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Clifford's circle chain theorem with n=5 circles

Following Penrose' notation, I omit letters "c" for circles and "P" for points,
to avoid cluttering. Here we see five (n=5)
blue circles c1, c2, c3, c4, c5 through O intersecting in
P12, P13, P14, P15, P23, P24, P25, P34, P35, P45
which again define
unique yellow circles like as c123, c125, c145, c234, c345, etc.
The construction of these yellow circles corresponds to applying the case
n=3
successively to three (blue) circles at a time, i.e. to c1,c2,c3 to give c123, etc.
According to the case with
n=4
circles through O, successively taking four yellow circles at a time yields
the five points P1234, P1235, P1245, P1345, P2345.
Clifford found that these five points all lie on one and the same (red) circle
c12345. Sir R. Penrose states that
this is basically a theorem due to Augueste Miquel
(France, Nantua, College des Castres.)

The bright red points can be moved interactively with the mouse pointer.

Taking O to infinity changes the above into

The five blue straight lines are now five circles with infinite radius intersecting at infinity O. You
can find a special case of the above for the points P12, P15, P23, P34, P45 situtated
on a circle by
Leo Dorst (Amsterdam).

It is interesting to note that Hongbo Li, *Automated Theorem Proving in the
Homogeneous Model with Clifford Bracket Algebra*,
in L. Dorst, C. Doran, J. Lasenby, Applications of Geometric Algebra in
Computer Sicnece and Engineering, Birkhaeuser, Boston, 2002, demonstates
a purely algebraic proof of the case n=5, based on the *conformal*
geometric algebra model of the Euclidean plane. With the help of the
geometric product of vectors he constructs an algebra of Clifford brackets,
with which he performs the proof.

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