Given a regular 10-gon inscribed in a circle with radius 1, we need to compute the sum of the squares of the distances between all pairs of vertices, i.e.,
\[
\sum_{1 \le i < j \le 10} (P_i P_j)^2.
\]
Each pairwise distance \(P_iP_j\) corresponds to a chord of the circle. The vertices \(P_1, P_2, \ldots, P_{10}\) are the 10th roots of unity in the complex plane, represented as \(1, \omega, \omega^2, \ldots, \omega^9\), where \(\omega = e^{2\pi i / 10}\).
The distance between two vertices \(P_i = \omega^i\) and \(P_j = \omega^j\) is given by the magnitude of the difference between their corresponding complex numbers:
\[
P_i P_j = |\omega^i - \omega^j|.
\]
To compute \((P_i P_j)^2\), we use:
\[
(P_i P_j)^2 = |\omega^i - \omega^j|^2.
\]
Since \(|z|^2 = z \overline{z}\) for any complex number \(z\), we have:
\[
|\omega^i - \omega^j|^2 = (\omega^i - \omega^j)(\overline{\omega^i} - \overline{\omega^j}).
\]
Because \(\omega\) is a root of unity, we know \(\overline{\omega} = \omega^{-1}\), so:
\[
\overline{\omega^i} = \omega^{-i}.
\]
Thus,
\[
|\omega^i - \omega^j|^2 = (\omega^i - \omega^j)(\omega^{-i} - \omega^{-j}) = 2 - \omega^{i-j} - \omega^{j-i}.
\]
This is equivalent to:
\[
|\omega^i - \omega^j|^2 = 2 - 2 \operatorname{Re}(\omega^{i-j}),
\]
where \(\operatorname{Re}(\omega^{i-j})\) denotes the real part of \(\omega^{i-j}\). Since \(\omega^{i-j}\) lies on the unit circle, \(\operatorname{Re}(\omega^{i-j}) = \cos\left(\frac{2\pi (i-j)}{10}\right)\).
Hence,
\[
|\omega^i - \omega^j|^2 = 2 - 2 \cos\left(\frac{2\pi (i-j)}{10}\right) = 4 \sin^2 \left(\frac{\pi (i-j)}{10}\right).
\]
The sum we are interested in is:
\[
\sum_{1 \le i < j \le 10} 4 \sin^2 \left(\frac{\pi (i-j)}{10}\right).
\]
To find this sum, we sum over all pairs \( (i, j) \) such that \( 1 \le i < j \le 10 \). Each \(\sin^2 \left(\frac{\pi k}{10}\right)\) where \( k = 1, 2, \ldots, 9 \) appears exactly 10 times because there are 10 vertices and each \( k \) corresponds to the distance between points \( k \) steps apart.
Thus, we have:
\[
\sum_{k=1}^9 \sin^2 \left(\frac{\pi k}{10}\right).
\]
Using the identity:
\[
\sum_{k=1}^{n-1} \sin^2 \left(\frac{\pi k}{n}\right) = \frac{n}{2},
\]
for \( n = 10 \), we get:
\[
\sum_{k=1}^9 \sin^2 \left(\frac{\pi k}{10}\right) = \frac{10}{2} = 5.
\]
Since each \(\sin^2 \left(\frac{\pi k}{10}\right)\) appears 10 times in our original sum, the total sum is:
\[
10 \times 4 \times 5 = 200.
\]
Therefore, the final answer is 200.
