TITLE: Interesting puzzle about a sphere and some circles
QUESTION [7 upvotes]: Suppose I have a sphere and I choose a point $P$ on it. Then I draw $N\ge 3$ circles on the sphere passing through that point in a manner such that all the intersection points of the final result involve $\ge 3$ circles. Why then must there be a point other than $P$ where all the circles pass through?
My thoughts are that we cannot have a polygonal decomposition of the sphere where every vertex has $\ge 6$ edges coming out of it (follows from Euler's relation). So there has to be an "object" with 2 edges only. But then I am not sure how to conclude that the above is true.
Added: I think @joriki 's assumption is right -- the circles don't just touch at $P$. Otherwise the question would be trivially wrong.

REPLY [5 votes]: This is equivalent to the Sylvester-Gallai problem in the Euclidean plane (by stereographic projection as in the other answer, then projective duality).  As suspected in the question there is a proof using Euler characteristic, which has been rediscovered many times. The extensive Wikipedia article on this subject dates it as far back as 1940 in a paper by E. Melchior.  http://en.wikipedia.org/wiki/Sylvester%E2%80%93Gallai_theorem  .  There are also metric proofs that consider an extremal piece of the diagram such as shortest point-line distance or minimal nonzero area of a triangle and show that no nonzero minimum can exist.