Needle probing for a convex body 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 such
  needle probe rays 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.
 A: If $V \gt 1/2$ then there must be some pair of points $(x,y,z)$ and $(-x,-y,-z)$ contained in the region, and the convex hull of these two points contains the origin. This generalizes to show that $3$ needles connecting the centers of opposite faces (the axes) suffice for $V\gt 1/4$: 
Consider the orbits of points under the volume-preserving action of changing signs. Generic points $(x,y,z)$ have orbits of size $8$: $(\pm x, \pm y, \pm z)$. If $V \gt 1/4$ then there must be some orbit so that $3/8$ of the points are contained in the region. Given any $3$ of these points, $2$ must have different signs in at least $2$ coordinates since if the first is $(x,y,z)$ and the second is $(x,y,-z)$, then the third point can't be only a single sign change away from both. If the signs differ in $2$ coordinates, suppose without loss of generality these are $(x,y,z)$ and $(x,-y,-z)$. The convex hull of these points contains the midpoint $(x,0,0)$ which is on the $x$-axis. If the points differ in $3$ signs, then their midpoint is $(0,0,0)$ on all axes. 
There is a limit to how much this generalizes, but perhaps the full action of the cubic symmetries will restrict the volumes of regions which avoid the diagonals, too.
