Difference between revisions of "2013 AMC 12A Problems/Problem 24"
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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>. |
Revision as of 13:08, 12 February 2013
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 ,
,
,
,
,
.
Note that there are segments of each length of
,
,
,
, respectively, and
segments of length
. There are
segments in total.
Now, Consider the following inequalities:
- : Since
- .
- is greater than
but less than
.
- is greater than
but equal to
.
- is greater than
.
- . Then obviously 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
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 .