proof of Pascal’s mystic hexagram

proof of Pascal’s mystic hexagram

We can choose

homogeneous coordinates

(

x

,

y

,

z

)

such that the equation
of the given

nonsingular

conic is

y

⁢

z

+

z

⁢

x

+

x

⁢

y

=

0

, or equivalently

z

⁢

(

x

+

y

)

=

-

x

⁢

y

(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

⋅

A

4

⁢

A

2

=

(

x

1

⁢

z

2

,

z

1

⁢

y

2

,

z

1

⁢

z

2

)

A

5

⁢

A

3

⋅

A

2

⁢

A

6

=

(

x

2

⁢

x

3

,

y

2

⁢

x

3

,

x

2

⁢

z

3

)

A

3

⁢

A

4

⋅

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

=

|

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

|

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

)

≠

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

.pdf

Hirst
(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#words

cut-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

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