Proving the count of symmetric configurations of pentagon
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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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add a comment |
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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
add a comment |
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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
$endgroup$
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
edited 41 mins ago
JonMark Perry
18.9k63891
18.9k63891
asked 3 hours ago
Sierra SorongonSierra Sorongon
365
365
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Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
2 hours ago
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
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Here are 5 symmetric pentagons on a $3times3$ grid:
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.
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I have a proof, but I'm on mobile. I think I'll just bail and give this one to you. Good luck ! 😀
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– Hugh
2 hours ago
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So this is an example of proof by exhaustive search.
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– Dr Xorile
2 hours ago
add a comment |
$begingroup$
@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:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities
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2 Answers
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2 Answers
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$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
$endgroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
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
add a comment |
$begingroup$
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
add a comment |
$begingroup$
@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:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities
$endgroup$
add a comment |
$begingroup$
@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:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities
$endgroup$
add a comment |
$begingroup$
@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:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities
$endgroup$
@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:
- Orthogonal axis of symmetry through the center vertex: 1 possibility.
- Diagonal axis of symmetry through the center vertex: 1 possibility
- Orthogonal axis of symmetry through the edge vertex: 1 possibility
- Diagonal axis of symmetry through the corner vertex: 2 possibilities
edited 1 hour ago
answered 1 hour ago
Dr XorileDr Xorile
11.8k22566
11.8k22566
add a comment |
add a comment |
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$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
2 hours ago