# 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

$$D=\left|\begin{array}{ccc}\hfill {x}_{1}{z}_{2}\hfill & \hfill {z}_{1}{y}_{2}\hfill & \hfill {z}_{1}{z}_{2}\hfill \\ \hfill {x}_{2}{x}_{3}\hfill & \hfill {y}_{2}{x}_{3}\hfill & \hfill {x}_{2}{z}_{3}\hfill \\ \hfill {y}_{3}{x}_{1}\hfill & \hfill {y}_{3}{y}_{1}\hfill & \hfill {z}_{3}{y}_{1}\hfill \end{array}\right|$$ |

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 $({x}_{1}+{y}_{1})({x}_{2}+{y}_{2})({x}_{3}+{y}_{3})\ne 0$ and

$S$ | $=$ | $({x}_{1}+{y}_{1})({y}_{2}{x}_{3}-{x}_{2}{y}_{3})$ | ||

$+$ | $({x}_{2}+{y}_{2})({y}_{3}{x}_{1}-{x}_{3}{y}_{1})$ | |||

$+$ | $({x}_{3}+{y}_{3})({y}_{1}{x}_{2}-{x}_{1}{y}_{2})$ | |||

$=$ | $0$ |

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 |