Suppose there is an unknown closed convex body $K$ of volume vol$(K) = V$ inside the unit cube $[-\frac{1}{2}, \frac{1}{2}]^d$ in $\mathbb{R}^d$. You are permitted to probe with a (one-dimensional) ray $r$, which detects whether $r$ includes some point of $K$, or instead if $r \cap K = \varnothing$. My question is:

Q1. What is the fewest number of suchneedle proberays that guarantee hitting $K$, as a function of its volume $V$?

I am especially interested in $\mathbb{R}^3$.
For example, for $V \ge \frac{1}{2}$,
one ray through the origin parallel to an axis suffices.
Here such a ray touches the boundary of a body $K$ with
vol$(K)=\frac{1}{2}$:

To be specific:

Q2. How many needle probes suffice for $V \ge \frac{1}{4}$ in $\mathbb{R}^3$?

Four parallel rays in a grid pattern do not suffice, but nine do suffice (I believe):

Q3. In $\mathbb{R}^3$, is it sometimes more efficient to use nonparallel rays?

This feels like a classic problem, but I am not finding literature. Thanks for pointers or ideas!

**Update 1** (*28May13*). Here is Yoav Kallus' example showing that dropping the middle of the $9$ points of
the grid permits a triangle with area larger than $\frac{1}{4}$ to avoid detection:

**Update 2** (*30May13*). **Q2** and **Q3** are now answered by Benjamin Dickman and Douglas Zare: for $V=\frac{1}{4}$, three needles suffice, and more are needed for parallel needles. **Q1**, in its full generality, is difficult, and so I've added the "open problem" tag.

Q2, those three needles suffice. (Clearly it is alsonecessaryto have at least three.) This should indicate yourQ3is answered in the affirmative. SurelyQ1is a bit tougher. $\endgroup$ – Benjamin Dickman May 28 '13 at 0:517more comments