proof a was to dualize, then calculate the Euler characteristic.
proof b: Among all triples of non-collinear points, let (p, q, r) be a
triple which minimizes the distance from p to the line through q and r.
Claim: the line through q and r contains no other point s. For
if q, r, and a point s are all on a single line, then two of the
three --- say, q and s in that order --- are on one side of the foot
of the perpendicular from p. Then (q, p, s) has a smaller
distance.
proof c: Transform projectively so that one point x is at infinity,
and all lines through it are horizontal lines. Among all non-horizontal
lines determined by two points in the set, let l be one of minimal
slope. Then if l contains three points p, q and r in that order,
the line xq cannot meet any other points.
2. There have been many generalizations of this. Kelly and Moser,
Canadian J. Math 16(1958) showed that for a set of n points, there are
at least 3n/7 determined by a pair of the points not meeting any other
point. Beck (Combinatorica 3(3-4), 1983, pp.281--297 showed that there
is a constant c such that if no line contains more than n-k points,
the total number of lines is at least ckn.
3. Here is the unsolved problem: Given a subset S of the plane, a
beam detector for S is a curve or union of curves D such that every straight
line which intersects S also intersects D.
Claim: If S is a convex closed curve in the plane, then the
minimum length for a beam detector for S is at least half the length of S.
Proof: there is a canonical measure on the set of lines in the plane: in
local coordinates given by (t, theta) where t is the intersection with a
line L, and (0 < theta < pi) is the angle, the measure is d t d theta.
This measure does not depend on L. The total measure of lines intersecting
a curve D is at most pi(length D). The total measure of lines
intersecting S is pi/2 length(S), since a.e. line which hits S hits twice.
QUESTION: Is there a lower bound strictly greater than pi
for the length of a beam detector for the unit circle?
(There are examples of beam detectors for the unit circle which have length
less than 2 pi. For instance, there is a ``bow-and-arrow'' configuration
inscribing the circle in a square: use a quarter of the circle, extended
by two half-sides of the square, together with half of a main diagonal of
the square.)
Any beam detector D for the circle
with length near pi would have to be mainly inside the
square, and it would have to have the property that most lines which
intersect D intersect it just once. Could D consist of a lot of tiny
short arcs, so that when you stand on any of the arcs and look at
all the others, they all tend to line up as in an orchard, but when
you view D from outside the circle, you only see trees?
Bill Thurston princeton!wpt