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 just really 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 just really 5. Can anyone enlighten me?
combinatorics
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$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
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– Hugh
1 hour ago
add a comment |
$begingroup$
In a 3 × 3 dot grid, there are 5 configurations of symmetric pentagons. I am confused about how to prove that it is just really 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 just really 5. Can anyone enlighten me?
combinatorics
combinatorics
edited 6 mins ago
Dr Xorile
11.8k22566
11.8k22566
asked 2 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
1 hour ago
add a comment |
$begingroup$
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
add a comment |
2 Answers
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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.
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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
1 hour ago
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So this is an example of proof by exhaustive search.
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– Dr Xorile
1 hour ago
add a comment |
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@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.
$endgroup$
$begingroup$
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
1 hour 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.
$endgroup$
$begingroup$
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
1 hour 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.
$endgroup$
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.
edited 8 mins 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
1 hour 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
1 hour ago
$begingroup$
So this is an example of proof by exhaustive search.
$endgroup$
– Dr Xorile
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
$begingroup$
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
1 hour ago
$begingroup$
So this is an example of proof by exhaustive search.
$endgroup$
– Dr Xorile
1 hour 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 9 mins ago
answered 52 mins 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
1 hour ago