Prove that if each of $m$ points in one straight line be joined to each of $n$ in a parallel straight line by straight line segments terminating at the points, then excluding the given points, the line segments will intersect $\frac14mn(m-1)(n-1)$ times. Note: The count is of intersections, not of points of intersection. E.g., in the picture below for the case $m=n=3$, each red dot represents one intersection, but the large black dot in the centre represents three intersections, one for each pair of segments intersecting at that point. (I did not bother to include the original straight lines on which the points lie; they’re pretty obvious and just clutter up the picture.)

enter image description here

  • $\begingroup$ i have given the wording exactly as given in the question $\endgroup$ – maths lover Aug 16 '13 at 7:51
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    $\begingroup$ Okay, but I think it is hard to understand, can you write the question in your own words? $\endgroup$ – Dan Brumleve Aug 16 '13 at 7:54
  • $\begingroup$ Are the two straight lines parallel, or do you consider the cases when the straight lines intersect? $\endgroup$ – peterwhy Aug 16 '13 at 8:06
  • $\begingroup$ A suggestion for rewording: paragraph 1, "Let two (parallel?) lines $p$ and $q$ be given. On line $p$ (describe points) are given, and on line $q$ (describe points) are given. Between these points (describe what)." Paragraph 2, "Prove that (use terms from paragraph 2) have (TeX-written formula; mathematics lovers should make an effort to learn at least basics of TeX)." $\endgroup$ – Vedran Šego Aug 16 '13 at 8:45

There's a problem if we allow the lines to intersect, for example:

Problem with intersecting lines

results in no intersections.

If the two lines are parallel, we have a unique intersection for any pair of $2$-element subsets: one from each line. This is illustrated below:

Parallel lines illustration

Conversely, any such intersection occurs between such pairs. The number of such pairs is $$\binom{m}{2}\binom{n}{2},$$ as required.

  • $\begingroup$ How does this take care of the problem that you might have two pairs of four that intersect at the same place? $\endgroup$ – narcissa Aug 19 '13 at 15:45

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