If every orthogonal projection of a manifold $M \subset \mathbb R^3$ on a plane is a disk, then $M$ is a sphere? Given  a 2-d manifold $M$ in $\mathbb R^3$ and a plane $\alpha$ in $\mathbb R^3$. Assume that $M$ is smooth and has convex, simply-connected inner part.
If we project $M$ orthogonally onto $\alpha$ we always get a round disk (maybe in different diameter as $\alpha$ changes), can we prove or disprove that $M$ is a sphere?
 A: Partial answer
One can prove that the radius is the same for all planes.
Take any plane $\alpha_1$. the projection $\pi_{\alpha_1}(M)$ being a closed disk $D_1$ means that the $M$ is a subset of the (solid) cyliner $C_1 = \pi_{\alpha_1}^{-1}(D_1)$ prependicular to $\alpha_1$ with radius $R_1$, as that of the $D_1$. Take any other nonparallel plane $\alpha_2$ and its cylinder $C_2 = \pi_{\alpha_2}^{-1}(D_2)$. The projection of $C_1$ on $\alpha_2$ is an infinite strip of width $2R_1$. This projection should contain the disk $D_2$, i.e. $\pi_{\alpha_2}(M) \subset \pi_{\alpha_2}(C_1)$, since $M \subset C_1$. This implies $2R_1 \ge 2R_2$. Similarly $2R_2 \ge 2R_1$.
Now if the radius of these projection is $R$ then any point $p$ not in the sphere $S^2$, with radius $R$, will be a distance $l \neq R$ from the center $O$ of $M$, which is found by finding the intersection of two lines prependicular to the centers of two nonparallel disks $D_1, D_2$.

*

*If $l \ge R$ then there is a plane $\alpha$ through $p$ and $C$ where the projection $\pi_\alpha(p)$ of $p$ is not in the disk $D$, i.e. $\|\pi_\alpha(p) - \pi_\alpha(O)\| = l \ge R$, a contradiction.


*It remains to show a contradiction for the case $l \le R$. ...
Better Argument:
One can conclude that $M \subset B_R = \bigcap_{\alpha: \text{ 2-d plane}} \pi_\alpha^{-1}(\pi_\alpha(M))$, where $B_R$ is the closed ball of radius $R$. Since $M$ is a 2d-manifold then necessarily $\partial B_R \subset M$. Otherwise $\exists p(\partial B_R \ni p \notin M)$ and a plane $\alpha$ through the center and $p$ and the $\pi_\alpha(M)$ is not a closed disk.
Since $M$ is, as $\partial B_R$, connected and 2-d, then with by the fact $\partial B_R \subset M$ there "must" be some theorem from topology that concludes that they are same.
A: Note that the question was raised on mathoverflow and an answer was validated, although somewhat terse and with hypotheses less precise than here.
https://mathoverflow.net/questions/39127/is-the-sphere-the-only-surface-with-circular-projections-or-can-we-deduce-a-sp
