proof of Pascal’s mystic hexagram

We can choose homogeneous coordinates $(x,y,z)$ 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 $A_{1}A_{5}A_{3}A_{4}A_{2}A_{6}$ are

 $A_{1}=(x_{1},y_{1},z_{1})$ $A_{4}=(1,0,0)$ $A_{2}=(x_{2},y_{2},z_{2})$ $A_{5}=(0,1,0)$ $A_{3}=(x_{3},y_{3},z_{3})$ $A_{6}=(0,0,1)$

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

 $A_{1}A_{5}:x_{1}z=z_{1}x$ $A_{4}A_{2}:y_{2}z=z_{2}y$ $A_{5}A_{3}:x_{3}z=z_{3}x$ $A_{2}A_{6}:y_{2}x=x_{2}y$ $A_{3}A_{4}:z_{3}y=y_{3}z$ $A_{6}A_{1}:y_{1}x=x_{1}y$

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

 $A_{1}A_{5}\cdot A_{4}A_{2}=(x_{1}z_{2},z_{1}y_{2},z_{1}z_{2})$
 $A_{5}A_{3}\cdot A_{2}A_{6}=(x_{2}x_{3},y_{2}x_{3},x_{2}z_{3})$
 $A_{3}A_{4}\cdot A_{6}A_{1}=(y_{3}x_{1},y_{3}y_{1},z_{3}y_{1})$

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

 $D=\left|\begin{array}[]{ccc}x_{1}z_{2}&z_{1}y_{2}&z_{1}z_{2}\\ x_{2}x_{3}&y_{2}x_{3}&x_{2}z_{3}\\ y_{3}x_{1}&y_{3}y_{1}&z_{3}y_{1}\end{array}\right|$

is zero. We have

 $D=$ $x_{1}y_{1}y_{2}z_{2}z_{3}x_{3}-x_{1}y_{1}z_{2}x_{2}y_{3}z_{3}$ $+z_{1}x_{1}x_{2}y_{2}y_{3}z_{3}-y_{1}z_{1}x_{2}y_{2}z_{3}x_{3}$ $+y_{1}z_{1}z_{2}x_{2}x_{3}y_{3}-z_{1}x_{1}y_{2}z_{2}x_{3}y_{3}$

Using (1) we get

 $(x_{1}+y_{1})(x_{2}+y_{2})(x_{3}+y_{3})D=x_{1}y_{1}x_{2}y_{2}x_{3}y_{3}S$

where $(x_{1}+y_{1})(x_{2}+y_{2})(x_{3}+y_{3})\neq 0$ and

 $\displaystyle S$ $\displaystyle=$ $\displaystyle(x_{1}+y_{1})(y_{2}x_{3}-x_{2}y_{3})$ $\displaystyle+$ $\displaystyle(x_{2}+y_{2})(y_{3}x_{1}-x_{3}y_{1})$ $\displaystyle+$ $\displaystyle(x_{3}+y_{3})(y_{1}x_{2}-x_{1}y_{2})$ $\displaystyle=$ $\displaystyle 0$

QED.

Remarks: For more on the use of coordinates in a projective plane, 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 plane in mind. A proof restricted to that case, based on Menelaus’s theorem, can be seen at http://www.cut-the-knot.org/Curriculum/Geometry/Pascal.shtml#wordscut-the-knot.org.

Title proof of Pascal’s mystic hexagram ProofOfPascalsMysticHexagram 2013-03-22 13:53:02 2013-03-22 13:53:02 mathcam (2727) mathcam (2727) 5 mathcam (2727) Proof msc 51A05