Chemistry - Pappus's hexagon theorem

Pappus's hexagon theorem (attributed to Pappus of Alexandria) states that given one set of collinear points A, B, C, and another set of collinear points a, b, c, then the intersection points x, y, z of line pairs Ab and aB, Ac and aC, Bc and bC are collinear. (Collinear means the points are incident on a line.)

The dual of this theorem states that given one set of concurrent lines A, B, C, and another set of concurrent lines a, b, c, then the lines x, y, z defined by pairs of points resulting from pairs of intersections A∩b and a∩B, A∩c and a∩C, B∩c and b∩C are concurrent.

A generalization of this theorem is Pascal's theorem, which was discovered by Blaise Pascal at the age of 16.

Contents

1 Proof of Pappus's Hexagon Theorem

1.1 First restatement
1.2 Second restatement
1.3 Third restatement
1.4 Fourth restatement
1.5 Fifth restatement
1.6 Sixth restatement
1.7 Seventh restatement

2 External links

3 See also

Proof of Pappus's Hexagon Theorem

Let there be six lines on a projective plane: U, V, W, X, Y, and Z. Then the theorem can be stated thus:

If
(1) the points equal to the intersections of U with V, X with W, and Y with Z are collinear,
and if
(2) the points equal to the intersection of U with Z, X with V, and Y with W are collinear, then
it must be true that
(3) the points equal to the intersections of U with W, X with Z, and Y with V are collinear.

Symbolically, Pappus's theorem can be stated as follows:
If

$\langle U \times V, X \times W, Y \times Z \rangle = 0$

and if

$\langle U \times Z, X \times V, Y \times W \rangle = 0$

then

$\langle U \times W, X \times Z, Y \times V \rangle = 0.$

First restatement

The symbolic statement above is equivalent to the following one:
If

$(U \times V) \cdot (\langle X,W,Z\rangle Y - \langle X,W,Y\rangle Z) = 0$

and if

$(U \times Z) \cdot (\langle X,V,W\rangle Y - \langle X,V,Y\rangle W) = 0$

then

$(U \times W) \cdot (\langle X,Z,V\rangle Y - \langle X,Z,Y\rangle V) = 0$ .

Second restatement

The first restatement above is equivalent to this one:
If

$\langle X,W,Z\rangle \langle U,V,Y\rangle - \langle X,W,Y\rangle \langle U,V,Z\rangle = 0$

and if

$\langle X,V,W\rangle \langle U,Z,Y\rangle - \langle X,V,Y\rangle \langle U,Z,W\rangle = 0$

then

$\langle X,Z,V\rangle \langle U,W,Y\rangle - \langle X,Z,Y\rangle \langle U,W,V\rangle = 0$ .

Third restatement

The second restatement above is equivalent to the following one:
If

$\left| \begin{matrix} X \cdot U & X \cdot V & X \cdot Y \\ W \cdot U & W \cdot V & W \cdot Y \\ Z \cdot U & Z \cdot V & Z \cdot Y \end{matrix} \right| = \left| \begin{matrix} X \cdot U & X \cdot V & X \cdot Z \\ W \cdot U & W \cdot V & W \cdot Z \\ Y \cdot U & Y \cdot V & Y \cdot Z \end{matrix} \right|$

and if

$\left| \begin{matrix} X \cdot U & X \cdot Z & X \cdot Y \\ V \cdot U & V \cdot Z & V \cdot Y \\ W \cdot U & W \cdot Z & W \cdot Y \end{matrix} \right| = \left| \begin{matrix} X \cdot U & X \cdot Z & X \cdot W \\ V \cdot U & V \cdot Z & V \cdot W \\ Y \cdot U & Y \cdot Z & Y \cdot W \end{matrix} \right|$

then

$\left| \begin{matrix} X \cdot U & X \cdot W & X \cdot Y \\ Z \cdot U & Z \cdot W & Z \cdot Y \\ V \cdot U & V \cdot W & V \cdot Y \end{matrix} \right| = \left| \begin{matrix} X \cdot U & X \cdot W & X \cdot V \\ Z \cdot U & Z \cdot W & Z \cdot V \\ Y \cdot U & Y \cdot W & Y \cdot V \end{matrix} \right|.$

Fourth restatement

The third restatement above is equivalent to the following one:
If $(X \cdot U) (W \cdot V) (Z \cdot Y) + (X \cdot V) (W \cdot Y) (Z \cdot U) + (X \cdot Y) (W \cdot U) (Z \cdot V)$ $- (X \cdot U) (W \cdot Y) (Z \cdot V) - (X \cdot V) (W \cdot U) (Z \cdot Y) - (X \cdot Y) (W \cdot V) (Z \cdot U)$ $= (X \cdot U) (W \cdot V) (Y \cdot Z) + (X \cdot V) (W \cdot Z) (Y \cdot U) + (X \cdot Z) (W \cdot U) (Y \cdot V)$ $- (X \cdot U) (W \cdot Z) (Y \cdot V) - (X \cdot V) (W \cdot U) (Y \cdot Z) - (X \cdot Z) (W \cdot V) (Y \cdot U)$ and if $(X \cdot U) (V \cdot Z) (W \cdot Y) + (X \cdot Z) (V \cdot Y) (W \cdot U) + (X \cdot Y) (V \cdot U) (W \cdot Z)$ $- (X \cdot U) (V \cdot Y) (W \cdot Z) - (X \cdot Z) (V \cdot U) (W \cdot Y) - (X \cdot Y) (V \cdot Z) (W \cdot U)$ $= (X \cdot U) (V \cdot Z) (Y \cdot W) + (X \cdot Z) (V \cdot W) (Y \cdot U) + (X \cdot W) (V \cdot U) (Y \cdot Z)$ $- (X \cdot U) (V \cdot W) (Y \cdot Z) - (X \cdot Z) (V \cdot U) (Y \cdot W) - (X \cdot W) (V \cdot Z) (Y \cdot U)$ then $(X \cdot U) (Z \cdot W) (V \cdot Y) + (X \cdot W) (Z \cdot Y) (V \cdot U) + (X \cdot Y) (Z \cdot U) (V \cdot W)$ $- (X \cdot U) (Z \cdot Y) (V \cdot W) - (X \cdot W) (Z \cdot U) (V \cdot Y) - (X \cdot Y) (Z \cdot W) (V \cdot U)$ $= (X \cdot U) (Z \cdot W) (Y \cdot V) + (X \cdot W) (Z \cdot V) (Y \cdot U) + (X \cdot V) (Z \cdot U) (Y \cdot W)$ $- (X \cdot U) (Z \cdot V) (Y \cdot W) - (X \cdot W) (Z \cdot U) (Y \cdot V) - (X \cdot V) (Z \cdot W) (Y \cdot U).$

