TITLE: Plain shape invariant with respect to two rotations
QUESTION [1 upvotes]: Suppose that a plane shape $\phi$ is invariant with respect to two rotations, $R_{A,\alpha}$ and $R_{B,\beta}$:
$$R_{A,\alpha}(\phi)=\phi$$ 
$$R_{B,\beta}(\phi)=\phi$$ 
$$ 0 \lt \alpha , \beta < 2\pi$$
Prove that $A\equiv B$
I have a nice proof but my son who is dealing with isometric transformations does not like it at all. :) He has a hint for me: "Use the fact that a plain figure cannot have two distinct centers of symmetry". But I don't know how to use it.
EDIT: Clarification:
1) The shape has finite dimensions.
2) Central symmetry = Rotation for 180 degrees.

REPLY [0 votes]: Ok, this is the first time that I'm answering my own question, I apologize for that. But I'm curious to learn what's wrong with this quite simple proof.

Consider center of rotation $A$ and shape $\phi$ (not shown in this drawing). Find point $M\in\phi$ that has the maximum distance from the center of rotation $A$. This point is not unique. If you rotate that point around $A$ for angle $\alpha$, you'll get the point $M_1$ that also belongs to shape $\phi$ with the same distance from the center of rotation $A$. You can repeat the whole process once again with point $M_1$ and obtain point $M_2$ that also belongs to shape $\phi$ and has the same distance from point $A$. You can repeat the same process as many times as you want. All these points $M_i$ lie on the same circle $k$.
For angle $\alpha=\pi$ there are two such points and for $0\lt\alpha\lt\pi$ there are at least three. For $\pi\lt\alpha\lt 2\pi$, you can replace rotation $R_{A,\alpha}$ with $R_{A,2\pi - \alpha}$ and reach the same conclusion:
Points $M_i$ ($i=1,...,n$) are equidistant and $n$ must be at least 2. This also means that every semicircle of circle $k$ contains at least one point $M_i$.
Now rotate the shape $\phi$ around point B for some angle $\beta$ (not shown). Each point $M_i\in k,\phi$ travels to point $M'_i\in k',\phi$ where $R_{B,\beta}(k)=k'$
In the worst case, when circles $k,k'$ intersect, only the (red) part of the circle $k'$ will be outside of circle $k$. But the central angle of that part is always greater than $\pi$ which means that at least one point $M'_i\in\phi$ must be in the red section. For such point:
$$AM'_i\gt AM $$ 
which is a contradiction.
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