Pappus Configuration for Circles
Let A, B, C, D, E, and F be any six points in the inversive plane. Draw the circles
ACE and BDF. Choose a point A* on the circle ACE.
Draw the circle ABA* and let B* be the point besides B where
that circle intesects the circle BDF.
Draw the circle BCB* and let C* be the other point where
it intesects ACE.
Draw the circle CDC* and let D* be the other point where
it intesects BDF.
Draw the circle DED* and let E* be the other point where
it intesects ACE.
Draw the circle EFE* and let F* be the other point where
it intesects BDF.
Draw the circle FAF*. Then A* lies on FAF*.
Let X and X* be the points where the circles ABA*B* and DED*E* meet, if
they do,
and Y and Y* where BCB*C* and EFE*F* meet, and Z and Z*
where
CDC*D* and DED*E* meet. Then X, X*, Y, Y*, Z, and Z* are concyclic.
The Euclidean case
When the six the starred points, A* through F* are all the same point, the point at
infinity, then all the circles are straight lines, and the
configuration reduces to the standard Pappus configuration of Euclidean and projective geometry.
The hyperbolic case
If the two circles ACE and BDF are orthogonal to a given circle G, and the point
A* is the inversion of A in the circle G, then the remaining starred points will
be inversions of the unstarred points in G, and the cirlce XYZ will also be orthogonal
to G. Thus, the hyperbolic version of Pappus configuration is a special case of the inversive
version.
The elliptic case
If the two circles ACE and BDF each meet a given circle G at antipodal points,
and the point A* is the elliptic inversion of A in the circle G, then so will
the remaining starred points be elliptic inversions of the unstarred points in G, and the
cirlce XYZ will meet G at antipodal points. Thus, the elliptic version of Pappus
configuration is also a special case of the inversive version.
1998
David E. Joyce
Department of Mathematics and Computer Science
Clark University
Worcester, MA 01610
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