Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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+ | == 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? | ||
+ | |||
+ | <math> \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}</math> | ||
+ | |||
+ | ==Solution== | ||
+ | |||
Suppose <math>p</math> is the answer. We calculate <math>1-p</math>. | Suppose <math>p</math> is the answer. We calculate <math>1-p</math>. | ||
− | Assume that the circumradius of the 12-gon is <math>1</math>, and the 6 different lengths are <math>a_1</math>, <math>a_2</math>, <math>\ | + | Assume that the circumradius of the 12-gon is <math>1</math>, and the 6 different lengths are <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_6</math>, in increasing order. Then |
<math>a_k = 2\sin ( \frac{k\pi}{12} )</math>. | <math>a_k = 2\sin ( \frac{k\pi}{12} )</math>. | ||
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<math>a_6 = 2</math>. | <math>a_6 = 2</math>. | ||
− | + | ||
Now, Consider the following inequalities: | Now, Consider the following inequalities: | ||
− | + | <math>a_3>2a_1 > a_2</math> | |
− | |||
− | |||
− | + | <math>a_4> a_1 + a_2>a_3</math> | |
− | + | <math>a_4<a_1 + a_3=a_5</math> | |
− | + | <math>a_1 + a_4 > a_6</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>: | ||
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1-3-5, 1-3-6, | 1-3-5, 1-3-6, | ||
2-2-6 | 2-2-6 | ||
+ | |||
+ | Note that there are <math>12</math> segments of each length of <math>a_1</math>, <math>a_2</math>, <math>\cdots</math>, <math>a_5</math>, respectively, and <math>6</math> segments of length <math>a_6</math>. There are <math>66</math> segments in total. | ||
In the above list there are <math>3</math> triples of the type a-a-b without ''6'', <math>2</math> triples of a-a-6 where a is not ''6'', <math>3</math> triples of a-b-c without ''6'', and <math>2</math> triples of a-b-6 where a, b are not ''6''. So, | In the above list there are <math>3</math> triples of the type a-a-b without ''6'', <math>2</math> triples of a-a-6 where a is not ''6'', <math>3</math> triples of a-b-c without ''6'', and <math>2</math> triples of a-b-6 where a, b are not ''6''. So, | ||
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So <math>p = 223/286</math>. | So <math>p = 223/286</math>. | ||
+ | |||
+ | ==Video Solution by Richard Rusczyk== | ||
+ | https://artofproblemsolving.com/videos/amc/2013amc12a/364 | ||
+ | |||
+ | ~dolphin7 | ||
+ | |||
+ | == See also == | ||
+ | {{AMC12 box|year=2013|ab=A|num-b=23|num-a=25}} | ||
+ | {{MAA Notice}} |
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 |
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