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.

  • $\begingroup$ Perhaps there was supposed to be an assumption that all intersection points between the circles are "real intersections", i.e. no two circles merely touch? $\endgroup$ – joriki Apr 27 '12 at 9:04
  • $\begingroup$ @joriki: I think so :) $\endgroup$ – Jon M Apr 27 '12 at 9:06
  • $\begingroup$ The question should ideally be self-contained and not rely on the comments, i.e. it would be preferable to state all assumptions explicitly in the question. $\endgroup$ – joriki Apr 27 '12 at 9:46
  • $\begingroup$ @yasmar: I don't see it. The obvious inductive hypothesis fails miserably, at least for me. $\endgroup$ – TonyK Apr 27 '12 at 18:00
  • $\begingroup$ Can't we first stereographically project the circles onto a plane such that they all still intersect transversally at $P$? Then we can invert the whole picture w.r.t a circle centered at $P$ converting all the circles to lines. We're left with the claim: There are $N$ non-parallel lines on the plane such that at least 3 lines meet at each point of intersection. Show that all the lines meet at the same point. $\endgroup$ – Jyrki Lahtonen Apr 27 '12 at 20:11

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.

  • $\begingroup$ As Jyrki said, very well spotted!! $\endgroup$ – Jon M Apr 28 '12 at 15:39
  • $\begingroup$ For completeness I'd like to add that the theorem by Melchior seems to only prove the version in which any tangency at $P$ of pairs of circles is forbidden, which translates into forbidding parallel lines. The stronger form I suggested, where tangency/parallel lines are allowed but just case where all circles are tangent/lines are parallel is included among the trivial solutions, does not follow, at least not from the Euler characteristic aregument. That produces at least three "ordinary intersections", but these could conceivably all be at infinity (i.e., concern pairs of parallel lines). $\endgroup$ – Marc van Leeuwen May 14 '12 at 8:30

Counterexample: consider 3 circles whose centres are on the same longitude and which all pass through the North Pole (P). Then the only intersection point is P, and it involves all 3 circles.

  • $\begingroup$ Thank you, but I think the question implicitly excludes the case where there are no intersections apart from that at $P$. And I think it is assumed that all the circles must intersect some other circle, otherwise this question would be trivially wrong! $\endgroup$ – Jon M Apr 27 '12 at 9:08
  • $\begingroup$ @Jon: That all circles must intersect some other circle already follows from the fact that the circles all pass through $P$. The case where there are no intersections apart from that at $P$ is excluded by the assumption that you've now added to the question, so there's some hope that we've now got a complete statement of the problem. $\endgroup$ – joriki Apr 27 '12 at 9:45
  • $\begingroup$ @joriki: You are right. I am sorry for the ambiguities. $\endgroup$ – Jon M Apr 27 '12 at 9:49
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    $\begingroup$ Each touching at $P$ may be considered a double intersection; so this example is "an exception that proves the rule". $\endgroup$ – John Bentin Apr 27 '12 at 12:31
  • $\begingroup$ @JohnBentin: Sneaky ;) but I think the question isn't intended to be interpreted this way. $\endgroup$ – Jon M Apr 28 '12 at 0:59

This is only a partial answer that I cannot see how to complete easily, but may help others finding one. Applying a stereographic projection with center the north pole $P$ onto a plane parallel to the tangent plane at $P$ (for instance the plane of the equator), the circles that pass throught $P$ become precisely the lines of the plane. Therefore the question gives rise to the following one:

Suppose a finite set $S$ of lines the plane has the property "whenever two lines of $S$ intersect, there is at least a third line of $S$ that passes through the point of intersection". Must it then be the case that the lines of $S$ either are all be parallel or all pass through one fixed point of the plane?

The condition added to the original question (non-tangency of circles at $P$) amounts to excluding parallel lines in the set (whose circles would be tangent at $P$); however I believe the even the question above to have an affirmative answer, which is a stronger statement (one can allow parallelism, provided one just claims that there cannot be more than one point of intersection between some lines of $S$). I think this should even be a fairly classical result, but I could not see any simple argument (based on counting lines and interections, or on considering the convex hull of all intersection points, and such) that proves it. Probably I'm overlooking something simple.

One can however get fairly close to a configuration that would contradict the property, for instance the sides and diagonals of a paralellogram, which give just one forbidden intersection of only two lines (at the center). Or a triangle with its medians and the lines joining the midpoints of the sides (there are three spurious intersections). Of course there are infinite collections of lines that disprove the statement without "finite", for instance the lines in a three directions on a triangular grid or even simpler the set of all lines of the plane.

Since almost-counterexamples often involve parallel lines, it might be easier with the extra assumption of no parallel lines in $S$, but I still can't see an easy proof.


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