Consider the following statement:
Suppose that $C$ is a smooth simple closed curve in $\mathbb R^2$ with positive curvature. Let $O$ be the mass center of the region bound by $C$. If $C$ is invariant under the rotation on $\mathbb R^2$ by $\pi/2$ about $O$, then $C$ is a circle.
I obtain this statement by applying nontrivial results by Simon Brendle, regarding the existence of diffeomorphism between uniformly convex domains that satisfies certain conditions.
(Recall that the positivity of curvature implies that $C$ bounds a uniformly convex region.)
However, the statement looks so elementary, so I believe that some easier proof exists.
But I am not capable of coming up with one.
Could anyone help?
Notice that the positivity of curvature excludes curves like $x^4+y^4=1$, which seems to be a counterexample at first glance.
Best Answer
Counterexample. The curve defined by the cartesian equation $$ x^4+y^4+x^2y^2=1 $$ has rotational symmetry of order $4$ and its curvature is always positive. Here's a plot of the curve:
And here's a plot ot the curvature as a function of polar angle: