Let $k$ and $n$ be two fixed integers. Let $C$ denotes the circle with radius $4n$ (in the plane $\mathbb{R}^2$). Suppose $\{C_1,C_2\}$ shows the set of two arbitrary tangent circles with radius $2n$ in $C$. Also, let $\{C_{11},C_{12}\}$ and $\{C_{21},C_{22}\}$ be the sets of two arbitrary tangent circles with radius $n$ in $C_1$ and $C_2$, respectively. Is there a finite number of points with size $k$ in the circle $C$ such that each $C_1$ and $C_2$ contains the odd number of points and each $C_{11}$, $C_{12}$, $C_{21}$ and $C_{22}$ contains the even number of points?

In the following I drew a time-fixed picture of the problem, since actually the circles can revolve in the original circles by the condition of being tangent in each time:

enter image description here

Motivation: Actually, one of my friends is working on the effects of critical points in the bounded area in the plane. He is engineer and do not like abstract mathematics. Based on his problem, I determine some special points in the bounded area and I give some properties to these points. After that, we define these points with their properties in the computer and we compute some special parameters of the phenomenon on that bounded area by some simple line integral and other mathematical tools. Actually, I helped him for all of this procedure. But in my private time, I abstracted this problem to the version which you can see.

In my opinion, for arbitrary $k$ the answer is no, and if these points exist, I think they have symmetry in the plane.

I think in general, the bellow claim is true:

Let $S$ be a set of points which is distributed in the circle $C$ such a way that any two tangent circles $C_1$ and $C_2$ in $C$ contain the even number of points of $S$. Then $S$ contains even number of points.

I appreciate any answer or helpful comment.

  • $\begingroup$ Does a circle $C$ "contain" a point $x$ when $x$ lies directly on $C$? Or do you mean "strictly contain," i.e., $x$ must lie in the interior of $C$? $\endgroup$ May 30, 2015 at 15:52
  • $\begingroup$ I mean $C$ contain strictly the point $x$, i.e, I like the interior case. $\endgroup$
    – Shahrooz
    May 30, 2015 at 17:40
  • 2
    $\begingroup$ To get a parity condition while the circles rotate, you can arrange for an odd number of points to enter the circle whenever an odd number of points leave. However, it seems like choosing a generic tilt for $C_1$ should prevent the points from being arranged that way for $C_{11}$. $\endgroup$ May 30, 2015 at 21:59

2 Answers 2


This is not a answer, just an animation illustrating the challenge. The blue circle $C$ contains $9$ points, and the red circles $C_1$ and $C_2$ each contain $2$ points, except at four discrete times (times corresponding to rotations that are multiples of $\pi/4$) when they contain just $1$ each (e.g., at the start/end).

          (Reload the page to repeat the animation.)

  • $\begingroup$ Nice animation dear Joseph, thanks. $\endgroup$
    – Shahrooz
    Jun 7, 2015 at 21:26

Do I misunderstand something, or such an easy argument works?

1. Assume that such set $S$ of $k$ points exists. Choose an arbitrary position of $C_1$ such that its center $O$ is not in $S$. The rest points inside $C_1$ are split into (open) semicircles (of possible circles $C_{11}$) joining $O$ with the boundary of $C_1$ and `facing clockwise'. Each such semicircle $c$ should contain an even number of points of $S$, otherwise the position of $C_{11}$ containing $c$ and that just a bit clockwise tp it contain numbers of points having different parities. Thus the total number of points inside $C_1$ is also even.

REMARK. The condition of each semicircle $c$ containing an even number of points in $S$, along with the similar condition for counterclockwise semicircles, is equivalent to that of every $C_{11}$ containing a number of points in $S$ of fixed parity. Such examples can be easily constructed in many ways [here was a not-quite-correct explanation], one of which is shown in the picture below. A dashed circle contains an even number of green points, and while it rotates around the center it gets and loses an even number of them at every moment.

Points distribution

2. The same argument works for the general question about $C$, $C_1$, and $C_2$, unless the center $O$ of $C$ lies in $S$. As far as I understand, in this question you do not fix the radii of $C_1$ and $C_2$, otherwise an example from the Remark above can be completed by $O$ (In the picture below, there is an odd number of green points, counting the center).

If the radii may vary and $O$ is in $S$, then one may choose $C_1$ and $C_2$ of equal radii so that they do not pass through other points of $S$, and then increase one radius and decrease the other so as to include $O$ inside $C_1$.

  • $\begingroup$ The circles $C_i$ is in the circle $C$ and also all of them are tongent. I can not understand how you claim the radii of circles $C_1$ and $C_2$ are not fix. When you fix $n$, all radiuces are fixed. Can you draw a picture for such distribution of points? $\endgroup$
    – Shahrooz
    Aug 27, 2015 at 19:13
  • $\begingroup$ In the claim at the end of your post, you say nothing about radii. If you meant that the radii of the $C_i$ are fixed then this claim seems to be wrong. $\endgroup$ Aug 27, 2015 at 19:21
  • $\begingroup$ On the other hand, I was also a bit wrong in describing the example, sorry. I'll correct it and make a picture quite soon. $\endgroup$ Aug 27, 2015 at 19:22
  • $\begingroup$ @Bogdanov, I mentioned about the raduces at the begining of the question. I am so curious to see your picture. $\endgroup$
    – Shahrooz
    Aug 27, 2015 at 19:41
  • $\begingroup$ Here you are; sorry again for that mess in the initial remark. $\endgroup$ Aug 27, 2015 at 20:57

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