Honeycomb Lattice Line Passing Through A plane is tiled with regular hexagons. This tiling is sometimes called the “honeycomb” lattice. Show that if a line passes through two points that are vertices of hexagons in the tiling, then the line passes through infinitely many such points
 A: Suppose we put a coordinate system such that the origin coincides with a hex center. 


*

*We can then find vectors $U$ and $V$ such that any hex center will have
coordinates $mU + nV$ for some integers $m, n$. $U$ and $V$ can for
example be the coordinates of two hexes adjacent to the hex at the
origin and adjacent to each other.

*We can also find integers $m, n$ such that if $A$ and $B$ are
coordinates of hex centers, then $(A - B) = mU + nV$.

*If $p$ is a vertex, then so is $p + mU + nV$ for integers $m$ and $n$.


Notice that the coordinates of a vertex is the average of the three cell centers surrounding it, that is, any vertex $p$ can be written as $\dfrac{A + B + C}{3}$, where $A, B, C$ are the coordinates of the hexes that surround $p$.
Suppose $p$ and $q$ are vertices. We will show that $p + (q - p)3k$ is a vertex.
Write $p = \dfrac{A + B+ C}{3}$ and $q = \dfrac{A' + B' + C'}{3}$. 
Then $p - q = \dfrac{A - A' + B - B' + C - C'}{3}$. 
Since $A$ and $A'$ are faces, there exist integer $a$ and $a'$ such that $(A - A') = aU + a'V$. Similarly, $(B - B') = bU + b'V$ for some integer $b$ and $b'$, and $(C - C') = cU + c'V$ for some integer $c$ and $c'$.
Thus, $p - q = \dfrac{(a + b + c)U + (a' + b'+ c')V}{3}$, and then $(p - q)3k = (a + b + c)kU + (a' + b' + c')kV$.
Now $(a + b + c)k$ and $(a' + b' + c')k$ are both integers, and therefor, $p + (p - q)3k$ is a vertex.
