wpt@princeton.UUCP (William Thurston) (11/06/86)
There is an old problem, first posed by Sylvester in the 1890's and
solved in the late 1940's, concerning lines and points in the plane:
1. Let F be a finite set of points in the plane, such that every
line which passes through at least two points of F passes
through at least three. Is F contained in a line?
As evidenced by the gap between the date it was posed and the date it
was first solved, it is not as easy as it might first sound.
However, there is a neat simple solution, as we rediscovered at lunch yesterday.
I think lots more people have heard the problem than have heard the solution,
so I'm posting it. There are some related unsolved problems, which
I'll post later.
The problem has a dual:
2. Let f be a finite set of lines in the plane, such that every
intersection point of two lines from f is an intersection
point of three or more. Show that the lines of f meet in
a single point.
Problems 1 and 2 are equivalent, by a standard construction:
think of the plane as the plane z=1 in 3-space {(x,y,z)}. Each line
determines a unique plane through the origin, and each point determines
a line through the origin. Given the configuration F of points in the
plane, you get a configuration of F' of lines in space. We may assume
that the z-axis is not in F', by moving the plane z=1 sideways a
bit if necesary. Let f' be the set of perpendiculars at the origin to
members of F', and let f be the intersections with the plane z=1.
Collinearity for a set of elements of F is equivalent to the corresponding
set of f meeting at a point.
Solution to Problem 2:
Consider the great circles where planes of f' intersects the unit sphere.
These create a subdivision of the sphere into spherical polygons.
If not all elements of f meet in a single point, then the each polygon has
three or more sides. Since at least three lines cross at every vertex,
the number of vertices is not more than 1/6 the number of ends of edges,
or 1/3 the number of edges. Similarly, if the lines do not all meet at a
single point, then the number of cells is not more than 1/3 the number of
sides of edges, or 2/3 the number of edges. But this contradicts Euler's
formula for the sphere, which says
|vertices| - |edges| + |2-dimensional cells| = 2 > 0
(where |vertices| is the number of vertices, etc.)
Bill Thurston princeton!wpt Princeton math department.