Consider a set of $n$ lines in $\mathbb{R}^3$ all concurrent in a point (let's call it center). Is it always possible to place a finite number of isometric copies of this set of lines in $\mathbb{R}^3$ so that the centers are all distinct and each lie on at least $n+1$ lines ($n$ of which are its own lines, also, two lines that perfectly overlap must still be counted as two), and each line intersects at least $2$ centers (one of which is its own center)?
I asked myself this question and I don't know how to tackle it. Intuitively, I would say that the answer is no. The only idea I came up with is trying to prove that if $n$ is big enough (surely greater than $2$, since the answer for $n=2$ is yes, but the proof of this case isn't very enlightening for the general case) the sets of lines which may be placed so and so has measure zero. However, there are some difficulties with this approach.
It goes without saying that for some sets of lines such a configuration exists. Take for example the cartesian axes. Eight copies might be placed so as to form a cube, with each line containing a side of the cube.
Any idea is very welcome.
For those who might be interested, this is the solution for the analogous problem in the plane (in this case the answer is 'yes'). We prove something stronger. Lines may be substitued with alf-lines originating from their center contained in a half-plane.
Let $l_1,\dots,l_k$ be half-lines originating from a point $P$ and contiained in an open half-plane. Place $P$ on a circle $\gamma$ centered at the origin so that the half-lines all intersect $\gamma$ in another point apart from $P$, call this point $P^1$. Call the intersection points $Q^1_1,\dots, Q^1_k$ in counterclockwise order. Now, let $n\in\mathbb{N}$ be such that, if $P^1,\dots,P^n$ are in counterclockwise order the vertices of a regular $n$-gon inscribed in $\gamma$, $P^2$ comes before (in counterclockwise order) $Q^1_1$. For $i=2,\dots, n$, define $Q^i_1,\dots, Q^i_k$ as the image of the points $Q^1_1,\dots, Q^1_k$ under the rotation that maps $P^1$ to $P^i$. Notice that for $i=1,\dots n$ and $j=1,\dots, k$, if $P^h$ is the vertex of the regular $n$-gon immediately preceeding $P^i_j$ (or it is the point itself), the angles that $P^i_j$ and $Q^h$ form with $Q^h_1,\dots, Q^h_k$ are the same (as they subtend the same arcs).
Yes, this is possible.
A straightforward algorithmic construction should yield a sparse set of fractal points similar to "Cantor Dust" that are aligned in such a way as to satisfy all of your intersection/overlap requirements.
The result won't be pretty but it will fit all requirements with a finite number of points (in a finite volume).
(I'm going to provide an example in $\mathbb{R}^2$ for ease of explanation; however it should be possible to generalize to $\mathbb{R}^3$ without any massive changes.)
Choose an arbitrary ordering $\{\ell_1,\ell_2,\ell_3,\dots \ell_k,\dots \ell_{n-1},\ell_n \}$ for the $n$-many lines crossing through point $P_0$.
Then arbitrarily choose which side of the $P_0$ intersection constitutes a positive direction for each line.
For $k=0$ place the point $P_0$.
For $k=1$ we want to incorporate the $\ell_1$ line. Take all currently-defined points (at this time its only the single $P_0$ point) and copy them in place. Take all the newly generated copies (at this time only a single new point $P_1$) and translate them along their respective $\ell_1$ lines in the positive direction by distance $d_1$.
(This should leave you with points $P_0$ and $P_1$ separated by a line segment of length $d_1$. All line segments connect exactly two points, and all points are adjoined by at least one non-empty line segment along $\ell_1$.)
For $k=2$ we want to incorporate the $\ell_2$ line. Take all currently-defined points ($P_0$ and $P_1$) and copy them in place. Take all the newly generated copies (call them $P_2$ and $P_3$) and translate them along their respective $\ell_2$ lines in the positive direction by distance $d_2$.
(This should leave you with a parallelogram of side lengths of $d_1$ and $d_2$. Both $\ell_1$ line segments connect two points each, both $\ell_2$ line segments connect two points each, and all points are each individually connected to a $\ell_1$ and $\ell_2$ segment.)
For $k=3$ we want to incorporate the $\ell_3$ line. Take all currently-defined points and copy them in place. Take all the newly generated copies and translate them along their respective $\ell_3$ lines in the positive direction by distance $d_3$.
(This should leave you with a parallelogram of side lengths of $d_1$ and $d_2$ with lines of length $d_3$ connecting all corners to the corresponding corners of the parallelogram's copy. Again, all $\ell_1$ line segments connect two points each, all $\ell_2$ line segments connect two points each, all $\ell_3$ line segments connect two points each, and all points are each individually connected to a $\ell_1$, $\ell_2$, and $\ell_3$ segment.)
Then continue the process for $k=4$ through $k=n$.
At each $k$ step the copy process doubles the number of points. Each line segment connecting a point to it's corresponding new copy satisfies the new $\ell_k$ requirement, while all $\ell_{<k}$ requirements were satisfied in previous steps for that point.
"But wait!" you say, "The $k=2$ parallelogram had provably distinct points for any possible lines $\ell_1$ and $\ell_2$ and any choice of lengths $d_1$ and $d_2$... but once we move up to $k=3$ that is no longer the case because, as a counterexample, $\ell_3$ aligned precisely along the parallelogram's diagonal could result in a shared corner if $d_3$ matches the length of the diagonal! So how can you make the claim that the points will be distinct?"
That is quite correct, and the answer is that we don't allow the lengths to be completely arbitrary. Instead, each successive $d_k$ is made so much larger than the previous that it would be impossible for new points to be generated anywhere even remotely close to any points from a previous generation. How? We draw a circle around all points in the $k-1$ step and make sure that $d_k$ is a large enough multiple of $r_{k-1}$ such that overlap becomes impossible.
At the minimum, the radius of $r_k$ (for $k>2$) would need to fit side-by-side copies of $r_{k-1}$ and so $r_k\geq2\cdot r_{k-1}$. But since we're just proving existence we might as well round things up to a nice big $r_k= 4\cdot r_{k-1}$ which clearly holds two $r_{k-1}$ circles with room so spare. If we define $d_k=4\cdot r_{k-1}$ then $d_k=r_k$ and $d_k= 4\cdot d_{k-1}$ follow.
If one sets $d_2=d_1$ and then iteratively applies the above, then a formula of $d_k=4^{k-2}\cdot d_1$ can be derived (for all $k>1$).
With the copy-and-translate process satisfying all connection requirements and the exponentially-growing expansion ruling out overlap, the above proves that it is always possible to generate examples satisfying your requirements in $\mathbb{R}^2$.
Generalizing the problem to $\mathbb{R}^3$ is nearly identical. The only difference being how the maximum diagonal for a parallelepiped depends on $\{d_1,d_2,d_3\}$ which will require care when stitching your choice of initial conditions to the $d_k=4\cdot d_{k-1}$ induction.