Proving the count of symmetric configurations of pentagon












4












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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?






geometry combinatorics







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  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    Feb 2 at 3:27
















4












$begingroup$


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?






geometry combinatorics







share|improve this question











$endgroup$












  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    Feb 2 at 3:27














4












4








4





$begingroup$


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?






geometry 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?






geometry combinatorics




geometry combinatorics






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Feb 2 at 7:31









Omega Krypton

4,5151440




4,5151440










asked Feb 2 at 2:48









Sierra SorongonSierra Sorongon

465




465












  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    Feb 2 at 3:27


















  • $begingroup$
    Hint: each pentagon has either a straight or diagonal line of symmetry.
    $endgroup$
    – Hugh
    Feb 2 at 3:27
















$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
Feb 2 at 3:27




$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
Feb 2 at 3:27










2 Answers
2






active

oldest

votes


















6












$begingroup$

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




symmetric pentagons

A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.







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
    Feb 2 at 3:25












  • $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    Feb 2 at 4:01



















3












$begingroup$

@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
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    6












    $begingroup$

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




    symmetric pentagons

    A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.







    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
      Feb 2 at 3:25












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      Feb 2 at 4:01
















    6












    $begingroup$

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




    symmetric pentagons

    A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.







    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
      Feb 2 at 3:25












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      Feb 2 at 4:01














    6












    6








    6





    $begingroup$

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




    symmetric pentagons

    A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.







    share|improve this answer











    $endgroup$



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




    symmetric pentagons

    A proof is to notice that there are only two lines of mirror symmetry on the grid, vertical (or horizontal by rotation) and diagonal. A line of symmetry can only contain 1 vertex, because the other vertices are mirrored, and 3 in a row is impossible (a vertex would have 3 edges). There are only 3 variations for each of the two possible LoS vertices, giving only 12 cases to test, and 12 ways to draw each one (12345 is the same as 23451 is the same as 54321), so there are only 144 possibilities to check.








    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Feb 2 at 11:04

























    answered Feb 2 at 3:23









    JonMark PerryJonMark Perry

    19.3k63991




    19.3k63991












    • $begingroup$
      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
      Feb 2 at 3:25












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      Feb 2 at 4:01


















    • $begingroup$
      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
      Feb 2 at 3:25












    • $begingroup$
      So this is an example of proof by exhaustive search.
      $endgroup$
      – Dr Xorile
      Feb 2 at 4:01
















    $begingroup$
    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
    Feb 2 at 3:25






    $begingroup$
    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
    Feb 2 at 3:25














    $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    Feb 2 at 4:01




    $begingroup$
    So this is an example of proof by exhaustive search.
    $endgroup$
    – Dr Xorile
    Feb 2 at 4:01











    3












    $begingroup$

    @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











    $endgroup$


















      3












      $begingroup$

      @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











      $endgroup$
















        3












        3








        3





        $begingroup$

        @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 Feb 2 at 5:04

























        answered Feb 2 at 4:20









        Dr XorileDr Xorile

        12.9k22569




        12.9k22569






























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