Given a set of straight lines, let each intersection define a vertex. Note that the lines must extend indefinitely, and cannot stop at a vertex (thus forcing every vertex to have even degree).

I aim to prove something along the lines of:

Suppose that there are $n$ sets of parallel lines, each with a distinct gradient. Then every vertex has degree $6$ if and only if there are $3$ sets of equally spaced parallel lines.

Informally, I'm trying to show that there cannot be any configuration of lines such that exactly three lines meet at every intersection, unless the lines "look like" the following figure (only a portion of the lines are shown).

enter image description here

Of course, there is an infinite number of lines in this case.

We can dispose of the case with two sets of parallel lines since no vertex has three lines passing through it. Analysis of small cases doesn't lead me to a method that I can use in general, since (after disposing of the $n=2$ case) I want to prove that if I have more than $3$ distinct gradients, there is always a point of intersection with exactly two lines passing through it.

  • $\begingroup$ @MishaLavrov We are not, but while your link is close there are still some intersections of only two lines. It is required that every intersection that arises has exactly three lines passing through it. $\endgroup$ – Bill Wallis Aug 22 '18 at 16:06

I have assumed in this proof that the grid points don't have an accumulation point. Proof is by contradiction. Suppose there are at least four gradients.

First we show that there are four grid points $A,B,C,D$ with the property that $AB,AC,AD,BC$ are all grid lines and $D$ is on the line segment connecting $B$ to $C$. Since there are at least three gradients, one can find grid points $A,B,C$ such that $AB,AC,BC$ are grid lines. Among all such triangles, choose the one with the smallest area. The fourth gradient must intersect all extended sides of this triangle. Without loss of generality, suppose a grid line parallel to the fourth gradient intersects the extended sides of the triangle in $X,Y,Z$ such that $C-A-X$, $B-A-Y$, and $B-C-Z$. Degree of $Y$ is 6, so there must exist a grid line through $Y$. It cannot intersect $AC$ (this would give a smaller triangle that $\triangle ABC$). So it either intersects $XA$ or $CZ$ at $E$. If it intersects $XA$, then $Y,X,E,A$ is the configuration we wanted. If it intersects $CZ$, then $Y,B,E,Z$ is the configuration we wanted.

Suppose among all such configurations of points in the grid, we have chosen $A,B,C,D$ such that $\triangle ABC$ has the least area. The third grid line through $D$ must intersect either $AB$ or $AC$ at a point $E$. In either case, we have a new configuration (either $A,D,C,E$ or $A,B,D,E$) with a smaller area triangle. This is a contradiction and so there are three gradients.

To show the lines in each gradient class are equidistant, again take the smallest area triangle $\triangle ABC$. Clearly the triangular tiling of the plane generated by $\triangle ABC$ gives rise to equidistant group of parallel lines. If there was any other line parallel to a gradient added to the group, it will create smaller area triangles.

  • $\begingroup$ This is exactly what I was after, thank you. I have just two questions: 1. Does $D$ lie between the points $B$ and $C$ (so that $AD$ lies in the interior of the triangle $\triangle ABC$)? 2. Is the notation $C - A - X$ used to indicate that $C, A, X$ are collinear? $\endgroup$ – Bill Wallis Aug 22 '18 at 15:36
  • $\begingroup$ C-A-X means A is between C and X and yes they are collinear. Also B-D-C. $\endgroup$ – Marco Aug 22 '18 at 15:38
  • $\begingroup$ Perfect, thank you so much! $\endgroup$ – Bill Wallis Aug 22 '18 at 15:41

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