Cutting a convex body into two congruent pieces

This question is related to How to make a sandwich from just one piece of bread?, asked on Feb 23 '17 by erz, and it goes as follows:

Question. If a convex closed and bounded region $C$ in the plane $\mathbb{R}^2$ can be cut along some straight line into two congruent pieces, must $C$ have either axial or central symmetry?

To be specific about what $cutting$ means, a $piece$ consists of the closure of the set of all points of $C$ that lie on the same side of the cutting line. Obviously, if $C$ has either axial or central symmetry, then it can be cut so.

The question can be phrased generally in $\mathbb{R}^n$, where we would cut a convex body along a hyperplane and consider $n$ kinds of symmetry, for example central, axial, and mirror in $\mathbb{R}^3$. I believe the answer is $yes$. Is it perhaps known already?

In retrospect (see Wlod AA's very nice answer): my intuition was way off.

• It might be more revealing to consider two congruent convex bodies glued together on a common shaped face. The merged body may not seem as symmetric if an appropriate twist is given before gluing. Gerhard "Even If One Maintains Convexity" Paseman, 2018.06.18. – Gerhard Paseman Jun 19 '18 at 0:17
• @GerhardPaseman, Yeah, there may be a counterexample in $\mathbb{R}^3$ already. – Wlodek Kuperberg Jun 19 '18 at 0:41
• If I am right then there are counter-examples in all dimensions $\ n>1$ (similar to the one in dimension $2$; it'd take more writing though). – Wlod AA Jun 19 '18 at 1:05

Here is a counter-example (in the complex plane $\mathbb C=\mathbb R^2.)\$ Let

$$\ P\ := \ \{ (x\ y)\in\mathbb C : 0\le x\le 1\quad\&\quad 0\le y\le 1-x^2\}$$

Then,

$$C\,\ :=\,\ P\,\cup\, i\!\cdot\! P$$

The imaginary line is the requested cut.

• Do pay attention to the difference between $\ \mathbb C\$ and $\ C.\$ Sorry for this notational near-confusion. – Wlod AA Jun 19 '18 at 1:00
• Włodek, dziękuję ("thank you" for those who have never read Falski). – Wlod AA Jun 19 '18 at 1:07
• Elementary, my dear Watson. – Wlodek Kuperberg Jun 19 '18 at 1:10
• There is even a pentagonal counter-example in $\mathbb{R}^2$ that your idea produces. – Wlodek Kuperberg Jun 19 '18 at 1:26
• One can see automatically a bunch of similar examples but pentagonal? I'd like to see it! Oh, of course, now I see it too. :) Thank you, Włodek, for telling me. Indeed, 5-gon sounds strange at first. – Wlod AA Jun 19 '18 at 8:28