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 confuse of how to prove that it is just really 5. Can anyone enlighten me?
combinatorics
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
$endgroup$
add a comment |
$begingroup$
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?
combinatorics
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
$endgroup$
$begingroup$
Hint: each pentagon has either a straight or diagonal line of symmetry.
$endgroup$
– Hugh
1 hour ago
add a comment |
$begingroup$
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?
combinatorics
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
$endgroup$
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
combinatorics
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
asked 1 hour ago
Sierra SorongonSierra Sorongon
365
365
$begingroup$
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
2
active
oldest
votes
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
$endgroup$
$begingroup$
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
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So this is an example of proof by exhaustive search.
$endgroup$
– Dr Xorile
37 mins 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.
- 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.
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
$endgroup$
$begingroup$
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
37 mins ago
add a comment |
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
$endgroup$
$begingroup$
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
37 mins ago
add a comment |
$begingroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
$endgroup$
Here are 5 symmetric pentagons on a $3times3$ grid:
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.
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
edited 1 hour ago
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
answered 1 hour ago
JonMark PerryJonMark Perry
18.9k63891
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
37 mins 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
37 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
37 mins ago
$begingroup$
So this is an example of proof by exhaustive search.
$endgroup$
– Dr Xorile
37 mins 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.
- 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.
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
$endgroup$
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.
- 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.
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
$endgroup$
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.
- 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.
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
$endgroup$
@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.
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
answered 18 mins ago
Dr XorileDr Xorile
11.8k22566
answered 18 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