Generalizing Pappus' Theorem



We define the following binary operation on pairs of points in the projective plane. Given x=(X1,X2) and y=(Y1,Y2), we define \(x \ast y = (\overline{X_{1} Y_{1}} \cap \overline{X_{2} Y_{2}},\overline{X_{1} Y_{2}} \cap \overline{X_{2} Y_{1}} ) \).
This operation is pictured below with \(x \ast y = z\).

It is easily checkable that this operation is non-associative, commutative, and has the additional property that \( (x \ast y) \ast x = y \). This final property entails that all repeated applications of \(\ast\) to x and y yield one of x, y, or \(x \ast y\).

Now choose three unrelated pairs of points x, y, and z. The following applet displays the orbit of points when repeatedly applying \(\ast x\),\(\ast y\), or \(\ast z\) to each of x, y, and z.

The following applet displays orbits of several iterations of this construction. The red, blue, and magenta colored points signify the pairs (X1,X2), (Y1,Y2), and (Z1,Z2). By clicking ``SPHERE'', you can view a hemispherical projection of the plane onto the sphere (which is then projected onto a circle) that gives you a global view of the nature of the orbits.
 

alt="Your browser understands the <APPLET> tag but isn't running the applet, for some reason." Your browser is completely ignoring the <APPLET> tag! Get Netscape. 


 

It turns out that every such orbit lies on a degree three polynomial, a cubic. This of course needs to be prooven, but perhaps the applet above makes it believable. Anyway, this can be prooven with some intense algebra and I am currently looking for a better proof.

We know that 9 points in the plane determine a cubic polynomial runing through them. Let C be a cubic curve (a degree three projective algebraic variety). This gives us the following theorem:

Theorem 1   Given a degree three projective algebraic variety C, and an octahedron with vertices lying on C. If any three of this octahedron's opposing edges intersect on C then all six of the octahedron's opposing edges intersect on C.

Proof: The six vertices of the octahedron determine three pairs of points in the plane (each point is paired with its opposite). Now we can perform our construction on these pairs, call them x, y, and z. As a consequence of what was discovered above, each point in the pairs \(x \ast y\),\(y \ast z\), and \(z \ast x\) lie on a single cubic curve. But 9 points are sufficient to determine a cubic curve, thus if three points of \(x \ast y \cup y \ast z \cup z \ast x\)lie on a cubic curve C as well as the points in \(x \cup y \cup z\), then they all must. $\diamondsuit$

Now we have the tools to prove Pappus' Theorem and Pascal's Theorem quite simply.
 
 
Theorem 2 (Pappus' Theorem)   Given points X1, Y1, and Z1 on a line l and points X2, Y2, and Z2 on a line m, the points \( A=\overline{X_{1} Y_{2}} \cap \overline{X_{2} Y_{1}} \), \( B=\overline{Y_{1} Z_{2}} \cap \overline{Y_{2} Z_{1}} \), and \( D=\overline{Z_{1} X_{2}} \cap \overline{Z_{2} X_{1}} \) are all colinear.

Proof: Let C be the degree three projective algebraic variety \( l \cup m \cup \overline{A B} \). Let \(O=l \cap m = \overline{X_{1} Y_{1}} \cap \overline{X_{2} Y_{2}}\). Then the nine points X1, Y1, Z1, X2, Y2, Z2, O, A, B determine a unique degree three projective algebraic variety which runs through each of these points. Consequently the point D must also lie on this curve C. Clearly D does not lie on l or m, hense D lies on \( \overline{A B} \). $\diamondsuit$

Pascal's Theorem can also be shown to follow directly from this result.

 
 

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