To prove that it is geometric invariant I need to find some others. I was thinking about proving it by the Pythagorean theorem, using the fact that in all cases the distance from the vertex to the center is the same, but I need one more invariant. What will that be? Thank you.
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Is this true? On a square with unit length sides, a line passing through the diagonal will have sum of distances $\sqrt{2}$, but taking a line bisecting an edge will give $2$. – Joppy Oct 06 '17 at 11:33
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For the sum of the signed distances it works and the sum is zero. – Hellen Oct 06 '17 at 11:35
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If you are using a vector sum, then apply a symmetry argument. Whatever the sum of vertices is, it should be unchanged under a $180/n$ degree rotation about the centre, where $n$ is the number of vertices. – Joppy Oct 06 '17 at 11:42
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The question amounts to show that $$\sum_{k=0}^{n-1}\cos^2\left(\frac{2k\pi}n+\phi\right)$$ is independent of $\phi$.
Differentiating on $\phi$, $$-2\sum_{k=0}^{n-1}\cos\left(\frac{2k\pi}n+\phi\right)\sin\left(\frac{2k\pi}n+\phi\right)=-4\sum_{k=0}^{n-1}\sin\left(\frac{4k\pi}n+\phi\right)=0.$$
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Assume $n\geq3$ and put $\omega:=e^{2\pi i/n}$. The complex numbers $w_k:=\omega^k$, $\> 0\leq k\leq n-1$, then form a regular $n$-gon in standard position. Let an $\alpha\in{\mathbb R}$ be given and consider the real parts of the slightly rotated $w_k$, i.e., the numbers $$x_k:={\rm Re}(e^{i\alpha}w_k)={1\over2}\bigl(e^{i\alpha}\omega^k+e^{-i\alpha}\bar \omega^k\bigr)\qquad(0\leq k\leq n-1)\ .$$ One has $$x_k^2={1\over4}\bigl(e^{2i\alpha}\omega^{2k}+2+e^{-2i\alpha}\bar\omega^{2k}\bigr)\ .$$ Since $\ {\displaystyle\sum_{k=0}^{n-1}\omega^{2k}={\omega^{2n}-1\over\omega^2-1}=0}$ we obtain $\ {\displaystyle\sum_{k=0}^{n-1}x_k^2={n\over2}}$, independently of $\alpha$.
Christian Blatter
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