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Let $Q\subset \mathbb{R}^3$ denote the cube $[-1,1]\times[-1,1]\times[-1,1]$.

Let $\xi\in\mathbb{S}^2$ be a vector on the unit sphere centered at the origin.

Let $\xi^{\perp}$ denote the linear subspace that is orthogonal to the vector $\xi$.

Denote $Q\mid_{\xi^{\perp}}$ as the projection of $Q$ onto the plane $\xi^{\perp}$.

Question: What shape is $\bigcap_{\xi\in\mathbb{S}^2}Q\mid_{\xi^{\perp}}$?

VShaw
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I don't think you've "mathematized" this correctly. Without even bothering with projecting $Q$, isn't the intersection of all the $\xi^{\perp}$ just the origin?

So you probably want to consider all $SO(3)$ rotations of the cube, and project them all onto a fixed plane, say the $x−y$ plane.

Once you've fixed this issue, the resulting intersection will be rotation invariant, as any element of $SO(3)$ can be composed with a suitable rotation around the $z$-axis. Additionally, rotating and projecting a convex set yields a convex set, and intersecting a bunch of convex sets yields a convex set. That doesn't seem to leave a lot of possibilities (hint, hint).

If you already knew all that, and were just curious about the resulting radius, take a look at the largest sphere you can inscribe in your cube.

JonathanZ
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Another possibility (which may or not be what the OP meant) …

For each given $\xi$, let $E_\xi$ be the Minkowski sum of $Q$ and the infinite line $L_\xi$ through the origin in the direction of $\xi$. Informally, $E_\xi$ is the shape that you get by sweeping $Q$ along $L_\xi$. It’s what CAD folks would call an extruded shape.

Then the intersection of all the $E_\xi$ is something interesting. Clearly it contains the unit sphere (since this is a subset of each $E_\xi$). I’d guess that the intersection is actually equal to the unit sphere. Given a point outside the unit sphere, it seems fairly clear that you can construct an $E_\xi$ that doesn’t contain it.

Something more general might be true: the intersection obtained from any convex body (not just a cube) might be the largest sphere contained within that body.

Edit:
Based on the comment below, the smallest enclosing sphere seems more likely. Willing to put more thought into this if the OP confirms that I’m solving the right problem.

bubba
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    Isn't $Q$ contained inside each $E_{\xi}$? – JonathanZ Nov 16 '23 at 06:56
  • Yes, you’re right. So my statement about the largest enclosed sphere is wrong. The smallest enclosing sphere seems more likely. I didn’t put much thought into this because I wasn’t even sure I was answering the right question. – bubba Nov 16 '23 at 23:58
  • Yeah, I (unsurprisingly) think that my answer is more likely what they intended, so discussing your version probably shouldn't take place here. But yours is none the less more interesting. For instance, if you start off with a 2-d disk centered on the origin, you get the same set back. There seem to be subtleties involving both taking the convex hull and considering the interior, and I feel like this may hang around in my brain for a while. thx. – JonathanZ Nov 17 '23 at 01:09
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    Getting on my computer to rescue you here - the intersection of all $E_\xi$ is the same as the intersection of bounding half-spaces of $Q$. https://math.stackexchange.com/questions/2130746/convex-sets-as-intersection-of-half-spaces Hence if $Q$ is closed and convex, your intersection equals $Q$. – Dustan Levenstein Nov 18 '23 at 02:05