What is the maximum number of sum (over all vertices) of number of distinct circles passing through at least three vertices of a convex polygon ($n$-gon), if the center of each circle required to belong to the set of vertices of the polygon?

In other words,

If we define a "centroid" by a quadruple of vertices $(a,b,c,d)$ such that $a$ is the center of a circle and three other vertices $b$, $c$, and $d$ are on circumference of this circle (so, $|ab|=|ac|=|ad|$), then what we want is the maximum number of centroids in a convex $n$-gon. (centroids are defined here https://arxiv.org/abs/1009.2218)

Any suggestion? I guess it should be of order $O(n)$. first i think there exists some 'circular order' for all the centroids around the polygon so it may be of order of n,secondly some results like Bose theorem ( see paper:"The Extremal spheres theorem " by O.Musin et al https://www.sciencedirect.com/science/article/pii/S0012365X10003997 ) suggests that number of some class of circles passing through three vertices is of order n-2. but I have no idea how to prove it.

Help me, thanks.

  • $\begingroup$ You should always try to include some of your thoughts about a problem. (What's the basis of your guess of order $O(n)$?) Even knowing where the problem came from (textbook exercise? contest? online challenge? your own brain?) can be helpful, as well as some idea of what tools are expected to be used. The more you can say, the better. This information helps answerers avoid wasting time (theirs and yours) telling you things you already know, duplicating your effort, or using techniques with which you are not familiar. (Edit your question to add clarifications. Comments are easily overlooked.) $\endgroup$ – Blue Jul 22 at 20:50

If $n$ is even, and $m=n/2$ is odd, there can be $m$ circles. Put $m$ points in a regular polygon, then each point is the same distance from the two furthest points.
Draw a circular arc connecting the furthest points, and add another point anywhere on the arc. Do that for each of the $m$ initial points and you get $n=2m$ points and $m$ circles.

  • $\begingroup$ what about a given convex polygon , its irregular .... $\endgroup$ – Mehrdad Jul 26 at 10:11
  • $\begingroup$ may be we can use this method for irregular polygon too, start with furthest points?? suggestion? $\endgroup$ – Mehrdad Jul 26 at 10:13

Subsume first that the 3 given vertices are not on a single line.

Then any pair of those 3 points defines a unique mid line. Any 2 of those will intersect. Btw. the third one will run through that very point too. This provides a single unique circle for this set of 3 given points. Whether or not this midpoint will be a further vertex of the given polygon has to be checked separately. Therefore the requested number could either be 1 or 0.

The maximum therefore, running over all possible polygons, would be simply 1.

Wrt. the excluded case above, it depends onto your geometry of infinity. If each line would have 2 points of infinity (on either end), then the orthogonal to the line through those 3 collinear points would provide 2 different centers of thus 2 different circles of infinite radius. Again those ends have to be checked to be incident to the vertex set of the given polygon. OTOH, if you'd identify those ends of each line, then you clearly will have a single circle again.

--- rk

  • $\begingroup$ thanks but i mean sum over all vertices of the polygon $\endgroup$ – Mehrdad Jul 22 at 20:07
  • $\begingroup$ in some references these circles are called "centroids" but i avoid calling centroid because it may confuse with the center of mass. $\endgroup$ – Mehrdad Jul 22 at 20:10
  • $\begingroup$ a centroid is characterized by a quadruple of vertices (a,b,c,d)such that a is center of a circle and three other vertices are on circumference of this circle so ab=ac=ad $\endgroup$ – Mehrdad Jul 22 at 20:15

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