Q: ??? 13 points in the plane are such that given any three
some fourth one lies on the same circle as those three. Show that all 13
[Julian Gilbey, 13 Jun 1996]
A: Partial solution: Label the points A..M. Call the fourth point the friend of the triple. There are 13C3 = 286 triples, so some point, wlog A, is the friend of at least 286/13 = 22 triples. These 22 triples involve 66 points (counting with multiplicity) so some point, wlog B, is in at least ceil(66/12) = 6 of them. These 6 triples involve 12 other points (counting with multiplicity), so some point, wlog C, is in at least ceil(12/11) = 2 of them. Thus we have two triples, wlog BCD and BCE, whose friend is A, and hence ABCDE are cocircular. What next?
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