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Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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== Problem==
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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?
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<math> \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}</math>
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==Solution==
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Suppose <math>p</math> is the answer. We calculate <math>1-p</math>.
 
Suppose <math>p</math> is the answer. We calculate <math>1-p</math>.
  
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So <math>p = 223/286</math>.
 
So <math>p = 223/286</math>.
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== See also ==
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{{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}}

Revision as of 17:55, 22 February 2013

Problem

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

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 $a_1$, $a_2$, $\cdots$, $a_6$, 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 $p = 223/286$.

See also

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by
Problem 23 		Followed by
Problem 25
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 12 Problems and Solutions
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