Nice question, classical. Throw this onto the complex plane, with the circle as the unit circle. Then $P_i=\omega^i$, where $\omega$ is a primitive tenth root of unity. We want
$$\sum_{1\le i
Note $|\omega^j-\omega^i|^2=|\omega^{j-i}-1|^2=(\cos(2\pi (j-i)/10)-1)^2+\sin(2\pi (j-i)/10)^2=2-2\cos(2\pi (j-i)/10)=2-2\Re(\omega^{j-i})$. The sum is then
$$2\binom{10}{2}-2\Re\left(\sum_{1\le i By symmetry, the inner sum is
$$\frac{1}{2}\sum_{i\ne j}\omega^j\omega^{-i}.$$
If we add in terms where $i=j$, this is $\sum \omega^j \sum \omega^{-i} = 0$, because the roots of unity sum to $0$. Hence the sum is $\frac{1}{2}(0-10)=-5$. Therefore, our final summation is
$$2\binom{10}{2}-2\cdot 5=90-10=\boxed{80}.$$
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