Let M lie on the arc overparenA_nA_1 of the circle…

Let $M$ lie on the arc $\overparen{A_nA_1}$ of the circle circumscribing a regular $n$-gon $A_1A_2\cdots A_n$ (the short arc between the two consecutive vertices $A_n$ and $A_1$, containing no other vertex). Prove: $\ 1^\circ$ if $n$ is even, $\displaystyle\sum_{k\text{ even}}|MA_k|^2=\sum_{k\text{ odd}}|MA_k|^2$; $\ 2^\circ$ if $n$ is odd, $\displaystyle\sum_{k\text{ even}}|MA_k|=\sum_{k\text{ odd}}|MA_k|$.

Part $1^\circ$ · $n$ even, squared distances

Step 1 · Set up and take the difference

Center the polygon at the origin, $A_k=Re^{2\pi i(k-1)/n}$, $M=Re^{i\alpha}$. Expanding each $|M-A_k|^2$ as in Problem 116, the even/odd difference is:

$$ S_{\text{even}}-S_{\text{odd}}=-2\,\text{Re}\left(M\left(\sum_{k\text{ even}}\bar A_k-\sum_{k\text{ odd}}\bar A_k\right)\right) $$

(the $|M|^2+|A_k|^2$ parts cancel in the difference since there are equally many even- and odd-indexed vertices when $n$ is even).

Step 2 · The even- and odd-indexed vertices each sum to zero

For $n=2m$ even, the odd-indexed vertices $A_1,A_3,\ldots,A_{2m-1}$ are $R,\ Re^{4\pi i/n},\ Re^{8\pi i/n},\ldots$ — themselves the vertices of a regular $m$-gon (scaled by $R$), so:

$$ \sum_{k\text{ odd}}A_k=R\sum_{j=0}^{m-1}e^{4\pi ij/n}=0 $$

(a geometric sum of the $m$-th roots of unity). Likewise $\sum_{k\text{ even}}A_k=0$ (since the total sum is $0$ and the odd part is already $0$).

Step 3 · Conclusion

Both sums in Step 1 vanish, so $S_{\text{even}}-S_{\text{odd}}=0$ — and this holds for any $M$ (the restriction to the arc $\overparen{A_nA_1}$ is not actually needed for this part):

$$ \boxed{S_{\text{even}}=S_{\text{odd}}=nR^2} $$

(using $S_{\text{even}}+S_{\text{odd}}=2nR^2$ from Problem 116 when $M$ is on the circle).

Part $2^\circ$ · $n$ odd, unsquared distances

Step 4 · Write each distance as a genuine (sign-free) sinusoid

Let $A_k$ sit at angle $\frac{2\pi(k-1)}n$ and let $M=Re^{i\psi}$ with $\psi$ chosen so $M$ lies strictly between $A_n$ and $A_1$ on the short arc. Then:

$$ |MA_k|=2R\left|\sin\left(\frac{\psi-\frac{2\pi(k-1)}n}2\right)\right| $$

Because $M$ is confined to the arc between $A_n$ and $A_1$ (and no other vertex), the quantity $\psi-\frac{2\pi(k-1)}n$ stays within a single interval of width less than $2\pi$ as $k$ runs from $1$ to $n$, so the half-angle stays within $(0,\pi)$ throughout — the sine never changes sign, and the absolute value can be dropped:

$$ |MA_k|=2R\sin\theta_k, \qquad \theta_k=\theta_1-(k-1)\frac\pi n \ \in(0,\pi) $$

Step 5 · Sum the alternating series via a geometric sum

We must show $\displaystyle\sum_{k=1}^n(-1)^{k+1}\sin\theta_k=0$. Writing $\sin\theta_k=\text{Im}\big(e^{i\theta_1}e^{-i(k-1)\pi/n}\big)$ and substituting $m=k-1$:

$$ \sum_{k=1}^n(-1)^{k+1}\sin\theta_k=\text{Im}\left(e^{i\theta_1}\sum_{m=0}^{n-1}(-1)^me^{-im\pi/n}\right)=\text{Im}\left(e^{i\theta_1}\sum_{m=0}^{n-1}r^m\right), \qquad r=-e^{-i\pi/n} $$

Step 6 · The geometric sum vanishes exactly when $n$ is odd

$$ \sum_{m=0}^{n-1}r^m=\frac{r^n-1}{r-1}, \qquad r^n=(-1)^ne^{-i\pi}=(-1)^n\cdot(-1)=(-1)^{n+1} $$

For $n$ odd: $(-1)^{n+1}=1$, so $r^n=1$, making $r^n-1=0$ — the entire geometric sum is $0$ (and $r\neq1$ for $n\ge3$, so no division-by-zero issue).

Step 7 · Conclusion

$$ \sum_{k=1}^n(-1)^{k+1}\sin\theta_k=\text{Im}(e^{i\theta_1}\cdot0)=0 \quad\Longrightarrow\quad \sum_{k=1}^n(-1)^{k+1}|MA_k|=2R\cdot0=0 $$

$$ \boxed{\sum_{k\text{ odd}}|MA_k|=\sum_{k\text{ even}}|MA_k|} \qquad(n \text{ odd}) $$

Verification

Numerically tested for $n=5$ and $n=7$, with $M$ swept across several positions on the arc between $A_n$ and $A_1$: in every case, $D_{\text{even}}$ and $D_{\text{odd}}$ matched to machine precision, while the corresponding squared-distance sums $S_{\text{even}}\neq S_{\text{odd}}$ (confirming Part $1^\circ$'s squared-distance identity genuinely requires $n$ even, and Part $2^\circ$'s unsquared identity genuinely requires $n$ odd).