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More precisely, in a regular sphere packing, either the HCP or FCC lattice packing, does there exist a line $L$ disjoint from every sphere, i.e., not touching any sphere? If so, one could "look through" the packing along the line-of-sight $L$.

Image from a (much!) earlier posting:

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  • $\begingroup$ Originally posted on MSE, but garnered no interest over a week. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 29 at 23:20
  • $\begingroup$ What if the cannonballs are reflective? Is there a ray starting in one face of the pyramid and ending in another? $\endgroup$
    – Carl-Fredrik Nyberg Brodda
    Commented Nov 30 at 1:37
  • $\begingroup$ @Carl-FredrikNybergBrodda: That was the suggestion that led to that 14 year-old Christmas-tree posting. The answer there does not resolve your more fundamental question. $\endgroup$
    – Joseph O'Rourke
    Commented Nov 30 at 2:26

1 Answer 1

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Yes. View the FCC packing as a series of stacked square packings, with spheres of unit radii centered at the points $(2a,2b,2\sqrt{2}c)$ and $(2a+1,2b+1,(2c+1)\sqrt{2})$ for all $a,b,c,\in\mathbb Z$:

                                   enter image description here

Then we can take the line $\{(x,1,0.01)\ |\ x\in\mathbb R\}$, which lies just above the contact points of spheres in the $z=0$ layer but isn't high enough on the $z$-axis to risk contacting any other layer.

Here's a diagram of the "holes" one can see through when viewing the packing along the $x$ axis:

enter image description here

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    $\begingroup$ Is this the same packing? This looks on the picture as square-grid based, while the original question was about triangle-grid based. $\endgroup$
    – domotorp
    Commented Nov 30 at 8:33
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    $\begingroup$ @domotorp: It is, but viewed from a different direction. $\endgroup$
    – user2357112
    Commented Nov 30 at 11:21
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    $\begingroup$ @JosephO'Rourke this argument only considers two adjacent layers of the packing, right? If so then it applies equally to FCC and HCP. $\endgroup$
    – Sam Hopkins
    Commented Nov 30 at 16:17
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    $\begingroup$ I suspect a similar construction will work for every known lattice packing in dimension greater than $3$. If we take a lattice packing of spheres of radius $r$ in $\mathbb R^n$ and consider a vector $v$ in the lattice of length $2r$ (so that the corresponding pair of spheres is touching), the projection of the lattice to the orthogonal complement of $v$ will have a density equal to the density of the original packing times the volume of an $n-1$-dimensional sphere of radius $r$ times $2r$ divided by the volume of an $n$-dimensional sphere of radius $r$. $\endgroup$
    – Will Sawin
    Commented Dec 1 at 0:52
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    $\begingroup$ Up to an (approximately) constant factor, this is the density of the original packing times $\sqrt{n}$. As long as the density of the original packing is less than $1$ over a constant times $\sqrt{n}$, the projection will have density less than $1$ and hence can't be a covering. But the known sphere packings have densities that are exponentially small in $n$, so the projection will not be a covering (i.e. you can see through) for large $n$. But since it happens for the best known packings for $n=3$ the "large n" probably starts already at $n=3$. $\endgroup$
    – Will Sawin
    Commented Dec 1 at 0:54

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