2014 AIME II Problems/Problem 12
Problem
Suppose that the angles of satisfy Two sides of the triangle have lengths 10 and 13. There is a positive integer so that the maximum possible length for the remaining side of is Find
Solution 1
Note that . Thus, our expression is of the form . Let and .
Using the fact that , we get , or .
Squaring both sides, we get . Cancelling factors, .
- Notice here that we cancelled out one factor of (x-1) and (y-1), which implies that (x-1) and (y-1) were not 0. If indeed they were 0 though, we would have
For this we could say that A must be 120 degrees for this to work. This is one case. The B case follows in the same way, where B must be equal to 120 degrees. This doesn't change the overall solution though, as then the other angles are irrelevant (this is the largest angle, implying that this will have the longest side and so we would want to have the 120 degreee angle opposite of the unknown side).
Expanding, .
Simplification leads to .
Therefore, . So could be or . We eliminate and use law of cosines to get our answer:
NOTE: This solution forgot the case of dividing by 0
Solution 2
As above, we can see that
Expanding, we get
Note that , or
Thus , or .
Now we know that , so we can just use the Law of Cosines to get
Note: This solution also forgets that or might be 120 when dividing by and
Solution 3
If , then , , so ; otherwise, so either or , i.e., either or . In all cases, one of the angles must be 120, which opposes the largest side. Final result follows.
-Mathdummy
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
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