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
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 $n=2^k$.