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

A1A5A4A2=(x1z2,z1y2,z1z2)
A5A3A2A6=(x2x3,y2x3,x2z3)
A3A4A6A1=(y3x1,y3y1,z3y1)

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

D=|x1z2z1y2z1z2x2x3y2x3x2z3y3x1y3y1z3y1|

is zero. We have

D= x1y1y2z2z3x3-x1y1z2x2y3z3
+z1x1x2y2y3z3-y1z1x2y2z3x3
+y1z1z2x2x3y3-z1x1y2z2x3y3

Using (1) we get

(x1+y1)(x2+y2)(x3+y3)D=x1y1x2y2x3y3S

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

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

QED.

Remarks: For more on the use of coordinates in a projective planeMathworldPlanetmath, 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 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 http://www.cut-the-knot.org/Curriculum/GeometryMathworldPlanetmath/Pascal.shtml#wordscut-the-knot.org.

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