Question

Three distinct segments are chosen at random among the segments whose endpoints 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?

$\dfrac{553}{715}$

$\dfrac{443}{572}$

$\dfrac{111}{143}$

$\dfrac{81}{104}$

$\dfrac{223}{286}$

Solution:

Inscribe the $12$ -gon in a unit circle. The segment lengths are $d_k = 2\sin(15k^\circ)$ for $1 \le k \le 6,$ with $12$ segments of each length $d_1, \ldots, d_5$ and $6$ of length $d_6.$

Comparing sums, the forbidden index triples $(a, b, c)$ with $d_a \le d_b \le d_c$ and $d_c \ge d_a + d_b$ are $(1,1,3),(1,1,4),(1,1,5),(1,1,6),(1,2,4),(1,2,5),(1,2,6),(1,3,5),(1,3,6),(2,2,6).$

Counting the corresponding segment selections and dividing by $\binom{66}{3}$ gives a failure probability of $\dfrac{63}{286},$ so the answer is $1 - \dfrac{63}{286} = \dfrac{223}{286}.$

Thus, the correct answer is E.

2013 AMC 12A Problem 24

Solution:

Problem 24 in Other Years