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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.
Proof
By the Fundamental Theorem, there is a projective transformation t which
maps A,B,C,D to X,Y,Z,U. Suppose that t(E) = W[u,v,w].
Observe that, since t preserves collinearity, no three of X,Y,Z,U,W can be
collinear. In particular, W is not on YZ or UX.
Let F(x,y,z) = ax2+bxy+cy2+fzx+gyz+hz2, and let C be the locus with
equation F(x,y,z) = 0.
The p-point X lies on C if and only if a = 0.
Likewise, Y, Z lie on C if and only if c = 0, h = 0.
Thus, F(x,y,z) must be of the form bxy+fzx+gyz.
The p-points U, W lie on C if and only if b+f+g = 0, buv+fwu+gvw = 0.
These two equations in b,f,g will have a one-parameter solution if nad only if
the equations are independent, i.e. the second is not a multiple of the first.
The second is a multiple of the first if and only if uv = wu = vw = 0.
These imply that u(v-w) = 0, so that u = 0 or v = w.
If u = 0, then W lies on YZ - the p-line x = 0.
Otherwise, v = w, so W lies on UX - the p-line y = z.
By an earlier remark, neither of these is true.
Thus, the p-points X,Y,Z,U,W lie on a unique conic C.
Applying the projective transformation t-1, we see that
t-1(C) is the unique
conic through A,B,C,D,E.
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