If there are $n$ straight lines in a plane, no two of which are parallel, and no three pass through the same point. Their points of intersection are joined.

How could we prove that the maximum number of fresh lines thus obtained is given by $$ \frac{n(n-1)(n-2)(n-3)}8$$


There are $\binom{n}{2}$ ways of choosing one point of intersection (pick two distinct lines from the set of $n$); there are $\binom{n-2}{2}$ ways of picking the second point of intersection to determine the new line (pick two distinct lines from among the remaining $n-2$). The two points uniquely determine a new line. But you've counted each line twice, since the order in which you select the two points does not matter. So the total number should be $$\frac{1}{2}\binom{n}{2}\binom{n-2}{2}.$$

None of those lines can be a line we already have, because that would require both intersection points to lie on the same line: if they do, then both $L_1\cap L_2$ and $L_3\cap L_4$ lie in some $L$ from our original set. If $L\neq L_1$ and $L\neq L_2$, then we get that $L$, $L_1$, and $L_2$ intersect a point, which is impossible; but if $L=L_1$ (or $L_2$), then $L_3\cap L_4$ lies in $L_1$, again giving three lines that meet at a point. So all of the lines counted above are "fresh lines".

However, some lines may be counted more than once by this process, even after taking into account the lack of order. For example, take $y=x$, $y=-x$, $y=2x+2$, $y=-2x+2$, $y=3x+3$, $y=-3x+3$. No two are parallel; no three intersect at the same point ($y=x$ intersects $y=2x+2$ at a point with $x=-2$, and $y=3x+3$ at a point with $x=-\frac{3}{2}$; intersects $y=-2x+2$ at a point with $x=2$, and $y=-3x+3$ at a point with $x=\frac{3}{2}$; $y=2x+2$ intersects $y=3x+3$ at a point with $x=-1$; and $y=2x+2$ intersects $y=-3x+3$ at a point with $x=\frac{1}{5}$; etc). But if you pick $L_1: y=x$, $L_2:y=-x$ (so you get $(0,0)$) and you pick $L_3:y=2x+2$, $L_4:y=-2x+2$ (so you get $(0,2)$) you'll get the same line as if you pick $L_3:y=3x+3$ and $L_4:y=-3x+3$ (which yields $(0,3)$).

  • $\begingroup$ Last paragraph was mangled; fixed now. $\endgroup$ – Arturo Magidin Feb 29 '12 at 21:28
  • 1
    $\begingroup$ Nice solution Arturo Magidin... $\endgroup$ – juantheron Jul 28 '16 at 9:48

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