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We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|BO|/|AO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to show (see details below) that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|B_1O|/|A_1O|>|BO|/|AO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tD$ for real $t>0$, where $D$ is the unit disk. Note that $C_t$ is smooth and has the same bisectors as $C$ -- which follows because, for any unit vector $u$ we have $\max(u\cdot C_t)=\max(u\cdot C)+\max(u\cdot(tD))=\max(u\cdot C)+t$, where $\cdot$ denotes the dot product and $u\cdot E:=\{u\cdot x\colon\, x\in E\}$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

Details on the inequality $|B_1O|/|A_1O|>|BO|/|AO|$: In the picture below, as in the setting above, $AB$ is a chord in $C$ through $O$, and $l_A$ and $l_B$ are the lines through the boundary points $A$ and $B$ supporting the smooth convex set $C$, and $l_A$ and $l_B$ are not parallel. The dashed line $l'_B$ is the line through $B$ parallel to $l_A$. The segment $A_1B_1$ is a chord of $C$ through $O$ close to the chord $AB$ of $C$, so that $A_1$ is almost on the line $l_A$ and $B_1$ is almost on the line $l_B$, since $C$ is smooth. Also, $B_1$ is "above" the line $l'_B$. The point $B'_1$ is the point of intersection of the line $l'_B$ with the line through $A_1$ and $B_1$. It is clear now that $|B_1O|/|A_1O|>|B'_1O|/|A_1O|=|BO|/|AO|$.

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|BO|/|AO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to show (see details below) that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|B_1O|/|A_1O|>|BO|/|AO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tD$ for real $t>0$, where $D$ is the unit disk. Note that $C_t$ is smooth and has the same bisectors as $C$ -- which follows because, for any unit vector $u$ we have $\max(u\cdot C_t)=\max(u\cdot C)+\max(u\cdot(tD))=\max(u\cdot C)+t$, where $\cdot$ denotes the dot product and $u\cdot E:=\{u\cdot x\colon\, x\in E\}$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

Details on the inequality $|B_1O|/|A_1O|>|BO|/|AO|$: In the picture below, as in the setting above, $AB$ is a chord in $C$ through $O$, and $l_A$ and $l_B$ are the lines through the boundary points $A$ and $B$ supporting the smooth convex set $C$, and $l_A$ and $l_B$ are not parallel. The dashed line $l'_B$ is the line through $B$ parallel to $l_A$. The segment $A_1B_1$ is a chord of $C$ through $O$ close to the chord $AB$ of $C$, so that $A_1$ is almost on the line $l_A$ and $B_1$ is almost on the line $l_B$, since $C$ is smooth. Also, $B_1$ is "above" the line $l'_B$. The point $B'_1$ is the point of intersection of the line $l'_B$ with the line through $A_1$ and $B_1$. It is clear now that $|B_1O|/|A_1O|>|B'_1O|/|A_1O|=|BO|/|AO|$.

Almost the same reasoning proves the natural generalization of this "planar" result to the corresponding one for any dimension.

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|BO|/|AO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to show (see details below) that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|B_1O|/|A_1O|>|BO|/|AO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tD$ for real $t>0$, where $D$ is the unit disk. Note that $C_t$ is smooth and has the same bisectors as $C$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

Details on the inequality $|B_1O|/|A_1O|>|BO|/|AO|$: In the picture below, as in the setting above, $AB$ is a chord in $C$ through $O$, and $l_A$ and $l_B$ are the lines through the boundary points $A$ and $B$ supporting the smooth convex set $C$, and $l_A$ and $l_B$ are not parallel. The dashed line $l'_B$ is the line through $B$ parallel to $l_A$. The segment $A_1B_1$ is a chord of $C$ through $O$ close to the chord $AB$ of $C$, so that $A_1$ is almost on the line $l_A$ and $B_1$ is almost on the line $l_B$, since $C$ is smooth. Also, $B_1$ is "above" the line $l'_B$. The point $B'_1$ is the point of intersection of the line $l'_B$ with the line through $A_1$ and $B_1$. It is clear now that $|B_1O|/|A_1O|>|B'_1O|/|A_1O|=|BO|/|AO|$.

