Let $A_1A_2....A_n$ be the regular polygon incribed a circle $(O)$ with radius $R$ what is value $k=\frac{\sum_{i

Question

Let $A_1A_2....A_n$ be the regular polygon incribed a circle $(O)$ with radius $R$, $\alpha$ is real number, let $$k=\frac{\sum_{i<j} A_iA_j^\alpha}{R^\alpha}$$

My question: I am looking for value of $k=?$

Answer 1

If $A_1A_2....A_n$ is a regular convex polygon incribed a circle $(O)$ with radius $R$, then find $$k=\frac{\sum_{i<j} A_iA_j^\alpha}{R^\alpha}$$ , where $\alpha\in\mathbb{R}$. Put $\alpha=1$ at the end.

Essentially, $k$ is just the sum of all the sides and diagonals in a regular convex polygon, each raised to the power of some $\alpha\in\mathbb{R}$, divided by $R^\alpha$.

By drawing all the possible lines of some such polygons, you can conclude$^1$ that the ``length types'' in a polygon of side $n$ are as follows:

CASE I: $n=2m-1,\ m\in\mathbb{N}-\{1\}$

\begin{array}{|c|c|c|} \hline \text{Type of line} & \text{Quantity} & \text{Length of line}\\ \hline \overline{A_1A_{1+l}},\ l=1,2,\ldots,m-1 & n & 2R\sin\left(\displaystyle\frac{l\pi}{n}\right)\\ \hline \end{array} \begin{align*} k&= \frac{\displaystyle\sum_{l=1}^{m-1}\left[n\times 2R\sin\left(\displaystyle\frac{l\pi}{n}\right)\right]^\alpha}{R^\alpha}\\ k&=(2n)^\alpha\displaystyle\sum_{l=1}^{m-1}\sin^\alpha\left(\displaystyle\frac{l\pi}{n}\right)\\ \text{For }\alpha=1,\ k&=2n\displaystyle\sum_{l=1}^{m-1}\sin\left(\displaystyle\frac{l\pi}{n}\right)\\ k&=n\displaystyle\cot\frac{\pi}{2n}\ \forall\ n=2m-1,\ m\in\mathbb N-\{1\}\tag{1} \end{align*} Proceed as here to get equation $(1)$. Nudge me should you face any difficulty simplifying.

CASE II: $n=2m,\ m\in\mathbb{N}-\{1\}$

\begin{array}{|c|c|} \hline \text{Type of line} & \text{Quantity} & \text{Length of line} \\ \hline \overline{A_1A_{1+l}},\ l=1,2,\ldots,m-1 & n & 2R\sin\left(\displaystyle\frac{l\pi}{n}\right)\\ \hline \overline{A_1A_{1+m}} & \frac n2 & 2R\sin\left(\displaystyle\frac{m\pi}{n}\right)=2R\\ \hline \end{array} Increasing just one side from the previous case adds only a term $(\frac n22R)^\alpha=(nR)^\alpha$ in the numerator as follows: \begin{align*} k&= \frac{\displaystyle\sum_{l=1}^{m-1}\left[n\times 2R\sin\left(\displaystyle\frac{l\pi}{n}\right)\right]^\alpha+(nR)^\alpha}{R^\alpha}\\ k&=n\displaystyle\cot\frac{\pi}{2n}\ \forall\ n=2m,\ m\in\mathbb N-\{1\} \end{align*} by similar procedure.

Finally, $$\boxed{k|_{\alpha=1}=n\displaystyle\cot\frac{\pi}{2n}\ \forall\ n\in\mathbb N-\{1,2\}}$$

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$^1$ One can prove it using permutations and combinations.