2013 AMC 12A Problems/Problem 24
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 .
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 |
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All AMC 12 Problems and Solutions |
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