For a regular n -gon with circumradius R , prove: 1^° the…
For a regular $n$-gon with circumradius $R$, prove: $\ 1^\circ$ the sum of the squares of all sides and diagonals is $n^2R^2$; $\ 2^\circ$ their sum is $nR\cot\dfrac\pi{2n}$; $\ 3^\circ$ their product is $n^{n/2}R^{n(n-1)/2}$.
Explanation of Method
Every side or diagonal has length $d_m=2R\sin\frac{m\pi}n$ for a "separation" $m=1,\ldots,n-1$ between vertex indices, and there are $n$ pairs at each separation $m<n/2$ (or $n/2$ pairs when $m=n/2$, $n$ even). Summing/multiplying $d_m$ (or $d_m^2$) over all $\binom n2$ pairs reduces to classical sums and products of $\sin\frac{m\pi}n$, each a known closed form.
Part $1^\circ$ · Sum of squared lengths
Step 1 · A general identity for sums of squared pairwise distances
For any points $z_1,\ldots,z_n$: $\ \displaystyle\sum_{j<k}|z_j-z_k|^2=n\sum_j|z_j|^2-\left|\sum_jz_j\right|^2$ (expand $\sum_{j\neq k}|z_j-z_k|^2=2n\sum|z_j|^2-2\left|\sum z_j\right|^2$ and halve).
Step 2 · Apply to the vertices of the regular $n$-gon
With $\sum_j|z_j|^2=nR^2$ and $\sum_jz_j=0$ (Problem 116, Step 3):
$$ \boxed{\sum_{j<k}|z_j-z_k|^2=n(nR^2)-0=n^2R^2} $$
Part $2^\circ$ · Sum of lengths
Step 3 · Sum of distances from one fixed vertex
From $z_0=R$ to all others: $\ \displaystyle\sum_{k=1}^{n-1}|z_0-z_k|=2R\sum_{k=1}^{n-1}\sin\frac{k\pi}n$. Using the classical identity $\displaystyle\sum_{k=1}^{n-1}\sin\frac{k\pi}n=\cot\frac\pi{2n}$ (derivable from the sine-sum formula $\sum_{k=1}^{n-1}\sin k\theta=\frac{\sin\frac{n\theta}2\sin\frac{(n-1)\theta}2}{\sin\frac\theta2}$ at $\theta=\frac\pi n$, which simplifies using $\sin\frac{n\theta}2=\sin\frac\pi2=1$):
$$ \sum_{k=1}^{n-1}|z_0-z_k|=2R\cot\frac\pi{2n} $$
Step 4 · Sum over all vertices, then correct for double-counting
Summing this quantity over all $n$ starting vertices counts every side/diagonal exactly twice:
$$ \boxed{\sum_{j<k}|z_j-z_k|=\frac{n\cdot2R\cot\frac\pi{2n}}2=nR\cot\frac\pi{2n}} $$
Part $3^\circ$ · Product of lengths
Step 5 · Product of distances from one fixed vertex
$$ \prod_{k=1}^{n-1}|z_0-z_k|=(2R)^{n-1}\prod_{k=1}^{n-1}\sin\frac{k\pi}n $$
Using the classical product identity $\displaystyle\prod_{k=1}^{n-1}\sin\frac{k\pi}n=\frac n{2^{n-1}}$ (see Problem 79, Part $1^\circ$'s method with $N=n$):
$$ \prod_{k=1}^{n-1}|z_0-z_k|=(2R)^{n-1}\cdot\frac n{2^{n-1}}=nR^{n-1} $$
Step 6 · Combine over all vertices, correcting for double-counting
Multiplying the above over all $n$ vertices counts each of the $\binom n2$ pairwise distances exactly twice, so:
$$ \left(\prod_{j<k}|z_j-z_k|\right)^2=\prod_{j=0}^{n-1}\left(nR^{n-1}\right)=\left(nR^{n-1}\right)^n $$
$$ \boxed{\prod_{j<k}|z_j-z_k|=\left(nR^{n-1}\right)^{n/2}=n^{n/2}R^{n(n-1)/2}} $$
Verification
For $n=5$, $R=1.7$: direct computation of all $\binom52=10$ pairwise distances gives $\sum d^2=72.25$, matching $n^2R^2=25(1.7)^2=72.25$ ✓; $\sum d=26.1603$, matching $5(1.7)\cot\frac\pi{10}=26.1603$ ✓; $\prod d=11269.75$, matching $5^{5/2}(1.7)^{10}=11269.75$ ✓. All three formulas were checked this way by direct brute-force computation of all $\binom n2$ pairwise distances for $n=3,\ldots,8$ and various $R$: exact agreement in every case.
Result
All three follow from the same strategy: reduce a symmetric sum or product over all $\binom n2$ pairwise distances to a sum or product over the $n-1$ distances from a single vertex, evaluated via classical sine identities, then correct for counting each pair twice.