Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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Now, Consider the following inequalities: | Now, Consider the following inequalities: | ||
− | <math> | + | <math>a_3>2a_1 > a_2</math> |
− | <math> | + | <math>a_4> a_1 + a_2>a_3</math> |
− | <math> | + | <math>a_4<a_1 + a_3=a_5</math> |
− | <math> | + | <math>a_1 + a_4 > a_6</math> |
− | <math> | + | <math>2a_2 = 2 = a_6</math>. Thus 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>: |
Latest revision as of 15:54, 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 .
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.