Consider $9n$ pencils through non-collinear points $p_1, \ldots , p_{9n}$ in $R^2$ each consisting of at most $n$ concurrent lines. Define the intersection $S$ of these pencils to be the set of points which lie on at least one line in each of the $9n$ pencils. Is it true that $|S|$ is $O(n)$?
$\begingroup$
$\endgroup$
4
-
$\begingroup$ I hope you don't mind, I added the arxiv tags. $\endgroup$– Gjergji ZaimiCommented Aug 26, 2011 at 7:45
-
$\begingroup$ As you pointed out my answer was incomplete. I will think about it some more and then un-delete it if I can make it work. $\endgroup$– Gjergji ZaimiCommented Aug 26, 2011 at 11:03
-
1$\begingroup$ Why 9n and not n? $\endgroup$– domotorpCommented Aug 28, 2011 at 10:00
-
$\begingroup$ Sure, $n$ would also be good. There are examples of $n$ pencils each with $n$ lines such that $|S| = 2n$. One way to see this example is as follows: Consider the regular 2n-gon. Let the pencils be centred at points at infinity in the directions joining the midpoints of opposite edges. This is not in the euclidean plane as described but can be modified to fit in it. But in general I don't know if $|S| = O(n)$. I think $9n$ should further restrict the configuration and wonder if there are examples with $|S| \geq cn$ in this case for any constant $c >1$. $\endgroup$– Sukhada FadnavisCommented Aug 28, 2011 at 11:13
Add a comment
|