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Clifford's circle chain theorem with n=7 circles

Following Penrose' notation, I omit letters "c" for circles and "P" for points,
to avoid cluttering.
Here we see seven (n=7)
blue circles c1, c2, c3, c4, c5, c6, c7 through a point O (taken to infinity)
intersecting in
P12, P13, P14, P15, P16, P17, P23, ..., P67
which again define
unique yellow circles like as c123, c124, ..., c567 (not all drawn, to
avoid cluttering).
The construction of these yellow circles corresponds to applying the case
n=3
successively to three circles each, i.e. to c1,c2,c3 to give c123, etc.
According to the case with
n=4
circles through O taking four yellow circles at a time yields
the small dark red points P1234, P1235, ..., P4567.
The case n=5 shows that always five of these points
at a time, e.g. P1234, P1235, P1245, P1345, P2345 lie on red
circles c12345, etc.
According to the case with
n=6
allways six such red circles, e.g. c12345,
c12346, c12356, c12456, c13456, c23456 will coincide in points like P123456.
The surprise in the case of n=7 below is that the total of seven such points
P123456, P134567, P124567, P123567, P123467, P123457, P123456 lie on one
unique (violet) circle c1234567.

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

[ circle chain theorem | GA with Cinderella ]

Soli Deo Gloria. Created with Cinderella
by Eckhard Hitzer (Fukui).