Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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==Solution 2== | ==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> | + | 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> |
− | |||
− | |||
Now continue as in Solution 1. | Now continue as in Solution 1. | ||
Revision as of 15:50, 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?
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 2
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
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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