Imagine that, somewhere inside an origin-centered, unit-radius sphere $S$ in $\mathbb{R}^3$, sits a convex body $K$ of volume vol$(K)=\alpha (\frac{4}{3} \pi)$, with $\alpha < 1$ the fraction of the volume of $S$. $K$ is inside $S$ at an unknown but fixed location and orientation. My question is: How many line-probes are needed to detect its presence? A line-probe is a line $L$ whose intersection with $K$ includes a point strictly interior to $K$. One might need many probes to certainly detect the presence of a small-volume $K$.
Let $f(k)$ be the volume fraction $\alpha$ such that (a) there is some body $K$ that is not detected by any fixed set of $k$ probes, and (b) every body with vol$(K) > \alpha$ is detectable by $k$ probes.
I believe $f(1)=\frac{1}{2}$: If $K$ fills a hemisphere, it could
"hide" in $S$ from any single probe. But any $K$ with more than half
the volume of $S$ necessarily includes the origin, and so a line
through
the origin would detect it.
It may be that $f(2)=\frac{1}{3}$ by two orthogonal probes that partition $S$
into two spherical caps and the sandwich between, each of
$\frac{1}{3}$
the volume of $S$.
And perhaps $f(3)=\frac{1}{4}$ via three probes through the origin.
But I am uncertain of these values of $f()$. If anyone can hide bodies of larger volumes
from these probes, please let me know!
This feels like a question that was likely considered before; if so, a pointer would be appreciated. Of course, the question generalizes to $\mathbb{R}^d$, with various dimensional probes. In $\mathbb{R}^1$ with point-probes, $f(k)=\frac{1}{k+1}$. Edit: Michael Biro suggests in the comments that the $f(2)$ example above could be generalized to establish that also $f(k)=\frac{1}{k+1}$ in $\mathbb{R}^3$.
Update.
Here is an illustration of Ilya Bogdanov's argument that my 2nd example does not
establish that $f(2)=\frac{1}{3}$:
$\{-1,1\}$
, not$\{0,1\}$
. $\:$ $\endgroup$