Three distinct segments are chosen at random among the segments whose end-points are the vertices of a regular 12-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?
$\textbf{(A)} \ \frac{553}{715} \qquad \textbf{(B)} \ \frac{443}{572} \qquad \textbf{(C)} \ \frac{111}{143} \qquad \textbf{(D)} \ \frac{81}{104} \qquad \textbf{(E)} \ \frac{223}{286}$
Suppose $p$ is the answer. We calculate $1-p$.
Assume that the circumradius of the 12-gon is $1$, and the 6 different lengths are $a_1$, $a_2$, $\cdots$, $a_6$, in increasing order. Then
$a_k = 2\sin ( \frac{k\pi}{12} )$. 
So $a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5$, 
$a_2=1$,
$a_3=\sqrt{2}\approx 1.4$, 
$a_4=\sqrt{3}\approx 1.7$,
$a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3$,
$a_6 = 2$.

Now, Consider the following inequalities:
$a_3>2a_1 > a_2$
$a_4> a_1 + a_2>a_3$
$a_4 a_6$
$2a_2 = 2 = a_6$. Thus any two segments with at least one them longer than $a_2$ have a sum greater than $a_6$.
Therefore, all triples (in increasing order) that can't be the side lengths of a triangle are the following. Note that x-y-z means $(a_x, a_y, a_z)$:

1-1-3, 1-1-4, 1-1-5, 1-1-6,
1-2-4, 1-2-5, 1-2-6,
1-3-5, 1-3-6,
2-2-6

Note that there are $12$ segments of each length of $a_1$, $a_2$, $\cdots$, $a_5$, respectively, and $6$ segments of length $a_6$. There are $66$ segments in total.
In the above list there are $3$ triples of the type a-a-b without 6, $2$ triples of a-a-6 where a is not 6, $3$ triples of a-b-c without 6, and $2$ triples of a-b-6 where a, b are not 6. So,
\[1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)\]
\[= \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286}\]
So $p = 223/286$.