Any symmetry that fixes three non-collinear points is the identity
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I am asked to finish the following sentence:
Let $sigma$ be an isometry on $mathbb{R}^2$, suppose it fixes the points $A$ and $B$
Suppose $sigma$ also fixes a third point $C$ which is not on the line $AB$.
Complete the
following sentence: If $P$ is a fourth point in the plane which is not fixed by $sigma$,
so $sigma(P) neq P$, then $A$, $B$ and $C$ are all on the $dots$ BLANK
This is a contradiction and
so $sigma(P) = P$. Since P is arbitrary, we conclude that $sigma$ is the identity.
I am not entirely sure what the author is going for. I think what they want is to make the case that $A,B$ and $C$ are on same reflection line, but I do not see how. This would contradict $C$ not being on the line through $AB$. It just seems like the argument skips a step
geometry euclidean-geometry isometry
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
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1
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I am asked to finish the following sentence:
Let $sigma$ be an isometry on $mathbb{R}^2$, suppose it fixes the points $A$ and $B$
Suppose $sigma$ also fixes a third point $C$ which is not on the line $AB$.
Complete the
following sentence: If $P$ is a fourth point in the plane which is not fixed by $sigma$,
so $sigma(P) neq P$, then $A$, $B$ and $C$ are all on the $dots$ BLANK
This is a contradiction and
so $sigma(P) = P$. Since P is arbitrary, we conclude that $sigma$ is the identity.
I am not entirely sure what the author is going for. I think what they want is to make the case that $A,B$ and $C$ are on same reflection line, but I do not see how. This would contradict $C$ not being on the line through $AB$. It just seems like the argument skips a step
geometry euclidean-geometry isometry
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
1
My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21
add a comment |
up vote
1
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favorite
up vote
1
down vote
favorite
I am asked to finish the following sentence:
Let $sigma$ be an isometry on $mathbb{R}^2$, suppose it fixes the points $A$ and $B$
Suppose $sigma$ also fixes a third point $C$ which is not on the line $AB$.
Complete the
following sentence: If $P$ is a fourth point in the plane which is not fixed by $sigma$,
so $sigma(P) neq P$, then $A$, $B$ and $C$ are all on the $dots$ BLANK
This is a contradiction and
so $sigma(P) = P$. Since P is arbitrary, we conclude that $sigma$ is the identity.
I am not entirely sure what the author is going for. I think what they want is to make the case that $A,B$ and $C$ are on same reflection line, but I do not see how. This would contradict $C$ not being on the line through $AB$. It just seems like the argument skips a step
geometry euclidean-geometry isometry
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
I am asked to finish the following sentence:
Let $sigma$ be an isometry on $mathbb{R}^2$, suppose it fixes the points $A$ and $B$
Suppose $sigma$ also fixes a third point $C$ which is not on the line $AB$.
Complete the
following sentence: If $P$ is a fourth point in the plane which is not fixed by $sigma$,
so $sigma(P) neq P$, then $A$, $B$ and $C$ are all on the $dots$ BLANK
This is a contradiction and
so $sigma(P) = P$. Since P is arbitrary, we conclude that $sigma$ is the identity.
I am not entirely sure what the author is going for. I think what they want is to make the case that $A,B$ and $C$ are on same reflection line, but I do not see how. This would contradict $C$ not being on the line through $AB$. It just seems like the argument skips a step
geometry euclidean-geometry isometry
geometry euclidean-geometry isometry
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
edited Nov 15 at 11:42
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
1,055321
1,055321
1
My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21
add a comment |
1
My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21
1
1
My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21
add a comment |
1 Answer
1
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oldest
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0
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accepted
You are correct, the term for the blank space would be bisection of the segment $P, sigma(P)$, since $d(P,X)=d(sigma(P), X)$ for each $Xin{A, B, C}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.
answered Nov 15 at 11:44
Berci
59.1k23671
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You are correct, the term for the blank space would be bisection of the segment $P, sigma(P)$, since $d(P,X)=d(sigma(P), X)$ for each $Xin{A, B, C}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.
answered Nov 15 at 11:44
Berci
59.1k23671
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
add a comment |
up vote
0
down vote
accepted
You are correct, the term for the blank space would be bisection of the segment $P, sigma(P)$, since $d(P,X)=d(sigma(P), X)$ for each $Xin{A, B, C}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.
answered Nov 15 at 11:44
Berci
59.1k23671
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You are correct, the term for the blank space would be bisection of the segment $P, sigma(P)$, since $d(P,X)=d(sigma(P), X)$ for each $Xin{A, B, C}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.
answered Nov 15 at 11:44
Berci
59.1k23671
You are correct, the term for the blank space would be bisection of the segment $P, sigma(P)$, since $d(P,X)=d(sigma(P), X)$ for each $Xin{A, B, C}$.
And it is indeed a contradiction, as the next line says, and that finishes the proof.
answered Nov 15 at 11:44
Berci
59.1k23671
answered Nov 15 at 11:44
Berci
59.1k23671
answered Nov 15 at 11:44
Berci
59.1k23671
answered Nov 15 at 11:44
Berci
59.1k23671
59.1k23671
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
add a comment |
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
Thank you, I had trouble finding the right word for it too. How considerate of you.
– WesleyGroupshaveFeelingsToo
Nov 15 at 11:48
add a comment |
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My initial instinct on reading this is that there is an error in the assignment, and "which is not on the line $AB$" should not have been there. Then you could fill "same line" into the BLANK.
– Mees de Vries
Nov 15 at 12:20
Yeah that's precisely the contradiction part.
– WesleyGroupshaveFeelingsToo
Nov 15 at 12:21