Fifth restatement

The first and fifth terms of each side of each equation of the fourth restatement can be canceled out. Also, some of the terms can have some internal rearrangements (through commutativity of dot product and scalar product), yielding what follows:
If

$(X \cdot V) (Y \cdot W) (Z \cdot U) + (X \cdot Y) (Z \cdot V) (U \cdot W)$

$- (X \cdot U) (Y \cdot W) (Z \cdot V) - (X \cdot Y) (Z \cdot U) (V \cdot W)$

$= (X \cdot V) (Y \cdot U) (Z \cdot W) + (X \cdot Z) (Y \cdot V) (Z \cdot W)$

$- (X \cdot U) (Y \cdot V) (Z \cdot W) - (X \cdot Z) (Y \cdot U) (V \cdot W)$

and if

$(X \cdot Z) (Y \cdot V) (U \cdot W) + (X \cdot Y) (Z \cdot W) (U \cdot V)$

$- (X \cdot U) (Y \cdot V) (Z \cdot W) - (X \cdot Y) (Z \cdot V) (U \cdot W)$

$= (X \cdot Z) (Y \cdot U) (V \cdot W) + (X \cdot W) (Y \cdot Z) (U \cdot V)$

$- (X \cdot U) (Y \cdot Z) (V \cdot W) - (X \cdot W) (Y \cdot U) (Z \cdot V)$

then

$(X \cdot W) (Y \cdot Z) (U \cdot V) + (X \cdot Y) (Z \cdot U) (V \cdot W)$

$- (X \cdot U) (Y \cdot Z) (V \cdot W) - (X \cdot Y) (Z \cdot W) (U \cdot V)$

$= (X \cdot W) (Y \cdot U) (Z \cdot V) + (X \cdot V) (Y \cdot W) (Z \cdot U)$

$- (X \cdot U) (Y \cdot W) (Z \cdot V) - (X \cdot V) (Y \cdot U) (Z \cdot W).$

Sixth restatement

There are twelve terms, out of fifteen possible ones given the six different letters in each term): six letters permute in 6! different ways, but then the three scalar factors permute in 3! ways, and each factor is a dot product which permutes in 2! ways: ${6! \over 3! \, 2! \, 2! \, 2!} = 15.$ Each equation has eight terms, so the three equations have 24 terms. Each of the twelve terms appears twice, and never in the same equation. Apply the following labels:

$t_1 = (X \cdot V) (Y \cdot W) (Z \cdot U),$

$t_2 = (X \cdot Y) (Z \cdot V) (U \cdot W),$

$t_3 = (X \cdot U) (Y \cdot W) (Z \cdot V),$

$t_4 = (X \cdot Y) (Z \cdot U) (V \cdot W),$

$t_5 = (X \cdot V) (Y \cdot U) (Z \cdot W),$

$t_6 = (X \cdot Z) (Y \cdot V) (U \cdot W),$

$t_7 = (X \cdot U) (Y \cdot V) (Z \cdot W),$

$t_8 = (X \cdot Z) (Y \cdot U) (V \cdot W),$

$t_9 = (X \cdot Y) (Z \cdot W) (U \cdot V),$

$t_{10} = (X \cdot W) (Y \cdot Z) (U \cdot V),$

$t_{11} = (X \cdot U) (Y \cdot Z) (V \cdot W),$

$t_{12} = (X \cdot W) (Y \cdot U) (Z \cdot V).$

Then Pappus's theorem can be restated as
If

t 1 + t 2 - t 3 - t 4 = t 5 + t 6 - t 7 - t 8

and if

t 6 + t 9 - t 7 - t 2 = t 8 + t 10 - t 11 - t 12

then

t 10 + t 4 - t 11 - t 9 = t 12 + t 1 - t 3 - t 5 .

Seventh restatement

Move the terms on the right side of the first two equations of the sixth restatement to the left side, and move the terms on the left side of the third equation to the right side, then arrange the terms into increasing numerical order:
If

$t_1 \ \ + t_2 - t_3 - t_4 - t_5 - t_6 + t_7 + t_8 \qquad \qquad \qquad \qquad = 0$

and if

$\qquad - t_2 \qquad \qquad \qquad + t_6 - t_7 - t_8 + t_9 - t_{10} + t_{11} + t_{12} = 0$

then

$0 = t_1 \ \qquad - t_3 - t_4 - t_5 \ \qquad \ \qquad \ \qquad + t_9 - t_{10} + t_{11} + t_{12}.$

When the first and second equations are added up, they result in the third equation.
Q.E.D.