TrilinearPolarInverse

1. Inscribed with given tripolar

Given a conic (c) and a line L(BO), not intersecting the conic at real points, to construct a triangle (AEF) satisfying the conditions:
(i) AEF is inscribed in the conic (c).
(ii) If P is the pole of line L with respect to (c), then L is the trilinear polar of P with respect to triangle AEF.
From these follows that P is the perspector of the conic (c) with respect to triangle AEF (see TriangleConics.html ) .

Consider an arbitrary point B on line L(BO) and the polar line AC of B with respect to the conic (c). Assume that the polar of B intersects the conic at real points {A,C}. If the points are complex, then work has to be done in the complex plane. Here I restrict myself to the real case.
Take a projective base consisting of {A,B,C,D}. D is on the conic and is determined as follows:
[1] Let G be the intersection of AC with L and P the harmonic conjugate of G with respect to {A,C}.
[2] D is one intersection point of line BP with the conic.
In this projective base the conic has an equation of the form:
[3] y² - xy = 0.
If E is a point on the conic with coordinates (x,y,z), then the intersection point H of BE with line AC has coordinates:
[4] H(x,0,z).
The pole I of line BE, being on line AC and harmonic conjugate to H with respect to {A,C} has coordinates:
[5] I(x,0,-z).
Next construct the harmonic conjugate G' of H with respect to AI. For this express H in the form:
[6] H = rA+sI , which gives H = 2xA - I, so that G' is:
[7] G' = 2xA+I, G'(3x,0,-z).
The triangle has the desired properties when G' coincides with G(1,0,-1), which implies:
[8] z = 3x.
In view of the equation y²-xz=0, this determines the position of E and F:

Later is in accordance with the fact that {E,F} are harmonic conjugate with respect to {B,H}. It is easy to see that for this position of {E,F} the resulting triangle satisfies all requested properties.

2. Structure of the solutions

Given the conic (c) and the line L consider a particular solution A₁B₁C₁ of the previous problem. Using the constructed triangle A₁B₁C₁ we can find all other solutions by taking the orbit of an arbitrary point A of the conic under the group generated by the recycler of A₁B₁C₁ with respect to the conic.
The recycler is defined as the projectivity (f) preserving the conic and mapping the vertices {A₁,B₁,C₁} cyclically to each other (f(A₁)=B₁, f(B₁)=C₁, f(C₁)=A₁).
L is the axis of f and is invariant under the map. P is the pole of the homography and remains fixed under f.
For each point A on the conic the triangle with vertices {A, B=f(A), C=f(B)} is a solution of the problem (1) i.e. it is inscribed in the conic and has perspectrix/perspector the same line/point L/P.

There is a basic homography (g) mapping the vertices {A₀,B₀,C₀} of an equilateral triangle to the vertices {A₁,B₁,C₁} and the center P₀ of the equilateral to the point P. The recycler (f) is conjugate under this map to the 120-degrees rotations of the equilateral i.e. g' = g^-1*f*g (* denoting composition of maps) is a 120-degrees rotation centered at P₀.
All triangles ABC have their sides tangent to a second conic c', which is the image of the incircle of the equilateral A₀B₀C₀ under g. P/L is again the perspector/perspectrix of c' with respect to all triangles ABC.

3. Conic plus line transitivity

The previous discussion shows that the group of projectivities of the plane acts transitively to the pairs (c,L) consisting of a conic (c) and a line (L) not intersecting the conic. In fact given two such pairs (c₁, L₁) and (c₂, L₂) we can construct, as in the previous section, projectivities {g₁, g₂} each mapping the pair (c₀, L₀) to (c_i, L_i). Here c₀ denotes the circle and L₀ the line at infinity. Then the composition g = g₂*(g₁)^-1 gives a projectivity as required.

1. Inscribed with given tripolar

2. Structure of the solutions

3. Conic plus line transitivity

See Also