PascalPoncelet

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.

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).

Pascal-Poncelet-Brianchon meeting

See Also