Centre of symmetry of a bound figure.

Question

While doing some interesting geometric questions I found the most interesting one, but the irony is that I m unable to do it.

The question is Prove that a bounded figure cannot have two centre of symmetry.

Though this fact seem quite obvious, but a mathematical proof is needed. You guys are always eager to see the work of OP. So, I would say that I know what is centre of symmetry and what is a bound figure, but I cannot relate these two.I think the proof can be done in standard manner(by assuming that there exist a bound figure which have two centre and symmetry and at the end we can prove by contradiction that such a figure cannot be a bound figure)

All the problem I m getting is how to interpret all my ideas on a paper and get a proof.

Thanks in advance for any help

Answer 1

Assume there are (at least) two distinct centers of symmetry $A \ne B$. Let $B_0 \equiv B$ and for $n \ge 1$ define $B_n$ recursively as the reflection across $B_0$ of the reflection of $B_{n-1}$ across $A_0$.

By symmetry all $B_n$ must belong to the figure, but it can be easily shown that $|B_nB| = 2n \,|AB|$ which contradicts the premise that the figure is bounded.

Answer 2

` 1. X is a central symmetric => closure(X) is central symmetric

2. for any coordinate: v1 is max of the coordinate value, v2 is a min of the coordinate value => center of symmetry must have the coordinate value (v1+v2)/2

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