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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.

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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.

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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.

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