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|BO|/|AO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to show (see details below) that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|B_1O|/|A_1O|>|BO|/|AO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tD$ for real $t>0$, where $D$ is the unit disk. Note that $C_t$ is smooth and has the same bisectors as $C$ -- which follows because, for any unit vector $u$ we have $\max(u\cdot C_t)=\max(u\cdot C)+\max(u\cdot(tD))=\max(u\cdot C)+t$, where $\cdot$ denotes the dot product and $u\cdot E:=\{u\cdot x\colon\, x\in E\}$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

Details on the inequality $|B_1O|/|A_1O|>|BO|/|AO|$: In the picture below, as in the setting above, $AB$ is a chord in $C$ through $O$, and $l_A$ and $l_B$ are the lines through the boundary points $A$ and $B$ supporting the smooth convex set $C$, and $l_A$ and $l_B$ are not parallel. The dashed line $l'_B$ is the line through $B$ parallel to $l_A$. The segment $A_1B_1$ is a chord of $C$ through $O$ close to the chord $AB$ of $C$, so that $A_1$ is almost on the line $l_A$ and $B_1$ is almost on the line $l_B$, since $C$ is smooth. Also, $B_1$ is "above" the line $l'_B$. The point $B'_1$ is the point of intersection of the line $l'_B$ with the line through $A_1$ and $B_1$. It is clear now that $|B_1O|/|A_1O|>|B'_1O|/|A_1O|=|BO|/|AO|$.

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|AO|/|BO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to see that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|A_1O|/|B_1O|>|AO|/|BO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tB$ for real $t>0$, where $B$ is the unit ball. Note that $C_t$ is smooth and has the same bisectors as $C$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

We should assume that $C$ is closed; otherwise, there are easy counterexamples. If $C$ is unbounded, then there is at most one width bisector, and hence easy counterexamples again.

So, assume that $C$ is a compact convex set with all width bisectors passing through a point $O$.

Without loss of generality, the interior of $C$ is nonempty (otherwise, the desired conclusion is obvious).

Note that $O$ is in the interior of $C$ -- otherwise, there will be a (straight) line $l$ non-strictly separating $O$ from $C$, and hence the width bisector of $C$ parallel to $l$ would not pass through $O$.

Now, suppose first that $C$ is smooth, in the sense that through each boundary point of $C$ there is only one line supporting $C$. Let $AB$ be a chord in $C$ through $O$ with the maximal value, say $r_{\max}$, of the ratio $|BO|/|AO|$. Suppose that $r_{\max}>1$. Then the lines, say $l_A$ and $l_B$, through the boundary points $A$ and $B$ supporting $C$ cannot be parallel (otherwise, the width bisector parallel to $l_A$ and $l_B$ would not pass through the point $O$). So, it is easy to show (see details below) that one can rotate the chord $AB$ slightly clockwise or counterclockwise about $O$ to get a new chord $A_1B_1$ of $C$ through $O$ with the ratio $|B_1O|/|A_1O|>|BO|/|AO|=r_{\max}$, a contradiction. So, $r_{\max}$ is $\le1$ and hence $=1$. That is, $C$ is centrally symmetric.

If now $C$ is not smooth, smooth it up by considering $C_t:=C+tD$ for real $t>0$, where $D$ is the unit disk. Note that $C_t$ is smooth and has the same bisectors as $C$. So, by what was shown, $C_t$ is centrally symmetric, for each real $t>0$. So, $C=\bigcap_{t>0}C_t$ is centrally symmetric. $\quad\Box$

Details on the inequality $|B_1O|/|A_1O|>|BO|/|AO|$: In the picture below, as in the setting above, $AB$ is a chord in $C$ through $O$, and $l_A$ and $l_B$ are the lines through the boundary points $A$ and $B$ supporting the smooth convex set $C$, and $l_A$ and $l_B$ are not parallel. The dashed line $l'_B$ is the line through $B$ parallel to $l_A$. The segment $A_1B_1$ is a chord of $C$ through $O$ close to the chord $AB$ of $C$, so that $A_1$ is almost on the line $l_A$ and $B_1$ is almost on the line $l_B$, since $C$ is smooth. Also, $B_1$ is "above" the line $l'_B$. The point $B'_1$ is the point of intersection of the line $l'_B$ with the line through $A_1$ and $B_1$. It is clear now that $|B_1O|/|A_1O|>|B'_1O|/|A_1O|=|BO|/|AO|$.