Let c be a conic and A a fixed point outside it. Draw from A lines in varying directions intersecting the conic
at points {B,B'}. Let also points {D,C} be the intersections with c of the polar of A with respect to c. The intersection point I of line BD and the tangent tB' to c at B' lies on a conic c'. Conic c' is bitangent to c at points D and C (see BitangentConics.html ).
The proof reduces to a trivial property of equilateral hyperbolas (see HyperbolaRelatedToEquilateral.html ). This is done by defining a homography which maps the conic c to a circle c0 and the point A to a point at
infinity. Such a homography is easily constructed by coordinates-correspondence with respect to self polar triangles, and
taking one of the sides of the autopolar triangle to pass through A.
The previous construction can be used to define a triangle inscribed in a conic c for which the conic has a
given pivot. In fact, assume F is a given point outside a conic and we want to construct a triangle inscribed in
the conic c and having the given point F as the pivot of the conic with respect to this triangle. For this it suffices
to fix a point D on the conic. There is then a unique triangle DD'D'' inscribed in the conic, which then has for
pivot the given point F.In fact, let A be the intersection point of the polar pF of F with the tangent tD of c at D. Draw from A lines intersecting the conic at points BB' and construct the locus-conic (d) of the previous
section. The intersection points {D1,D2} of this conic with the polar pF of F determine through lines {DD1, DD2} the other vertices of the desired triangle. See the file TrianglesGivenPivotCanonical.html for a special construction along these lines.
The locus of points I is invariant with respect to the harmonic perspectivity f with axis CD and center A. The polar
of I passes from B'. Similarly the polar of I', constructed as the intersection of B'D and the tangent at B (thus
belonging to the locus), passes through B. If follows easily that line II' passes through A and points I, I' are
conjugate with respect to the harmonic perspectivity f.
The case of circle and its center is easy to handle. Besides it delivers another way to handle the general case by
mapping the general case through a homography to this particular one.
In fact, given a point A on the circle and its tangent there, consider another point B variable on the circle
and draw from there the parallel BC to the tangent, C being the other intersection point of this parallel
with the circle. The tangent tC at C becomes parallel to AB exactly when the triangle ABC is
equilateral.This simple fact identifies the equilaterals inscribed in the circle as the unique triangles ABC
for which the center is the pivot of the circle with respect to ABC.