Let $ABC$ be a closed equilateral triangle (including its edges and vertices).

For each edge of $ABC$, place a red line orthogonal to it. Let $r^{AB}$ be the red line orthogonal to edge $AB$. Otherwise, placement of the three red lines is arbitrary: each may pass through the interior of $ABC$, just intersect it at a vertex, or not intersect it at all.

Within $ABC$, place blue points, enumerated by $1, 2, \ldots, n$, such that the (blue) line passing through any two such points runs parallel to an edge. Denote a line parallel to $AB$ by $b^{AB}_i$. Thus, each $b^{AB}_i$ intersects $r^{AB}$ orthogonally (possibly outside $ABC$).

Each red line, $r^{AB}$, partitions each (if any) of its corresponding lines, $b^{AB}_i$, into three elements: the intersection itself, the segment of $b^{AB}_i$ to its 'left', and the segment to its 'right'. For each $b^{AB}_i$, at most one blue point is allowed to lie in each of these elements. (Thus, if $r^{AB}$ and $b^{AB}_i$ intersect outside $ABC$, only one blue point may lie along $b^{AB}_i $inside $ABC$.)

What is the maximum value of $n$, the number of blue points?

I have an ugly, lengthy proof that it is $n=3$, but suspect that there must be a clean combinatorial result for which this is a special case.

  • 1
    $\begingroup$ I'm having trouble seeing what's going on. It seems to me that if there is at least one red line, the maximum is trivially $3$, since that red line can only have one blue point on its left side, one on its right side, and on on the red line itself. (EDIT: Ahh, or there may be any number of points on the line itself, is that it?) If there are no red lines, can't there be any number of blue points? $\endgroup$ – Brian Tung Dec 8 '17 at 18:55
  • $\begingroup$ Also, what is the purpose of the blue lines? If the blue points are discretionary, can't they be placed arbitrarily within the triangle? $\endgroup$ – Brian Tung Dec 8 '17 at 18:57
  • $\begingroup$ Thank you @BrianTung. Yes, any number of blue points may lie <i>along</i> a red line. Thanks also for catching my ambiguity about "no red lines", which I've now clarified in the problem description. $\endgroup$ – Colin Rowat Dec 8 '17 at 19:18
  • $\begingroup$ @BrianTung: I think with one red line you can also have arbitrarily many points, as your edit shows. As long as one is not forced to put blue points at every intersection of blue lines, it works. I also think this should be an answer. $\endgroup$ – Ross Millikan Dec 8 '17 at 19:19
  • $\begingroup$ @RossMillikan: No longer. Now, if any of the three red lines is not placed, the entire triangle (including its interior) is considered to be on one side of it. That means, essentially, that one has to place all three red lines in order to achieve the maximum, since only one point can be on one side of any red line. I still don't see the point of the blue lines. Is there any sense in which the placement of the blue points is not arbitrary inside the triangle? $\endgroup$ – Brian Tung Dec 8 '17 at 19:20

There are two possible configurations of three blue points:

  1. collinear, all lying on a $b^{AB}_i$, with $r^{AB}$ running through the middle point. It is not possible to place a fourth blue point on $b^{AB}_i$ as that would violate the 'one blue point per element' requirement. Try, instead, to place a fourth blue point elsewhere in the triangle. Join it by straight lines to each of the collinear blue points: these three new lines are neither parallel to each other (as they intersect at the fourth point) nor to the original line (as each connects a point on that line to one off of it). By the 'any two such points' requirement, each of these four non-parallel lines must be parallel to an edge of the triangle, a contradiction.

  2. non-collinear, not all lying on a $b^{AB}_i$. In this case, by the 'any two such points' requirement, they form an equilateral triangle with edges parallel to those of $ABC$. (For each $b^{AB}_i$ created, $r^{AB}$ intersects it at some point in the closed interval between its end points.) Again, try to place a fourth point. If it is collinear with two existing points, we return to the case above. Otherwise, the lines connecting it to the existing three points are not all parallel to an edge of $ABC$, a contradiction.

Thus, while $n=3$ is feasible, $n > 3$ is not.


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