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Let $A_1, A_2,\ldots, A_{2n}$ be $2n$ points on a circle centered at $O$ with the additional property that the centroid of this set of points coincides with $O$. In other words, the sum of the vectors $OA_1$, $OA_2,\ldots OA_{2n}$ is zero.

Prove or disprove:

there exist three lines $L_1$, $L_2$ and $L_3$ through $O$ with the following properties:

a). For each $1\le i\le 3$, $L_i$ is a halving line - this means that exactly $n$ points lie in each of the half-planes determined by $L_i$.

b). The angle between any two of these three lines is exactly 60 degrees.

Any help would be greatly appreciated.

Thanks,

Dan

Dan Ismailescu
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1 Answers1

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B is clearly false. Take, for example, a regular octagon. Lots of hexagons work too, if opposite vertices are diagonally opposite, and they are not regularly spaced out.

A is false, at least for odd $n$. Here's an easy counter example.
Let $\omega$ be a nth root of unity. Let $\theta$ be any complex number with norm 1 that is not the negative of an nth root of unity. Consider your points on the complex plane. The points $\omega^i, \theta\omega^i$ for $i=1$ to $n$ are $2n$ points that add up to 0 (since the individual sums do), and there are no 2 points which are diametrically opposite each other. Hence there are no candidates for $L_i$, and we don't even need to check if they could half the set.

It is easy to see how to extend this counter example for any $n$ with an odd divisor. I'm not certain how to proceed for $n=2^k$.

Calvin Lin
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  • Hi Calvin and thank you for your interest. I am afraid I do not quite understand your argument. If one has a centrally symmetric $2n$-gon then every single diameter is a halving line. So in this case any three diameters forming 60 degree angles will work. One thing that I should probably mention that one can discard the cases when the points are nicely spaced. For all I care one can perturb the points a little bit so that no circular arc determined by any two of them is a rational multiple of Pi. Also, I am interested in a quantitative answer rather than a qualitative one. I'll explain below. – Dan Ismailescu Jun 28 '13 at 20:28
  • I intend to prove a statement of the following kind: Given $2n$ points on a circle draw three diameters forming 60 degree angles with each other. There are six 60 degree arcs formed, let's call them $R_1$, $R_2,\ldots R_6$, labeled counterclockwise. Each of these arcs contains a certain number of points, say, $n_1$, $n_2,\ldots, n_6$. The type of result I am interested in is this: can you always draw these diameters such that $n_1$ and $n_4$ are about the same, $n_2$ and $n_5$ are about the same and $n_3$ and $n_6$ are about the same? Please let me know if this makes sense to you. Thanks. – Dan Ismailescu Jun 28 '13 at 20:38
  • @DanIsmailescu For the simple hexagon example, consider $0, 1, 2, 180, 181, 182$. Though the lines exist, none of them are 60 degrees apart. As for the second case, let me give you 10 points (in terms of angles/degrees) that don't satisfy your conditions. $0, 1, 72, 73, 144, 145, 216, 217, 288, 289$. They do sum up to 0 (due to roots of unity), but none of them are diametrically opposite to each other, more are they $60$ degrees apart. Hence, your statement is false. – Calvin Lin Jun 28 '13 at 21:15
  • Ok, here is the misunderstanding: the diameters do not have to have endpoints among the given points. So if the hexagon is 0,1,2,180,181 and 182 I can draw the diameters 3-183, 63-243 and 123-303 and each of them is a halving line. – Dan Ismailescu Jun 28 '13 at 21:48
  • @DanIsmailescu Even with that interpretation, there is no way for you to split apart the $0, 1, 2, ...$ set with the 60 degree requirement. The points $0,1$ or $1,2$ will always be in the same sector. – Calvin Lin Jun 28 '13 at 23:26
  • I do not try to split those points. I just want each diameter to be a halving line - same number of points on each side. Of course, the points can be bunched up close together but that's not the issue. – Dan Ismailescu Jun 29 '13 at 01:22