[alogo] Pascal-Poncelet-Brianchon meeting

The figure below illustrates three famous theorems from the domain of plane projective geometry. To construct it, start with a conic (a), a point A not on (a) and two points A1, A2 on (a).
1) Construct first the polar line a' of point A w.r. to the conic a.
2) Construct then the conjugate points A3 of A2 and A4 of A1.
3) A* is the intersection point of AA1 and the polar a'. A5, A6 are intersections of (a) with the lines A3A* and A2A*. The resulting hexagon p = A1...A6 is inscribed in the conic (c).
4) Pascal's theorem asserts the collinearity of A, A25, A36, which are the intersection points of "opposite" sides of the hexagon.
5) a'' is the polar line of A* with respect to the conic (a). By its construction, every line through A* intersects the polygon p in two points C, C' (not drawn) and the line a'' at a point C'', such that (CC'A*C'')=-1, i.e. a harmonic division.
6) B2, B5 are the intersection points of the polar a' of A with the sides A2A3, A5A6 of p. B36, B25 are the intersection points of the diagonals A3A6, A2A5 with a''. Joining them with B2, B5 we find the other points B1, B6, B3, B4 on the sides of p and define a second hexagon q = B1...B6.
7) The conic B through the points B1, ..., B5 passes also through B6 and is tangent to the sides of p at these points.
8) Thus, polygon p is simultaneously inscribed in a and circumscribed on b. By its construction, the lines joining opposite vertices of p meet at A*. In general Brianchon's theorem asserts, that the same is true for every polygon circumscribed on a conic.
9) p is an instance of a "Poncelet" polygon i.e. a polygon simultaneously inscribed and circumscribed on two conics. Poncelet's "great" theorem asserts that if for two given conics a, b, such a polygon exists, then there exist infinitely many.

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

Some additional remarks:
10) A, B15, B35 are collinear. This is another instance of Pascal's theorem.
11) There is a (involutive) projectivity F, interchanging points of the pair (A2,A3) and also of (A1,A4). F has A as an isolated fixed point. Also every point of the line a' is fixed point of F.
12) F leaves the conics a, b invariant. It leaves also invariant every family-member of the conic-family (J) generated by a and b.
13) A and a' are respectively pole-polar of every conic of the family (J).
14) A and A* singular points of (J). a'' considered as a double line is a singular member of J.
15) The transformation F maps a Poncelet polygon (s) to a Poncelet polygon (s'). There are exactly two such polygons that remain invariant under F. One is s1=p and the other (a kind of "dual" to s1), s2 has contact points with (b) the intersection points of (b) with the diagonals of p.
16) The construction could start with an involutive F leaving invariant a family of conics (J), pick a member (a) of (J) and consider one of the two F-invariant Poncelet hexagons inscribed in (a).

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