the five point theorem

It is well-known that if A,B,C are non-collinear points, then there is a
unique circle through the points. It is reasonable to ask how many points
are required to specify a conic. The best way to tackle this is to work in
projective geometry.

Recall that a projective conic C has an equation of the form

F(x,y,z) = ax2+bxy+cy2+fzx+gyz+hz2 = 0.    
A point on C gives an equation linear in the coefficients of F. for example,
if U[1,1,1] lies on C, then a+b+c+f+g+h = 0.

Observe that, for a non-zero k, kF defines the same conic as F, so we may
guess that five p-points should be sufficient to determine a conic. Given
five p-points, we get five equations in the six coefficients a,b,c,f,g,h. These
are consistent since they have a solution a=b=c=f=g=h=0, though this does
not give a conic. Since there are five equations in six variables, the general
solution will involve at least one parameter. To prove that there is a single
locus, we need to show that there is exactly one parameter, so that the
general solution has the form {k(a,b,c,f,g,h) : k real}.

Of course, we need to check that the solution does not yield a degenerate
locus. Since a degenerate locus with several points consists of at most two
p-lines, at least three of the five p-points must lie on one p-line. Thus, if
we choose five p-points with no three collinear, then the resultant locus
will be a (non-degenerate) conic.

Our proof uses the Fundamental Theorem of Projective Geometry to reduce
the problem to the case where four of the p-points are the standard p-points

The Five Point Theorem

If A,B,C,D,E are p-points, with no three collinear, then there is a unique
conic through all five p-points.


The CabriJava applet below illustrates the theorem. You can move any of
the points A,B,C,D,E and see the conic. With a steady hand, you may be able
to drag E onto the segment and see the locus degenerate into a line pair.
This is easier if you make AB hoizontal.

thre three point theorem