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?
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: $a_3>2a_1 > a_2$ $a_4> a_1 + a_2>a_3$ $a_4<a_1 + a_3=a_5$ $a_1 + a_4 > a_6$ $2a_2 = 2 = a_6$. 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, \[1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)\] \[= \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286}\] $\fbox{\frac{223}{286}}$.