[alogo] Isotomic transformation as a quadratic one

Quadratic transformations are related to families of conics. Given a family of conics, and a point P, all polars of P with respect to the conic-members of the family pass through another point P' = F(P). This defines the quadratic transformation of the family. In practice, P' is constructed as the intersection point of two polars L1, L2 of P with respect to two fixed members of the family.

A particular case of quadratic transformation is the isotomic conjugation Q' = F(Q) with respect to a triangle ABC. This is defined as follows: For each point P consider the intersections (traces) {PA, PB, PC} of the cevians through P with the opposite sides of the triangle. Take then the symmetrics {PA', PB', PC'} with respect to the middles of the corresponding sides. Using Ceva's theorem follows easily that lines {APA', BPB', CPC'} intersect at a point P' which is then defined as the isotomic conjugate of P.

Here we show that the isotomic transformation of ABC coincides with the quadratic transformation of a special family of conics consisting entirely of hyperbolas. The family namely consists of all conics passing through the four points {G,A',B',C'}, where G is the centroid and {A',B',C'} are the vertices of the antiparallel triangle of ABC (created by parallels to the sides from the opposite vertices). All these conics are hyperbolas. Their family is generated by the linear combinations of two degenerate members c1, c2, consisting of the two lines intersecting at A (degenerate conic c1(AA',AB')) and the two lines intersecting at B (degenerate conic c2(BA',BB')).

[0_0] [0_1] [0_2] [0_3]
[1_0] [1_1] [1_2] [1_3]

The proof, modulo the remarks made above, is trivial. Apply the definition of the quadratic transformation by taking the polars of P with respect to c1, c2. For example, the polar L1 = AP' of P with respect to c1(AA',AB') and line L = AP are harmonic conjugates with respect to AA', AB'. Project P, P' on BC to points PA, PA' correspondingly. Since AB' is parallel to BC, points PA, PA' are symmetric with respect to the middle A* of BC (see Harmonic_Bundle.html ). Analogous results are obtained by joining P, P' with the remaining vertices B and C. This completes the proof.
Although the property has a simple proof. The context involves several interesting ideas studied in the references given below.

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