Mixing
Does having two lines of symmetry $y=0$ and $y=-x$ imply that the shape also has lines of symmetry $x=0$ and...
$begingroup$
I have been wondering if a shape/curve that has a line of symmetry along the lines $y=0$ and $y=-x$ is guaranteed to also have lines of symmetry along the lines $x=0$ and $y=x$.
My gut feeling tells me that this is true. All the shapes that I can think of satisfy this. But I can't think of a way to prove this.
Also, is this relationship "if and only if" (with a $Leftrightarrow$ symbol) or just "implies" (with a $Rightarrow$ symbol).
I tried looking up similar questions here but I couldn't find any. Maybe I worded it too verbosely.
coordinate-systems symmetry
edited Jan 30 at 11:10
abc...
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
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add a comment |
$begingroup$
I have been wondering if a shape/curve that has a line of symmetry along the lines $y=0$ and $y=-x$ is guaranteed to also have lines of symmetry along the lines $x=0$ and $y=x$.
My gut feeling tells me that this is true. All the shapes that I can think of satisfy this. But I can't think of a way to prove this.
Also, is this relationship "if and only if" (with a $Leftrightarrow$ symbol) or just "implies" (with a $Rightarrow$ symbol).
I tried looking up similar questions here but I couldn't find any. Maybe I worded it too verbosely.
coordinate-systems symmetry
edited Jan 30 at 11:10
abc...
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
$endgroup$
add a comment |
$begingroup$
I have been wondering if a shape/curve that has a line of symmetry along the lines $y=0$ and $y=-x$ is guaranteed to also have lines of symmetry along the lines $x=0$ and $y=x$.
My gut feeling tells me that this is true. All the shapes that I can think of satisfy this. But I can't think of a way to prove this.
Also, is this relationship "if and only if" (with a $Leftrightarrow$ symbol) or just "implies" (with a $Rightarrow$ symbol).
I tried looking up similar questions here but I couldn't find any. Maybe I worded it too verbosely.
coordinate-systems symmetry
edited Jan 30 at 11:10
abc...
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
$endgroup$
I have been wondering if a shape/curve that has a line of symmetry along the lines $y=0$ and $y=-x$ is guaranteed to also have lines of symmetry along the lines $x=0$ and $y=x$.
My gut feeling tells me that this is true. All the shapes that I can think of satisfy this. But I can't think of a way to prove this.
Also, is this relationship "if and only if" (with a $Leftrightarrow$ symbol) or just "implies" (with a $Rightarrow$ symbol).
I tried looking up similar questions here but I couldn't find any. Maybe I worded it too verbosely.
coordinate-systems symmetry
coordinate-systems symmetry
edited Jan 30 at 11:10
abc...
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
edited Jan 30 at 11:10
abc...
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
edited Jan 30 at 11:10
abc...
3,237739
edited Jan 30 at 11:10
abc...
3,237739
edited Jan 30 at 11:10
abc...
3,237739
3,237739
asked Jan 30 at 10:51
SweeperSweeper
1526
asked Jan 30 at 10:51
SweeperSweeper
1526
asked Jan 30 at 10:51
SweeperSweeper
1526
1526
add a comment |
add a comment |
2 Answers
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Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
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Yes, this is true.
Consider a point $(x,y)$ in the $1^{text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.
answered Jan 30 at 11:09
abc...abc...
3,237739
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2 Answers
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2 Answers
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$begingroup$
Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
$endgroup$
add a comment |
$begingroup$
Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
$endgroup$
add a comment |
$begingroup$
Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
$endgroup$
Let $(u,v)$ be a point on the curve. Then:
- symmetry about $y=0$ implies that $(u,-v)$ is on the curve;
- symmetry about $y=-x$ implies that $(-v,-u)$ is on the curve.
Now we can apply these rules iteratively:
- reflect $(u,-v)$ about $y=-x$ to get $(v,-u)$
- reflect $(v,-u)$ about $y=0$ to get $(v,u)$
And $(v,u)$ is $(u,v)$ reflected about $y=x$. So $y=x$ is indeed a line of symmetry.
Similarly, reflecting $(-v,-u)$ about $y=0$ then $y=-x$ gives $(-u,v)$, which is $(u,v)$ reflected about $x=0$. So $x=0$ is also a line of symmetry.
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
edited Jan 30 at 11:53
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
answered Jan 30 at 11:09
TonyKTonyK
43.8k358137
43.8k358137
add a comment |
add a comment |
$begingroup$
Yes, this is true.
Consider a point $(x,y)$ in the $1^{text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.
answered Jan 30 at 11:09
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
Yes, this is true.
Consider a point $(x,y)$ in the $1^{text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.
answered Jan 30 at 11:09
abc...abc...
3,237739
$endgroup$
add a comment |
$begingroup$
Yes, this is true.
Consider a point $(x,y)$ in the $1^{text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.
answered Jan 30 at 11:09
abc...abc...
3,237739
$endgroup$
Yes, this is true.
Consider a point $(x,y)$ in the $1^{text{st}}$ quadrant. By line of symmetry $y=0$, the curve also contains $(-x,y)$. Then by the line of symmetry $y=-x$, the curve contains $(-x,-y)$ (since it contains $(x,y)$). Now, the curve must contain $(-x,y)$ since the line of symmetry $y=0.$
Therefore we have proved the $4$ points $(x,y),(x,-y),(-x,-y),(-x,y)$ are simultaneously on the curve or off the curve.
Then you can try prove it has lines of symmetry $x=0$ and $y=x$. $(*)$
Comment if you got stuck on how to prove $(*)$ or if I have made anything unclear.
answered Jan 30 at 11:09
abc...abc...
3,237739
answered Jan 30 at 11:09
abc...abc...
3,237739
answered Jan 30 at 11:09
abc...abc...
3,237739
answered Jan 30 at 11:09
abc...abc...
3,237739
3,237739
add a comment |
add a comment |
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