Any symmetry that fixes three non-collinear points is the identity
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
816217
add a comment |
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
816217
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
down vote
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
816217
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
816217
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
816217
edited Nov 15 at 11:42
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
816217
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
816217
asked Nov 15 at 11:25
WesleyGroupshaveFeelingsToo
816217
816217
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
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
58.8k23671
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
58.8k23671
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
58.8k23671
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
58.8k23671
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
58.8k23671
answered Nov 15 at 11:44
Berci
58.8k23671
answered Nov 15 at 11:44
Berci
58.8k23671
answered Nov 15 at 11:44
Berci
58.8k23671
58.8k23671
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 |
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2999574%2fany-symmetry-that-fixes-three-non-collinear-points-is-the-identity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
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