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Can $n$ circles on a plane generate $m$ intersection points where at least $k$ circles intersect?

For $k = 2$ the answer is obvious since we can always place circles so that every one of them intersects every other, generating in total at most $2 {n\choose{2}}$ intersection points of $2$ circles.

What can we say for $k = 3$? In particular I am interested in $n = 7$, $m = 12$.

It is known (see figure) that $8$ circles can generate $12$ intersection points of at least $3$ circles.

            enter image description here

The question is: can we generate $12$ intersection points of at least $3$ circles using only $7$ circles in total?

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Yes, we can. Consider the usual drawing of the Fano plane with 7 vertices, 6 lines, and a circle. Replace the circle with a line through two of the three vertices.

Now we have 7 lines with 6 triple intersections in the plane. Considering the plane as a subset of the real projective plane, we get 7 planes through the origin in $\mathbb{R}^3$ with 6 lines of triple intersection.

Now intersect this configuration with the unit sphere. We get 7 great circles and 12 triple intersection points (in six antipodal pairs). By stereographic projection onto the plane, we are done.

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    $\begingroup$ A faster route: choose three concurrent great circles on the sphere. Pick "a nice point" on one of the circles far from the two points of concurrency, and make that new point (and antipode) another two points of concurrency with two more great circles going through the point. This gives four more pairs of antipodal points through which one threads the two remaining circles. Now use the Fano plane or other means to show that the concurrency between these last two circles is (or can be, with judicious choices) a triple concurrency. Gerhard "Geodesic Routes Are Often Faster" Paseman, 2017.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2017 at 17:34
  • $\begingroup$ Actually, you may have to coordinate the choices of the last four circles, say two assigned to each of two of the original three circles, arranged so that they intersect in pairs on the third original. In any case, the choice can be motivated by the Fano plane design without using projective space as an intermediary. Gerhard "Saves Projecting For Other Domains" Paseman, 2017.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2017 at 17:45
  • $\begingroup$ Adam P. Goucher, thanks for your answer. $\endgroup$
    – myro
    Commented Feb 25, 2017 at 20:09
  • $\begingroup$ Gerhard Paseman, thanks for your construction. $\endgroup$
    – myro
    Commented Feb 25, 2017 at 20:10
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    $\begingroup$ I lost count. (Oops.) I thought we were going for 7 pairs of triple concurrency. (Actually I suspect a deeper level of confusion. Let's go instead with my losing count.) Gerhard "Gets Excited Around Projective Planes" Paseman, 2017.02.25. $\endgroup$
    – Gerhard Paseman
    Commented Feb 25, 2017 at 20:26

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