Anallagmatic cubics, part 3

Section 18

If the pole of a cubic moves on a straight line, the cubic moves through two fixed points. Its reciprocal cubic envelopes a curve which is the antitransverse of an inscribed conic.

The equation of this conic is

√(lx) + √(my) + √(nz) = 0

which is the inscribed conic with perspector (1/l:1/m:1/n).

I have left out a long computation, which I may show in a Mathematica page.

Section 19

We propose to show

All cubics which are circumscribed to ABC and to the antimedial triangle and which pass through the center of gravity of these triangles, is an anallagmatic cubic in the transformation of reciprocal points.

All cubics which are circumscribed to ABC and pass through the centers of inscribed and excribed circles, is an anallagmatic cubic in the transformation of inverse points.

(Remember that "reciprocal" refers to the isotomic conjugate, and "inverse" to the isogonic.)

Consider point M, N with reciprocals M', N'; and P the point of intersection of MM', NN'.

We say that all cubics which pass through 8 fixed points, pass through a 9th point. (this is a famous theorem only quoted here).

One says that all the anallagmatic cubics pass through the 7 points ABCGG_aG_bG_c M (or N) so that it also passes through M' or N'. Turning around P, the points M, M' engernder an anallagmatic cubic with pole (or pivot) P, which must be the mentioned cubic.

Section 20

The converse of the preceding theorem is not true. For example, the cubic

k xyz + l x(yy+zz) + m y(zz+xx) + ... = 0

is anallagmatic (since it is invariant when its points are changed to reciprocals) but does not pass through any of GG_aG_bG_c.

Section 21

The four tangents that can be drawn from a point on a cubic forms an anharmonic pencil of constant ratio.

This theorem, applied over reciprocal cubics, gives the equalities

(a) w₁(w₂ABC) = w₂(GG_aG_bG_c)
(b) w₂(w₁ABC) = w₁(GG_aG_bG_c)

a particular case of which is

The lines which join the same point w1 to the center of gravity and also the vertices of the antimedial triangle are in equal anharmonic ratio as that of the lines which join the reciprocal point w2 to w1 and the point ABC.

The lines which join the same point w1 to the centers of inscribed and excribed circles are in equal anharmonic ratio as that of the lines which join the inverse point w2 to w1 and the point ABC.

Remeber that a parallel theorem exists using the incenters and the inverse point (or isogonal conjugate) as this one.

In the isogonic case we have these results

H(OABC) = O(I_oI_aI_bI_c) , O(HABC) = H(I_oI_aI_bI_c)
G(KABC) = K(I_oI_aI_bI_c) , K(GABC) = G(I_oI_aI_bI_c)
W(W'ABC) = W'(I_oI_aI_bI_c) , W'(WABC) = W(I_oI_aI_bI_c)

where K is the symmedian point, O the circumcenter, K the symmedian point, and W, W' the Brocard points (which will be called eO, wO in The Triangle Book).

There are many others, suych as the following

K(OABC) = O(I_oI_aI_bI_c) , K(HABC) = H(I_oI_aI_bI_c)

from which we derive the conic HOABC, which is the Jerabek hyperbola.

Section 22

Each member of the relations (a) and (b) contains 5 points which define a conic.

This time we separate this theorem into its mirror isotomic and isogonic parts

At each point P in the plane of ABC there are three points such that X(PABC) = P(GGaGbGc) one of which is the reciprocal of P.

At each point P in the plane of ABC there are three points such that X(PABC) = P(IoIaIbIc) one of which is the inverse of P

Constructing the two conics defined by these inequalities, we have

P(GG_aG_bG_c) = P(I_oI_aI_bI_c)

Which meet in the four points P₁, P₂, P₃, P₄ and henceence the conic
X(PABC)

contains both the reciprocal and inverse of P. It is transformed by reciprocal points into the line PP' where P' is the reciprocal of P or is transformed by inverse points into PP" where P" is the inverse of P.

Section 23

Given 4 fixed points on a cubic, and a conic going through them, the chord which joins the two other intersections of the conic with the cubic, passes through a fixed point on the cubic. This point is called the coresidual of the 4 fixes points.

We now apply this to particular points on particular cubics

Section 24

Consider the conics passing through BCGG_a. In this particular case for the cubic Q, the line BC meets the cubic in A, and G_a in A₂, and the line AA₂ meets the cubic again in w₂ so that w₂ is the coresidual.

Condidering BG, CG_a which meet at G_b, resulting in the tangency of w₂G, w₂G_a, w₂G_b, w₂G_c at G and its harmonic associates.

Similarly w₁ is the coresidual for Q' and the points BCGG_a.

Section 25

Aw₁ is tangent at A to the cubic Q and BC meets the cubic at A₂. Hence w₂ is again the coresidual of ABCw₁.

The conics ABCw₁G, ABCw₁G_a,... are tangent to Q at the centroid and its harmonic associates.

Section 26

The lines BC, AG meet the cubic Q again at A₂ and G_a. The coresidual of the system of 4 points ABCG is the point on Q that is the concurrence of A₂G_a, B₂G_b, C₂G_c. Since A₂G_a is parallel to Aw₁, its coresidual is the antimedial w₁, or w₁ with respect to the antimedial triangle.

We then have this theorem

When a triangle is inscribed in ABC is in perspective to ABC, it is again in perspective to the antimedial triangle, and the second center of perspective is the antimedal of the reciprocal of the first.

When a triangle is inscribed in ABC is in perspective to ABC, it is again in perspective to the antisupplementary triangle, and the second center of perspective is the antisupplementary to the inverse of the first.

If, in the preceeding, one substitutes for G one of the points G_a, G_b, G_c, one finds that the triangle A₂B₂C₂ is perspective with each of these triangles GG_bG_c, G_cGG_a, G_bG_aG, and the these three centers of perspective are on the lines w₁G_a, w₁G_b, w₁G_c.

A₂B₂C₂ is inscribed in ABC and perspective to ABC, and is in perspective to 4 other triangles; and the centers of perspective form a system homologic with the system GG_aG_bG_c, the pole of homology being the reciprocal of the perspector of the triangles ABC and A₂B₂C₂.

A2B2C2 is inscribed in ABC and perspective to ABC, and is in perspective to 4 other triangles; and the centers of perspective form a system homologic with the system GGaGbGc, the pole of homology being the reciprocal of the perspector of the triangles ABC and A2B2C2.

A2B2C2 is inscribed in ABC and perspective to ABC, and is in perspective to 4 other triangles; and the centers of perspective form a system homologic with the system IoIaIbIc, the pole of homology being the inverse of the perspector of the triangles ABC and A2B2C2.

Hence:
1. The lines which join the vertices of the median triangle to the centers of the inscribed or exinscribed circles, meet three by three on the lines drawn form the symmedian point to these same centers.

2. The lines which join the vertices of the orthic triangle to the centers of the inscribed and exinscribed circles, meet three by three on the lines from the circumcenter to the centers of these same circles.

3. The lines which join the points of contact of the inscribed circle with the sides of ABC to the centers of the inscribed and exinscribed circles, meet three by three on the lines from the isogonic conjugate of the Gergonne point to the centers of these same circles.

--- on to part 4 --