Difference between revisions of "2013 AMC 12A Problems/Problem 24"

Revision as of 14:49, 8 November 2022

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

$a_2=1$ ,

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

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

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

$a_6 = 2$ .

Now, Consider the following inequalities:

$a_3>2a_1 > a_2$

$a_4> a_1 + a_2>a_3$

$a_4<a_1 + a_3=a_5$

$a_1 + a_4 > a_6$

$2a_2 = 2 = a_6$ . Thus 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

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.

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 $p = 223/286$ .

Solution 2

Just like Solution 1, we find $1-p$ . Let $a_n$ be the length of the diagonal that connects vertices that are $n$ sides apart. By the law of Cosines, we have $a_n^2=1+2a_n^2$ WLOG $a_1=1$ , then we can use the well known fact that $a_n=\sqrt{2^n-1}$ holds, thus $a_2=\sqrt{3}, a_2=\sqrt{7}, a_2=\sqrt{15}, a_2=\sqrt{31}, a_6=\sqrt{63}$ . Let the sides of the triangle be $p\le{r}\le{s}$ . It is obvious that if $x+y\ge{z}$ , then $a_x+a_y>a_z$ too. Thus, we need only check $p=1, p=2$ . If $p=1$ : The ordered pairs $(r,s) = (1,3), (1,4), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 5), (3, 6)$ do not form a triangle for a total of $9$ cases.

Now continue as in Solution 1.

Video Solution by Richard Rusczyk

https://artofproblemsolving.com/videos/amc/2013amc12a/364

~dolphin7

@@ Line 55: / Line 55: @@
 So <math>p = 223/286</math>.
+==Solution 2==
+Just like Solution 1, we find <math>1-p</math>. Let <math>a_n</math> be the length of the diagonal that connects vertices that are <math>n</math> sides apart. By the law of Cosines, we have <cmath>a_n^2=1+2a_n^2</cmath>  WLOG <math>a_1=1</math>, then we can use the well known fact that <math>a_n=\sqrt{2^n-1}</math> holds, thus <math>a_2=\sqrt{3}, a_2=\sqrt{7}, a_2=\sqrt{15}, a_2=\sqrt{31}, a_6=\sqrt{63}</math>. Let the sides of the triangle be <math>p\le{r}\le{s}</math>. It is obvious that if <math>x+y\ge{z}</math>, then <math>a_x+a_y>a_z</math> too. Thus, we need only check <math>p=1, p=2</math>.
+If <math>p=1</math>: The ordered pairs <math>(r,s) = (1,3), (1,4), (1,5), (1,6), (2, 4), (2, 5), (2, 6), (3, 5), (3, 6)</math> do not form a triangle for a total of <math>9</math> cases.
+Now continue as in Solution 1.
 ==Video Solution by Richard Rusczyk==

2013 AMC 12A (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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