Number of intersections of $n$ lines Suppose you have $n$ lines in a plane. No two are parallel. Let $W$ be the number of intersections of these $n$ lines. Is it possible that $1 < W < n$?
 A: Well, the lines do intersect in $n(n-1)/2$ points, question is with degeneration (points being congruent) what range of possibilities in distinct points is there.
1 is possible.  All lines through the same point.
Slide one line away from this common point, is going to intersect the other $n-1$ lines in $n-1$ points.
So no.
If you have multiple bundles of lines going through a multiple common points ($p$ through point A $q$ going through point B) again no, on account of $p*q >= p+q$ for all  $p,q >= 2$.  Meaning, the p lines that go through point A will intersect with the q lines that go through the point B in $p*q$ points, leaving with a total of $p*q+2$ intersection points.  If a line passes through A and B, the number of lines will be $p+q+1$, still >= than the number of intersections $p*q+2$ (equality reached for $p=q=1$). 
That means that attempting to decrease the intersections by having multiple degenerate points will also fail.
and so on...
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P.S.  To make it more rigorous ... induction!
Statement: 
"no n nonparallel lines meet in p distinct points $1<p<n$"
Base case:
For 3 lines we have either one intersection point or 3 intersection points. 
Inductive Step: "If true for n-1, true for n"
Assume true for n-1.  Then the number of intersection points for the arrangement is 1 or $>=n-1$.  
If the number of intersection points is 1, then an additional line will add 0 points (if it passes through the existing intersection point), or n-1 points if it does not pass through the common intersection point of the n-1 lines (because it has to intersect all n-1 existing lines, and each point has to be distinct from the common point of the $n-1$ lines why?).  So you end up with either 1 point or n points. 
If the number of intersections is >1, the way to avoid adding (too many) intersection points is by going through pre-existing intersection points. However, any intersection of say q lines, means that these q lines alone intersect the remaining $n-1-q$ lines in $q*(n-1-q)$ points. As $q>=2$, add 1 for the remaining (n-1-q) lines (they have to meet in at least one point) this alone is $>=n$ as soon as $n>4$.  For 4 points, the possibilities are 1 intersection, 4, and 6 intersection points (by direct inspection - homework!).
So bottom line 
if $1<intersectionpoints<n-1$ for n-1 lines 
then $1<intersectionpoints<n$ for n lines.
So the inductive step is proven so the statement is proven.
-this needs serious trimming and clarifying, but to best of my ability, it's technically correct.
A: Consider building up $n$ lines one at a time. Start with one line $\ell_{1}$ and consider a second $\ell_{2}$. If $\ell_{2}$ is not parallel to $\ell_{1}$, they must intersect, so $W=1$. Now consider a third line $\ell_{3}$. You can place it so that it intersects the intersection of $\ell_{1}$ and $\ell_{2}$ (such that $W=1$) or so that it intersects both $\ell_{1}$ and $\ell_{3}$ at separate points (such that $W=3$). If you continue placing lines up to $\ell_{n-1}$ so that they intersect at the same point, you can see that you can either place $\ell_{n}$ such that it passes through the intersection of all of the other lines ($W=1$) or intersects every line at a unique point ($W=n$).
