# 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. Jun 19, 2018 at 0:17
• @GerhardPaseman, Yeah, there may be a counterexample in $\mathbb{R}^3$ already. Jun 19, 2018 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). Jun 19, 2018 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. Jun 19, 2018 at 1:00
• Włodek, dziękuję ("thank you" for those who have never read Falski). Jun 19, 2018 at 1:07
• Elementary, my dear Watson. Jun 19, 2018 at 1:10
• There is even a pentagonal counter-example in $\mathbb{R}^2$ that your idea produces. Jun 19, 2018 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. Jun 19, 2018 at 8:28