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?

$\dfrac{553}{715}$

$\dfrac{443}{572}$

$\dfrac{111}{143}$

$\dfrac{81}{104}$

$\dfrac{223}{286}$

Answer: E

Concepts:regular polygon triangle inequality complementary counting

Difficulty rating: 2650

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 $\begin{gathered} (1,1,3),(1,1,4),(1,1,5), \\ (1,1,6),(1,2,4),(1,2,5), \\ (1,2,6),(1,3,5),(1,3,6), \\ (2,2,6). \end{gathered}$

The first three triples ending in $3,4,5$ contribute $3\binom{12}{2}12;$ the two repeated-length triples ending in $6$ contribute $2\binom{12}{2}6;$ the three triples of distinct non-diameter lengths contribute $3\cdot12^3;$ and the two remaining diameter triples contribute $2\cdot12^2\cdot6.$ Dividing their sum by $\binom{66}{3}$ gives failure probability $63/286,$ so the answer is $1-63/286=223/286.$

Thus, the correct answer is E.

2013 AMC 12A Problem 24

Answer: E

Difficulty rating: 2650

Solution:

Problem 24 in Other Years