Distances of points around unit circle $P_1,\cdots , P_{10}$ are ten points on the unit circle 
What  is the largest possible value of the quantity
$$\sum_{1\le i<j\le 10} |P_i-P_j|^2$$
 A: We may just use the identity:
$$\|P_1+P_2+\ldots+P_n\|^2 = \sum_{k=1}^{n}\|P_k\|^2+2\sum_{1\leq i < j\leq n}\langle P_i,P_j\rangle \tag{1}$$
from which it follows that:
$$\sum_{1\leq i<j\leq n}\|P_i-P_j\|^2 = 2(n-1)\sum_{k=1}^{n}\|P_k\|^2-2\sum_{1\leq i<j\leq n}\langle P_i,P_j\rangle\tag{2}$$ can be written as:
$$\sum_{1\leq i<j\leq n}\|P_i-P_j\|^2 =\color{red}{-\|P_1+\ldots P_n\|^2}+\color{blue}{(2n-1)\sum_{k=1}^{n}\|P_k\|^2}.\tag{3} $$
If every $P_k$ lies on the boundary of a unit ball, the blue term is just $\color{blue}{n(2n-1)}$.
The red term is always negative, unless the centroid of $P_1,\ldots,P_n$ is the centre of the previous ball.
Now the question is trivial and, interestingly, dimension-independent.
A: $\newcommand{\angles}[1]{\left\langle\,{#1}\,\right\rangle}
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\begin{align}
\sum_{i = 1}^{9}\sum_{j = i + 1}^{10}
\verts{\expo{\ic\theta_{i}} - \expo{\ic\theta_{j}}}^{\color{#f00}{2}} & =
\sum_{i = 1}^{9}\sum_{j = i + 1}^{10}\pars{\expo{\ic\theta_{i}} - \expo{\ic\theta_{j}}}\pars{\expo{-\ic\theta_{i}} - \expo{-\ic\theta_{j}}}
\\[5mm] & =
\sum_{i = 1}^{9}\sum_{j = i + 1}^{10}\bracks{2 - 2\cos\pars{\theta_{i} - \theta_{j}}}=
4\sum_{i = 1}^{9}\sum_{j = i + 1}^{10}\,\,\,
\sin^{2}\pars{\theta_{i} - \theta_{j} \over 2}
\\[5mm] & \leq
4\sum_{i = 1}^{9}\pars{10 - i} =
4\pars{9 \times 10 - \half \times 9 \times 10} = \color{#f00}{180}
\end{align}
