Let $D$ be a bounded simply connected region (open subset homeomorphic to the disc) 
in the plane, containing the origin. 
Suppose that for every line $L$ through the origin the intersection $L\cap\partial D$
consists of two points $z\_1$ and $z\_2$ such that
$|z\_1-z\_2|=\mathrm{diam} D$. Does it follow that $D$ is a disc?


[I can prove this](https://mathoverflow.net/questions/157587/2f0-sup-z-w-in-d-fz-fw-if-and-only-if-f-is-linear/157616#157616) with the additional restriction that $\partial D$ is smooth. [This problem](https://mathoverflow.net/questions/157740/maximal-regions-with-given-diameter) is also related to the present question.



Characterization of discs
Following the suggestion of Benoît Kloeckner, moderator, I have deleted my later answer, then edited and appended this (originally partial) answer to make it complete.


Pick an arbitrary direction. Then draw the line $L$ through the origin, perpendicular to the chosen direction. Since $L$ intersects the boundary of $D$ at two points, say $z\_1$ and $z\_2$, such that the segment $\overline{z\_1z\_2}$ is a diameter of $D$, the lines perpendicular to this segment and passing through $z\_1$ and $z\_2$, respectively, bound a parallel strip containing $D$ - otherwise the diameter of $D$ would be greater than the distance from $z\_1$ to $z\_2$. This proves that the width of $D$ in every direction is equal to the diameter of $D$.


Remark: This also proves that the closure $\bar{D}$ of $D$ is convex, being the intersection of a family of strips. in fact, $\bar{D}$ is strictly convex, as every set of constant width must be.


Another remark: The same proof works in every dimension; just replace the *arbitrary direction* by *arbitrary hyperplane* with respect to which we look at the width of $D$.


Thus far, this does not quite answer the question. Among all examples of convex bodies of constant width I know, only the ball has the property that all diameters have one common point. It remains to prove that no other such body exists.


It has been established that $\bar{D}$ is strictly convex, that is, every support line of $D$ contains exactly one boundary point of $D$. Also, each support line of $D$ has its "opposite" support line, forming a strip between them of width $d$ and containing $D$. Now, suppose $\bar{D}$ is not smooth. Specifically, let $x\_0$ be a boundary point of $D$ at which there are two intersecting support lines. Then the corresponding opposite to them support lines touch $\bar{D}$ at points $x\_1$ and $x\_2$, respectively, such that each of the segments $\overline{x\_0x\_1}$ and $\overline{x\_0x\_2}$ is a diameter of $D$. But since all diameters of $D$ meet at a single point, namely at the origin, $x\_0$ *must be* the origin, contrary to the assumption that the origin lies in the interior of $\bar{D}$. This, in view of Alexandre's comment at the end of his question, implies that $D$ is a circular disk.


By the way, it is not necessary to assume that the origin lies in the interior of $\bar{D}$, since this follows from the other assumptions. Namely, if the origin were a boundary point of $D$, then a line passing through it and penetrating the interior of $D$ would intersect the boundary of $D$ at another point, the two points forming a diameter. Then the line $L$ through the origin and perpendicular to the penetrating line would be a support line of $D$. But there should be another boundary point on $L$ at the diameter-distance from the origin - a clear contradiction: two perpendicular diameters meeting at their end points. Thus $D$ is a circle centered at the origin.


Final remark: The statement for the plane implies the same in higher dimensions, by taking all 2-dimensional cross-sections of the body through the origin: each of them is smooth and each satisfies the same assumptions on the diameters. Since all such cross-sections are congruent circles, the body is a ball.


