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 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
A1A5⋅A4A2=(x1z2,z1y2,z1z2) |
A5A3⋅A2A6=(x2x3,y2x3,x2z3) |
A3A4⋅A6A1=(y3x1,y3y1,z3y1) |
To say that these are collinear 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
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 |
---|---|
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 |