2013 AMC 12A Problem 24

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}

There are

6

possible lengths

d_k = 2\sin(15k^\circ);

count how many segments have each length

Use complementary counting: subtract the triples whose longest length is at least the sum of the other two

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

\frac{63}{286},

so the answer is

1-\frac{63}{286}=\frac{223}{286}.

Thus, the correct answer is E.

Problem 24 in other years