2013 AMC 12A Problems/Problem 24
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 .