Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

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

(→‎Video Solution by Richard Rusczyk)
m (→‎Solution)
Line 29: Line 29:
 
Now, Consider the following inequalities:
 
Now, Consider the following inequalities:
  
−
-  <math>a_1 + a_1 > a_2</math>: Since <math>6 > (1+\sqrt{2})^2</math>  
+
<math>\cdot a_1 + a_1 > a_2</math>: Since <math>6 > (1+\sqrt{2})^2</math>  
  
−
-  <math>a_1 + a_1 < a_3</math>.
+
<math>\cdot a_1 + a_1 < a_3</math>.
  
−
-  <math>a_1 + a_2</math> is greater than <math>a_3</math> but less than <math>a_4</math>.
+
<math>\cdot a_1 + a_2</math> is greater than <math>a_3</math> but less than <math>a_4</math>.
  
−
-  <math>a_1 + a_3</math> is greater than <math>a_4</math> but equal to <math>a_5</math>.
+
<math>\cdot a_1 + a_3</math> is greater than <math>a_4</math> but equal to <math>a_5</math>.
  
−
-  <math>a_1 + a_4</math> is greater than <math>a_6</math>.
+
<math>\cdot a_1 + a_4</math> is greater than <math>a_6</math>.
  
−
- <math>a_2+a_2 = 2 = a_6</math>. Then obviously any two segments with at least one them longer than <math>a_2</math> have a sum greater than <math>a_6</math>.
+
<math>\cdot a_2+a_2 = 2 = a_6</math>. Then obviously any two segments with at least one them longer than <math>a_2</math> have a sum greater than <math>a_6</math>.
  
 
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 <math>(a_x, a_y, a_z)</math>:
 
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 <math>(a_x, a_y, a_z)</math>:

Revision as of 14:45, 19 September 2021

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:

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

$\cdot a_1 + a_1 < a_3$.

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

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

$\cdot a_1 + a_4$ is greater than $a_6$.

$\cdot a_2+a_2 = 2 = a_6$. 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

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 $p = 223/286$.

Video Solution by Richard Rusczyk

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

~dolphin7

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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2013_AMC_12A_Problems/Problem_24&oldid=162459"