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 X[1,0,0],Y[0,1,0],Z[0,0,1],U[1,1,1]. 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.