Alteration: Partially solving IMO 2014/6
by yayups, Jul 7, 2017, 5:39 AM
We will use the method of alteration to give a partial solution to IMO 2014/6.
Solution
Remark: We can actually improve the constant to
without too much work. The key idea is to differentiate between triangular regions and non-triangular regions. The details are left to the reader.
IMO 2014/6 (partial problem) wrote:
Say we have
lines in the plane in general position, i.e. no two lines are parallel, and no three concur. Show that we can color at least
of the lines blue such that there is no finite region that has complete blue boundary.


We will use the method of alteration. The first step is to randomly color the lines blue, i.e. color a line blue with probability
(
will depend on
). Note that the expected number of blue lines is
. Now, we will remove blue lines in a deterministic way such that there are no more blue regions (this is the alteration part).
We will now find the expected number of blue regions. The first step is to count the number of total regions. The way we do this is consider a graph whose vertices are the intersection points. Note that we then have
vertices. The edges will be the finite line segments between adjacent vertices on a line. Then, there are
edges (since each line has
edges). Therefore, by Euler's formula we have that
, so
. Now, any region is picked with probability at most
since there are at least
lines bounding it, and
, so by linearity of expectation, the expected number of blue regions is at most
.
Now, the idea is to pick
. Now, for each blue region, we remove one line, so we remove at most
. Therefore, the expected number of blue lines left is at least
. This means that there is some coloring with at least
blue lines that has no blue regions since the expected number of blue lines in a properly chosen such coloring is
.




We will now find the expected number of blue regions. The first step is to count the number of total regions. The way we do this is consider a graph whose vertices are the intersection points. Note that we then have









Now, the idea is to pick





Remark: We can actually improve the constant to

This post has been edited 2 times. Last edited by yayups, Jul 9, 2017, 8:29 PM