

      4. Diagonal axis of symmetry through the corner vertex: 2 possibilities






      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 (call it $A$)

        • 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 vertex $A$: The center, the edge, and the corner.



        The center can have an orthogonal 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:




        1. Orthogonal axis of symmetry through the center vertex: 1 possibility.

        2. Diagonal axis of symmetry through the center vertex: 1 possibility

        3. Orthogonal axis of symmetry through the edge vertex: 1 possibility

        4. Diagonal axis of symmetry through the corner vertex: 2 possibilities






        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 (call it $A$)

        • 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 vertex $A$: The center, the edge, and the corner.



        The center can have an orthogonal 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:




        1. Orthogonal axis of symmetry through the center vertex: 1 possibility.

        2. Diagonal axis of symmetry through the center vertex: 1 possibility

        3. Orthogonal axis of symmetry through the edge vertex: 1 possibility

        4. Diagonal axis of symmetry through the corner vertex: 2 possibilities







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 1 hour ago

























        answered 1 hour ago









        Dr XorileDr Xorile

        11.8k22566




        11.8k22566






























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