Pappus's hexagon theorem

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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.

Statement and 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

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

Proof

Let

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

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

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

We need to show that if $alpha$ = 0 and $beta$ = 0, then $gamma$ = 0.

Step 1.

Using the identity

$langle\; A,B,Crangle\; =\; langle\; C,A,Brangle\; =\; langle\; B,C,Arangle$

we can express $alpha$, $beta$, and $gamma$ in the following equivalent forms:

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

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

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

Step 2.

We can apply the identities

$langle\; A,B,Crangle\; =\; A\; cdot\; (B\; times\; C)$

$A\; times\; (B\; times\; C)\; =\; (A\; cdot\; C)B\; -\; (A\; cdot\; B)C$

to get

$alpha\; =\; (U\; times\; V)\; cdot\; ((X\; times\; W)\; times\; (Y\; times\; Z))$

$beta\; =\; (Y\; times\; W)\; cdot\; ((U\; times\; Z)\; times\; (X\; times\; V))$

$gamma\; =\; (X\; times\; Z)\; cdot\; ((Y\; times\; V)\; times\; (U\; times\; W))$

and then

$alpha\; =\; (U\; times\; V)\; cdot\; (langle\; X,W,Zrangle\; Y\; -\; langle\; X,W,Yrangle\; Z)$

$beta\; =\; (Y\; times\; W)\; cdot\; (langle\; U,Z,Vrangle\; X\; -\; langle\; U,Z,Xrangle\; V)$

$gamma\; =\; (X\; times\; Z)\; cdot\; (langle\; Y,V,Wrangle\; U\; -\; langle\; Y,V,Urangle\; W)$

Step 3.

Using the distributive property of the dot product:

$alpha\; =\; langle\; X,W,Zrangle\; langle\; U,V,Yrangle\; -\; langle\; X,W,Yrangle\; langle\; U,V,Zrangle$

$beta\; =\; langle\; U,Z,Vrangle\; langle\; Y,W,Xrangle\; -\; langle\; U,Z,Xrangle\; langle\; Y,W,Vrangle$

$gamma\; =\; langle\; Y,V,Wrangle\; langle\; X,Z,Urangle\; -\; langle\; Y,V,Urangle\; langle\; X,Z,Wrangle$

Step 4.

Using the identities

$langle\; A,B,Crangle\; =\; langle\; C,A,Brangle\; =\; langle\; B,C,Arangle$

$langle\; A,B,Crangle\; =\; -langle\; A,C,Brangle\; =\; -langle\; C,B,Arangle\; =\; -langle\; B,A,Crangle$

We can permute the terms as follows:

$alpha\; =\; langle\; X,W,Zrangle\; langle\; U,V,Yrangle\; -\; langle\; X,W,Yrangle\; langle\; U,V,Zrangle$

$beta\; =\; -langle\; U,Z,Xrangle\; langle\; Y,W,Vrangle\; +\; langle\; X,W,Yrangle\; langle\; U,V,Zrangle$

$gamma\; =\; langle\; U,Z,Xrangle\; langle\; Y,W,Vrangle\; -\; langle\; X,W,Zrangle\; langle\; U,V,Yrangle$

Step 5.

We can now add these equations to get:

$alpha\; +\; beta\; +\; gamma\; =\; 0$

$gamma\; =\; -(alpha\; +\; beta)$

from which it follows that if $alpha$ = 0 and $beta$ = 0, then $gamma$ = 0.

Q.E.D.

See also

• Plane geometry
• Projective geometry
• Pappus of Alexandria
• Pappus configuration

External links

• Pappus' hexagon theorem at Cut-the-knot.org at cut-the-knot
• Dual to Pappus' hexagon theorem at Cut-the-knot.org at cut-the-knot

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