The classic Orchard planting problem asks for the maximum number of 3-point straight lines attainable from a configuration of $n$ points drawn on a plane.

Here we are interested in a variant of this problem. What is the maximum number of 4-point circles attainable from a configuration of 10 points drawn on a plane? Each attained circle must pass through at least 4 points.

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    $\begingroup$ For a lower bound, you can take a solution to the 9-point linear orchard planting problem, add the point at infinity, and apply a Möbius transformation on the whole thing to turn all the lines into circles. So that gives you 10 by itself, and it seems that you can wiggle some points around in the original solution to make more circles (which remain circles when Möbius-transformed). $\endgroup$ – Deusovi Aug 31 '20 at 5:41
  • $\begingroup$ Wow I didn't even think of that. Brilliant! $\endgroup$ – Dmitry Kamenetsky Aug 31 '20 at 5:45
  • $\begingroup$ The Wolfram's page you linked contains a table which says that for n=10 points and required k=4 points in a line one can achieve up to 5 lines. So at least 5 circles are possible. $\endgroup$ – CiaPan Aug 31 '20 at 8:06

I can do

$22$: enter image description here

Less aesthetically pleasing but more revealing version

above version was obtained from this by inversion in a circle enter image description here


enter image description here

The construction is as follows: two concentric regular pentagons: This gives $2$ circumscribed circles and $5\times 2 \times 2$ symmetric trapezoids each admitting a circumcircle by symmetry.

Here is a less busy picture---the full is obtained bv overlaying successive rotations by $72°$ and by adding the two circumcircles of the two pentagons. enter image description here

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    $\begingroup$ That is amazing. +1 $\endgroup$ – Jaap Scherphuis Aug 31 '20 at 19:23
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    $\begingroup$ I find it interesting that the sizes of the pentagons don't seem to matter so long as they're concentric. Any two parallel edges have four points that form an isosceles trapezium, which is always inscribable within a circle. Very cool indeed. $\endgroup$ – Ian MacDonald Aug 31 '20 at 20:38
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    $\begingroup$ @DmitryKamenetsky Yep, if I'm not mistaken multiples of $5$ should do nicely: $15\rightarrow 63,20\rightarrow 124\ldots$ $\endgroup$ – Paul Panzer Sep 1 '20 at 4:11
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    $\begingroup$ @DmitryKamenetsky Wait, probably you can squeeze out extra circles by chosing the pentagon radii correctly! I'd guess at least another $5$ for $15$ and another $10$ for $20$. $\endgroup$ – Paul Panzer Sep 1 '20 at 4:26
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    $\begingroup$ @PaulPanzer sounds like this would make a great integer sequence. I'll try to write a program that finds these point placements. $\endgroup$ – Dmitry Kamenetsky Sep 1 '20 at 4:27

The following answer follows the excellent idea by Deusovi in a comment to the question.

Start with a solution to the original 9-tree orchard problem, with 10 lines of 3 trees.

Then add the point at infinity to get 10 points, and 10 lines with 4 points on them, and use a Möbius transformation to change them all to circles with 4 points on them.
In particular, I used points at the following coordinates:
$$\begin{array}{|c|c|c|} \hline Point & Original & Transformed \\ \hline A & \infty & (0,0) \\ \hline B & (1,2) & (1/5,-2/5) \\ \hline C & (2,2) & (1/4, -1/4) \\ \hline D & (3,2) & (3/13,-2/13) \\ \hline E & (0,1) & (0,-1) \\ \hline F & (2,1) & (2/5,-1/5) \\ \hline G & (4,1) & (4/17,-1/17) \\ \hline H & (0,3) & (0,-1/3) \\ \hline I & (2,3) & (2/13,-3/13) \\ \hline J & (4,3) & (4/25,-3/25) \\ \hline \end{array}$$
The last column is the new coordinate after the $z \to 1/z$ transformation of the complex plane, which in cartesian coordinates is the map $(x,y) \to (x/s,-y/s)$ where $s=x^2+y^2$.

The original ten lines then become the ten circles ABCD, AEFG, AHIJ, AHBF, AHCG, AIBE, AICF, AIDG, AJCE, AJDF. I chose the original points such that no line goes through the origin, ensuring that after the transform they are circles rather than straight lines (the origin maps to the point at infinity, and would be contained on any straight line).

The original arrangement also has the circles BDEG, DBHJ, BDIF, EFHI, FGIJ, EGHJ, and they remain circles after the transformation, for a total of 16 circles.

enter image description here

  • $\begingroup$ This is truly superb work! Well done. I am very impressed. Do you think this is optimal? $\endgroup$ – Dmitry Kamenetsky Aug 31 '20 at 10:45
  • $\begingroup$ It took me a while to work out how the transformation works. I believe $(x,y)$ becomes $(x/s, -y/s)$, where s is $x^2+y^2$. Perhaps this is worth adding to the answer. $\endgroup$ – Dmitry Kamenetsky Aug 31 '20 at 11:02
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    $\begingroup$ @DmitryKamenetsky I don't know if this is optimal, but I don't see a way to get more circles from tweaking the original arrangement. Maybe one of the alternative arrangements to the original orchard problem works better. $\endgroup$ – Jaap Scherphuis Aug 31 '20 at 11:23
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    $\begingroup$ It's also possible that the optimal solution can't be derived directly from a 9-line orchard configuration. Maybe there's a 17-circle strategy that doesn't assign any particular point to 9 different circles -- if this is the case, then the inversion won't work. $\endgroup$ – Deusovi Aug 31 '20 at 11:26
  • $\begingroup$ @Deusovi Agreed (but it's a point on 10 circles). With 17 circles, there will be a point on at least 7 circles, but that still leaves many possibilities. $\endgroup$ – Jaap Scherphuis Aug 31 '20 at 11:33

I found

Two more solutions that produce 22 circles, but have a very different structure to the one found earlier.

(2.8,2.4) (3,1) (4,4) (3,2) (1.5,1.5) (3.411764706,1.647058824) (2.333333333,2.333333333) (2,0) (1.692307692,2.461538462) (2.461538462,1.692307692) enter image description here

(3.076923077,2.384615385) (2.068965517,2.172413793) (0.8,1.4) (0,7) (2.702702703,3.216216216) (2,1) (1.176470588,2.294117647) (1.333333333,3) (2,3) (3.529411765,1.117647059) enter image description here

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    $\begingroup$ They are indeed different from mine, but I'm not sure they are different from each other. Cross ratios are the same for both: 14 ⨉ 1/2;8 ⨉ 1/5 (my solution has nonrational Cross ratios which come in two groups of 10 and two groups of 5) $\endgroup$ – Paul Panzer Sep 16 '20 at 5:05

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