12
$\begingroup$

Problem. Several circles are drawn on the plane and all points of their intersection or touching are marked. For which $n$ it is possible that each circle contains exactly $n$ marked points and each point belongs to exactly $n$ circles?

Examples for $n=2,3$ are trivial. For $n=4$ and $5$ there only two known examples on the figures below (we apply an inversion with a center not lying on these lines to obtain the required configurations). Are there examples for $n>5$ and other examples for $n=4,5$?

UPD $n=5$ example arises from stereographic projection of $12$ vertices of an icosahedron and $12$ circles passing through any $5$ vertices incident to the same vertex. Ilya Bogdanov noticed that $n=4$ example arises from stereographic projection of the following configuration of $10$ point and $10$ circles on the sphere. Consider polyhedron which is a convex hull of the following $10$ points: a vertex of some octahedron, $4$ midpoints of edges incident with this vertex, $4$ centers of faces incident with this vertex and the center of the octahedron; and $10$ circles: $8$ circumscribed circles of all faces of the polyhedron, the circle through midpoints and the circle through face centers.

n=4n=5

$\endgroup$
4
  • $\begingroup$ Note that the $n=5$ example here arises from taking an icosahedron on the sphere and using circles centered at every vertex with a radius equal to the (spherical) edge length, then taking a stereographic projection to the plane. The uniqueness of the Platonic solids seems like some evidence that such configurations may be very rare, though as your $n=4$ example shows they don't have to derive from regular polyhedra. $\endgroup$ Aug 26, 2022 at 1:00
  • 2
    $\begingroup$ These are beautiful! I wonder if it might be useful to put them on a sphere and view them as arrangements of planes in space. $\endgroup$ Aug 26, 2022 at 2:09
  • 3
    $\begingroup$ Do you insist that the number of circles and points is finite? Otherwise there is a solution for $n=6$, e.g. https://www.vectorstock.com/royalty-free-vector/seamless-circles-and-hexagons-pattern-vector-2520168 $\endgroup$ Aug 30, 2022 at 13:48
  • 1
    $\begingroup$ Yes, the number of points/circles is finite ("several" means finite). $\endgroup$ Aug 30, 2022 at 17:10

2 Answers 2

3
$\begingroup$

This is just a partial answer.

Notice that in all your examples so far, for any given pair of points in your drawing, there are at most two circles containing both of them. I claim that there is no example with $n \geq 6$ that has this property.

Supposing there were such an example, draw it out (with all vertices drawn -- no points at infinity). You have a non-simple planar graph $G$, non-simple because some of your vertices have two different edges connecting them. We may suppose $G$ is connected, because every connected component of a valid configuration is also a valid configuration. For every pair of doubly connected vertices in $G$, delete one of the two edges connecting them, so that the result is a simple connected planar graph $H$.

If you have $v$ vertices, then you also have $v$ circles in your configuration. Each circle gives you exactly $n$ edges in $G$, for a total of $vn$ edges in $G$. This means there are $\geq\! vn/2$ edges in $H$. But it is well known that, in a planar graph, the number of edges must be $\leq\! 3v-6$. (This formula can be found, for example, here on Wikipedia.) If $n \geq 6$, then the number of edges in $H$ is $\geq\! vn/2 \geq 3v$, so $H$ violates this formula.

$\endgroup$
2
  • $\begingroup$ "For every pair of doubly connected vertices in G, delete one of the two edges connecting them" and then some circles will contribute fewer than $n$ edges, won't they? $\endgroup$ Sep 1, 2022 at 13:35
  • 2
    $\begingroup$ We have exactly $n$ edges in $G$ for each circle, $vn$ edges in total. Then after deleting we obtain new graph $H$ with at least $vn/2$ edges. $\endgroup$ Sep 1, 2022 at 13:42
0
$\begingroup$

You can use a "hypercube". Take n vectors with length 1. Now, make every combination as the sum of the vectors. In general, you now have 2ⁿ different points, which we can represent with "binary". The circles are every circle with radius 1 centred around one of the points. Every point has in general n points with distance 1, which are obtained by changing a single digit of the binary string. This means that every circle contains n points, in general. Analogously, every point lies on n different circles with radius 1, which are the circles centred around the neighbours.

I first came up with a more complicated way, extending the contructions of Clifford's circle theorems (there is a wikipage), also using 2ⁿ points and circles, but then I came up with the easy solution.

$\endgroup$
2
  • $\begingroup$ The question asked about circles in the plane. $\endgroup$ Dec 7, 2022 at 18:10
  • $\begingroup$ In the problem we consider ALL points of intersection (or touching) of circles. So your construction doesn't work. $\endgroup$ Dec 8, 2022 at 7:27

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.