Suppose that C and C' are p-conics, and that
(a) A lies on C, and A' on D, and
(b) P lies inside (resp. outside) C, and P' inside (resp. outside) C'.
Then there are precisely two projective transformations which map
C to C', A to A', and P to P'.
proof
(1) P outside C.
Since the chord RS is constructed using the ideas of polar and pole, the
Let R'S' be the chord of C' associated with A',P'.
If t is as in the statement of the theorem, then the above remarks show
If, conversely, t maps C to C', A to A', and {R,S} to {R',S'}, then, since
We have reduced the problem to that of finding transformations which map
Although there are differences in detail depending on whether P and P' lie
inside or outside their respective p-conics, the basic idea is the same. We
associate with the pair A,P a chord RS of C in such a way that P can be
recovered from A and the chord RS. Likewise, with A',P', we associate the
chord R'S'. We show that any suitable projective transformation must map
RS to R'S' . Since A,R,S lie on C and A',R'S' on C', the three point theorem
can be applied.
From our definition of outside, the polar of P with respect to C cuts C twice.
Let the intersections be R and S. Here, RS does not depend on A.
Suppose that we are given the chord RS, then we recover P as the pole of
the p-line RS.
(2) P inside C.
Let L be the polar of P with respect to C, and T the p-point where L meets
the p-line AP. As P is inside C, L does not meet C, so T is outside C. Thus,
the polar of T with respect to C cuts C twice, at R,S, say. Once again, we
have a chord RS associated with A,P. This time, it does depend on A. Note
that, as T lies on the polar of P, La Hire's theorem shows that P lies on the
polar of T, i.e. on RS. Thus P is the intersection of AT and RS.
Suppose that we are given the chord RS, and the p-point A. We recover T
as the pole of the p-line RS. Then we recover P as the intersection of the
p-lines AT and RS.
invariance theorem shows that, if t is any projective transformation,
then t(RS) will be the chord of t(C) associated with t(A), t(P).
that t(RS) = R'S', i.e. {R,S} maps to {R',S'} as sets.
A, RS determine P, and A', R'S' determine P', t maps P to P'.
C to C', A to A' and {R,S} to {R',S'}. The last condition can be satisfied in
two ways - we can map R to R' and S to S' or R to S' and S to R'. Finally, we
invoke the three points theorem which shows that there is exactly one map
of each type. Thus, we have precisely two suitable transformations.
