(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

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

Take a

[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] y

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 y

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.

The

L is the

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 {A

All triangles ABC have their sides tangent to a second conic c', which is the image of the incircle of the equilateral A

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TrilinearProjectivity.html

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