Suppose $B$ is an axis cube aligned in $\mathbb{R}^3$, i.e. the vertices are $0, e_1, e_2, e_3, e_1 + e_2, e_1 + e_3, e_2 + e_3, e_1 + e_2 + e_3$ where $e_1 = (1, 0, 0)$. Suppose $C$ is an infinite cylinder that is parametrized as $$p(\alpha, t)= c_0 + t b_0 + r\cos(\alpha) b_1 + r\sin(\alpha) b_2,$$ where $c_0$ is some base offset of the cylinder, $b_0$ is the main axis of the cylinder, and $b_1, b_2$ are chosen so that together with $b_0$ they form an orthonormal basis of $\mathbb{R}^3$. Here $t \in \mathbb{R}$ and for simplicity we will assume that $ \alpha \in [0, 2\pi)$. The cylinder could have any $c_0 \in \mathbb{R}$, $b_0 \in \mathbb{R}^3$ and radius $r>0$.
I've thought of the following hypothesis that I would like to prove or find a counter-example to:
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The gist of the hypothesis, informally - I claim that there exists a unique $\alpha_0$, such that the segment of the cylinder $s: t \mapsto p(\alpha_0, t)$ is in some sense the "closest" to the cube along all such segments for all fixed $t$ - at any fixed $t$, it's closer to the cube than the other segment.
Alternatively, we could consider the cylinder at any height fixing the $t'$, which forms a circle $\alpha \mapsto p(\alpha, t')$. I claim the $\alpha$ that minimizes the distance between this circle to the cube will be the same, regardless of $t'$.
More formally, suppose the cylinder does not overlap the cube and suppose for some $t^* \in \mathbb{R}$ there is a unique $\alpha^* \in [0, 2\pi)$ that satisfies $p(\alpha^*, t^*)$ is the closest point to $B$ along the circle $\alpha \mapsto p(\alpha, t^*)$. Then for any $t, \alpha$ it is true that $$d(p(\alpha^*,t), B) \leq d(p(\alpha, t), B),$$ where $d$ is the distance function, defined as $d(x,S) = \min_{y \in S} \|x-y\|$.
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I've tried various things, but nothing has worked so far. I think it could be true for convex objects in general.