Points in the plane Part 11

[jay_shark]

There are 4n points in the plane, no three are colinear . 2n are colored blue and 2n are colored white . Is it possible to divide the plane such that one side of the line contains n blue , n white and the other side contains n blue , n white ?

[jay_shark]

I'm very intrigued by this problem and I'm completely dumbfounded . I've solved similar problem to this but I can't quite figure this one out .

[thylacine]

Turn it around in your mind.

[jay_shark]

what are you talking about ?

[thylacine]

I was dropping a hint. Instead of thinking about the dividing line, think about a perpendicular line, and (orthogonally) project the 4n points onto that line. Now see what happens as you rotate THAT line around.

[Siegmund]

The answer is yes; I am not sure if this is a rigorous enough argument to convince everyone (but hey, when you aren't in college and you aren't writing a journal article, you can get away with that.)

First, consider 2n points of the same colour.

Pick any slope you wish, and there exists a line of that slope that has n points on each side of it (or has n-1 points each side of it and 2 points touching it.)

You may rotate this line through 180 degrees "smoothly" (this is the part that probably needs made more rigorous) - if it is not touching a point you may rotate it freely about any point along its length, until it as about to touch one point; if is touching only one point, rotate it about that point until it touches two points (one on each side of the line), then move it across those two points.

Now, pick one slope, and find a line that divides the blue points, and another line of the same slope that divides the white points. These will be parallel, but probably not coincide.

Rotate both lines smoothly through 180 degrees, as above, keeping them parallel. They must, somewhere in the course of that rotation, run into each other.

At this instant, you have either achieved your goal, or you have a line that touches two points of the same colour and no points of the other colour, and a slight further rotation will put those two points on opposite sides of the line.

[jay_shark]

Hey Siegmund , I don't quite get your solution . How many lines are there in total ? Also , about what point are these lines rotating on ?

I understand that it's possible to have two parallel lines such that there are n points of each color on either side . This will split the plane into three parts and at this point I'm not sure what to do .

Perhaps this is a much more difficult question than I thought .

[bigpooch]

Very nice. This gives it away.

[jay_shark]

I don't quite understand Thylacines hint . If anyone can solve this problem I'll appreciate it .

[bigpooch]

I'm just clarifying thylacine's hint, but there are a few
technical details you may have to show, but I assure you
that it's "not hard".

As stated, take a line L, just about any line that doesn't
intersect any of the 4n points. You can project all these
4n points onto this line L by perpendicular lines. Since no
more than two points can lie on a perpedicular by virtue of
the fact that no three points are colinear, the kinds of
"projected points" (images) can have AT MOST two "sources"
or points (preimages). It's possible that the images are
"nice" in that you can find a point P on the line L that
divides the number of preimages on L evenly, i.e., there are
the same number of preimages on one side of L (with respect
to P) as the other. If not, then, there is a point X that
is the image of two "sources", so you can rotate by a very
small delta radians this line L to produce a line M that
won't have such a "defect". (You can do so because there
are only a finite number of points.) In any case, let Q be
a proper dividing point such that Q is at the midpoint of
the two closest "image points" (in the case that a line L
doesn't divide "well", you take Q to be the "double image"
causing the problem). Thus, Q depends on M.

In any case, label one side of a "proper line" M with
respect to Q and call it the "primary half" of the line.

If you are very lucky and you chose M well, not only will Q
divide the images into 2n sources on each side of Q, but
there will be n blue and n white "preimages" on each half
of the line M. Then you're done.

Suppose you weren't so "lucky". Let b(M) be the number of
blue "sources" or preimages for the primary half of the
line.

Now, if you rotate this line M (think of doing it slowly and
call the line rotated by theta radians M(theta) ),
b(M(theta)) will experience jumps by only one (another
technical detail!), but since b(M(pi)) = 2n - b(M(0)) (when
you rotate through 180 degrees, you get the "opposite") and
since b(M(0)) is NOT n, then there is some angle rho for
which b(M(rho)) is n, which is strictly between b(M(0))
and b(M(pi)) by the fact that b doesn't jump around by more
than 1 at a time.

(I hope this is okay. I must have edited this three times;
it's 2:15 am here...zzz)