2013 AMC 12A Problems/Problem 24

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 $a_1$ , $a_2$ , $\cdot$ , $a_6$ , in increasing order. Then

$a_k = 2\sin ( \frac{k\pi}{12} )$ .

So $a_1=(\sqrt{6}-\sqrt{2})/2 \approx 0.5$ ,

$a_2=1$ ,

$a_3=\sqrt{2}\approx 1.4$ ,

$a_4=\sqrt{3}\approx 1.7$ ,

$a_5=(\sqrt{6}+\sqrt{2})/2 = a_1 + a_3$ ,

$a_6 = 2$ .

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.

Now, Consider the following inequalities:

- $a_1 + a_1 > a_2$ : Since $6 > (1+\sqrt{2})^2$

- $a_1 + a_1 < a_3$ .

- $a_1 + a_2$ is greater than $a_3$ but less than $a_4$ .

- $a_1 + a_3$ is greater than $a_4$ but equal to $a_5$ .

- $a_1 + a_4$ is greater than $a_6$ .

- $a_2+a_2 = 2 = a_6$ . Then obviously 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

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,

$1-p = \frac{1}{66\cdot 65\cdot 64} ( 3\cdot 3 \cdot 12\cdot 11\cdot 12 + 2\cdot 3 \cdot 12\cdot 11\cdot 6 + 3\cdot 6\cdot 12^3 + 2\cdot 6 \cdot 12^2 \cdot 6)$

$= \frac{1}{66\cdot 65\cdot 64} (12^2 (99+33) + 12^3(18+6)) = \frac{1}{66\cdot 65\cdot 64} (12^3 \cdot 35) = \frac{63}{286}$

So $p = 223/286$ .

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2013 AMC 12A Problems/Problem 24