1. There are n labeled points in the plane in general position, no 3 on a line. Draw a
directed line from the lower numbered point Pi to the higher numbered point Pj, j>i. Some of the
other points lie strictly to the left of the line, some strictly to its right. Let Eij be the # of
points on the left minus the # of points on the right. Let E = Sum(n >= j > i >= 1)Eij. How should
the points be arranged and labeled to maximize or minimize E (or make it exactly zero, if possible)?
Can E=0 for any odd n? (No for n=3.) There are papers on the number of lines which cut a point set
in half but this is different.
2. There are n points in the plane in general position: no 3 on a line, no 4 on a circle.
There are n(n-1)(n-2)/6 circles through all triplets of points. Each circle Ci, 1 <= i <=
n(n-1)(n-2)/6, strictly excludes Ti points and strictly encloses n-3-Ti. How should the n points be
arranged to maximize or minimize Sum(i) of Ti, and what are the maximum and minimum? For about 6
quick trials with 5 points, I get exclusion totals over all 10 circles of only 10, 11, and 12. Is it
always true that the total number of excluded points is at least equal to the number of circles?
3. Put n points are on a sphere and disallow any two being polar opposites, and define the
exteriors of all circles as larger than their interiors, and ask the same question.
4. Are there arrangements where some points are always enclosed or excluded? What is the
distribution of points enclosed or excluded, over all circles, and what property of the arrangement
does this depend on?
5. The same question can be asked of plane squares and ellipses through the appropriate
number of points.
Conjectures? Proofs?
Steve Gray
Writing Eij as sum_{k \neq i,j} {1, if Pk is left of Pi->Pj; else, -1},
we can rewrite E as sum_{i<j, k \neq i,j} {1, if Pk is left of Pi->Pj;
else -1}. In this formulation, each 3-element subset {a,b,c} of {1..n},
with a < b < c, contributes exactly three times to this sum: i = a, j =
b, k = c; i = a, j = c, k = b; and i = b, j = c, k = a.
If Pc is left of Pa->Pb, then this triplet contributes +1 to the sum
for E, getting +1 contributions when i = a, j = b, k = c, and i = b, j
= c, k = a, and a -1 for i = a, j = c, k = b. Call such a triplet
{a,b,c} positively oriented. Similarly if Pc is right of Pa->Pb, the
contribution of this triplet is -1 to the sum for E, and we can call
such a triplet negatively oriented.
With this in hand, we can write E as
E = sum_{{a,b,c} \in S(n,3)} {+1, if {a,b,c} is positively oriented
{-1, if {a,b,c} is negatively oriented
Thus it is clear that the maximum possible value for E is n choose 3,
which occurs when all triplets {a,b,c} are positively oriented, and
similarly that the minimum is the negative of n choose 3. These values
are both realizable... evenly distribute the n points along the
semicircle x^2+y^2=1, x>=0. Labelling the points 1 through n,
increasing in a counterclockwise fashion from the bottom yields a
labelled point-set where all triplets are positively oriented, and
reversing this labelling produces the opposite.
As far as E=0, since E is a sum of n choose 3 +/- 1's, it can only
yield a zero sum when n choose 3 is even, which requires n \neq 3 (mod
4). Don't know about the other cases.
Good luck with your investigations!