Here’s a nifty little math puzzle I came up with recently.

**Preliminary Information**

There are three sets of parallel lines on a 2D surface (plane). Each set contains an infinite number of lines, and each line is infinite.

The parallel lines in each set are spaced a constant perpendicular distance from each other. For the first set (blue), this constant distance is d1. For the second set (green), the distance is d2. For the third set (orange), the distance is d3. Also, the angular orientation of the second and third sets with respect to the first set is given by θ1 and θ2. These variables are shown in the “zoomed” illustration below.

Whenever one line crosses over another line, this is an intersection. When three lines cross at a single point, *this is considered ***three** intersections; line 1 intersects line 2, line two intersects line 3, and line 3 intersects line 1.

Note that if θ1 and θ2 equal each other, lines from sets 2 and 3 will be parallel and will either not intersect each other or will intersect each other at every point along their lines. If θ1 and θ2 equal 0**°** or 180**°**, their sets will be parallel to set 1, and we will have the same issue. Thus, for the purposes of this puzzle, **assume that 0****°** < θ1 < θ2 < 180°.

Also, assume that** d1, d2, d3 ≠ 0**.

**My puzzle is this:**

Find an expression that gives the average number of intersections per some area, A, for all possible cases of d1, d2, d3, θ1, and θ2. In other words, if area (A) is known, I should be able to plug it into the expression to get the average number of points in that area.

**A Truth ****(possibly helpful, possibly not)**:

For any finite area, A, a finite number of intersections exists. As this area increases, the average number of intersections within that area also increases. When area (and thus the number of intersections) approaches infinity, the ratio of intersections to area converges to the average value of intersections per area. This value is the solution to this puzzle.

If you need clarification, feel free to ask in the comments below.

Good luck! :)

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