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}$
Solution
Let $\ell_k=2\sin(15^\circ k)$ be the length of a segment joining two vertices $k$ steps apart, $k=1,\dots,6$: $\ell_1\approx0.518$, $\ell_2=1$, $\ell_3\approx1.414$, $\ell_4\approx1.732$, $\ell_5\approx1.932$, $\ell_6=2$. There are $12$ segments of each type $k=1,\dots,5$ and $6$ of type $6$, i.e. $66$ segments, so there are $\binom{66}3=45760$ choices.
Three segments fail to form a triangle with positive area exactly when the two shorter lengths add up to at most the longest. Checking the types $k_1\le k_2\le k_3$ shows that this happens only for $(1,1,3),(1,1,4),(1,1,5),(1,1,6)$, $(1,2,4),(1,2,5),(1,2,6)$, $(1,3,5),(1,3,6)$ and $(2,2,6)$ (note $\ell_1+\ell_3=\ell_5$ and $2\ell_2=\ell_6$ are degenerate cases).
Counting: $(1,1,k_3)$ with $k_3=3,4,5$ gives $\binom{12}2\cdot12=792$ each, and $k_3=6$ gives $\binom{12}2\cdot6=396$; $(1,2,4),(1,2,5),(1,3,5)$ give $12^3=1728$ each, and $(1,2,6),(1,3,6)$ give $12\cdot12\cdot6=864$ each; $(2,2,6)$ gives $396$. The total is $3\cdot792+396+3\cdot1728+2\cdot864+396=10080$.
The probability is $1-\dfrac{10080}{45760}=\dfrac{223}{286}$.
Three distinct segments are chosen at random among the segments whose endpoints are the vertices of a regular $10$-gon. What is the probability that the lengths of these three segments are the three side lengths of a triangle with positive area?
