proof of Pascal’s mystic hexagram
We can choose homogeneous coordinates such that the equationof the given nonsingular conic is , or equivalently
(1) |
and the vertices of the given hexagram are
(see Remarks below).The equations of the six sides, arranged in opposite pairs, are then
and the three points of intersection of pairs of opposite sides are
To say that these are collinear is to say that the determinant
is zero. We have
Using (1) we get
where and
QED.
Remarks: For more on the use of coordinates in a projectiveplane, see e.g. http://www.maths.soton.ac.uk/staff/AEHirst/ ma208/notes/project.pdfHirst(an 11-page PDF).
A synthetic proof (without coordinates) of Pascal’s theoremis possible with the aid of cross ratios or the related notionof harmonic sets (of four collinear points).
Pascal’s proof is lost; presumably he had only the real affine planein mind. A proof restricted to that case, based on Menelaus’s theorem,can be seen athttp://www.cut-the-knot.org/Curriculum/Geometry
/Pascal.shtml#wordscut-the-knot.org.