Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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==Solution 1 ALT== | ==Solution 1 ALT== | ||
− | 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+ | + | 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-1}^2</cmath> WLOG let <math>a_1 = 1</math>, then it is a well known fact that <math>a_n=\sqrt{2^n-1}</math> holds as an explicit formula. Thus we have <math>a_1=1, a_2=\sqrt{3}, a_3=\sqrt{7}, a_4=\sqrt{15}, a_5=\sqrt{31}, a_6=\sqrt{63}</math> |
Now continue as in Solution 1. | Now continue as in Solution 1. |
Revision as of 14:53, 8 November 2022
Contents
[hide]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?
Solution
Suppose is the answer. We calculate .
Assume that the circumradius of the 12-gon is , and the 6 different lengths are , , , , in increasing order. Then
.
So ,
,
,
,
,
.
Now, Consider the following inequalities:
. Thus any two segments with at least one them longer than have a sum greater than .
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 :
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 segments of each length of , , , , respectively, and segments of length . There are segments in total.
In the above list there are triples of the type a-a-b without 6, triples of a-a-6 where a is not 6, triples of a-b-c without 6, and triples of a-b-6 where a, b are not 6. So,
So .
Solution 1 ALT
Just like Solution 1, we find . Let be the length of the diagonal that connects vertices that are sides apart. By the law of Cosines, we have WLOG let , then it is a well known fact that holds as an explicit formula. Thus we have
Now continue as in Solution 1.
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 |
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