Let [tex]p_i[/tex] denote the vector starting at the origin and ending at the vertex [tex]P_i[/tex] of the 12-gon. There is an angle of (360/12)º = 30º between consecutive vectors.
Recall that for any two vectors [tex]u,v[/tex], we have
[tex]u\cdot v=\|u\|\|v\|\cos\theta[/tex]
with [tex]\theta[/tex] the angle between the two vectors. Also recall that
[tex]\|v\|^2=v\cdot v[/tex]
For [tex]1\le i<j\le12[/tex], [tex]P_iP_j[/tex] is the length of the vector [tex]p_i-p_j[/tex]. So
[tex](P_iP_j)^2=(p_i-p_j)\cdot(p_i-p_j)=p_i\cdot p_i-2p_i\cdot p_j+p_j\cdot p_j[/tex]
The 12-gon is inscribed in a circle of radius 1, which means each vector [tex]p_i[/tex] has length 1, and from this we have
[tex](P_iP_j)^2=2-2p_i\cdot p_j=2-2\cos\theta_{i,j}[/tex]
where [tex]\theta_{i,j}[/tex] is the angle between vectors [tex]p_i[/tex] and [tex]p_j[/tex] with [tex]i\neq j[/tex], and these angles are multiples of 30º.
There are [tex]\binom{12}2=66[/tex] terms in the sum (from 12 total vertices, you take 2 at a time).
- 11 of these terms are the squared distances between consecutive vertices and separated by 30º, equal to [tex](P_1P_2)^2[/tex];
- 10 of them are the squared distances between vertices that are two vertices apart, separated by 60º, equal to [tex](P_1P_3)^2[/tex];
- 9 of them are the squared distances between vertices that are three vertices apart, separated by 90º, equal to [tex](P_1P_4)^2[/tex];
- and so on, down to the 1 remaining uncounted squared distance between vertices that are ten vertices apart, separated by 330º, [tex](P_1P_{11})^2[/tex].
So we have
[tex]\displaystyle\sum_{1\le i<j\le12}(P_iP_j)^2=\sum_{n=1}^{11}(12-n)(P_1P_n)^2[/tex]
[tex]=\displaystyle2\sum_{n=1}^{11}(12-n)(1-\cos(30n)^\circ)=\boxed{144}[/tex]
