proof of Pascal’s mystic hexagram

We can choose homogeneous coordinatesMathworldPlanetmath (x,y,z) such that the equation of the given nonsingular conic is yz+zx+xy=0, or equivalently

z(x+y)=-xy (1)

and the vertices of the given hexagram A1A5A3A4A2A6 are

A1=(x1,y1,z1) A4=(1,0,0)
A2=(x2,y2,z2) A5=(0,1,0)
A3=(x3,y3,z3) A6=(0,0,1)

(see Remarks below). The equations of the six sides, arranged in opposite pairs, are then

A1A5:x1z=z1x A4A2:y2z=z2y
A5A3:x3z=z3x A2A6:y2x=x2y
A3A4:z3y=y3z A6A1:y1x=x1y

and the three points of intersection of pairs of opposite sides are


To say that these are collinearMathworldPlanetmath is to say that the determinant


is zero. We have

D= x1y1y2z2z3x3-x1y1z2x2y3z3

Using (1) we get


where (x1+y1)(x2+y2)(x3+y3)0 and

S = (x1+y1)(y2x3-x2y3)
+ (x2+y2)(y3x1-x3y1)
+ (x3+y3)(y1x2-x1y2)
= 0


Remarks: For more on the use of coordinates in a projective planeMathworldPlanetmath, see e.g. ma208/notes/project.pdfHirst (an 11-page PDF).

A synthetic proof (without coordinates) of Pascal’s theorem is possible with the aid of cross ratios or the related notion of harmonic sets (of four collinear points).

Pascal’s proof is lost; presumably he had only the real affine planeMathworldPlanetmath in mind. A proof restricted to that case, based on Menelaus’s theorem, can be seen at

Title proof of Pascal’s mystic hexagram
Canonical name ProofOfPascalsMysticHexagram
Date of creation 2013-03-22 13:53:02
Last modified on 2013-03-22 13:53:02
Owner mathcam (2727)
Last modified by mathcam (2727)
Numerical id 5
Author mathcam (2727)
Entry type Proof
Classification msc 51A05