2014 AIME II Problems/Problem 12
Contents
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:
Solution 2
WLOG, let C be the largest angle in the triangle.
As above, we can see that
Expanding, we get
CASE 1: If or ,
This implies one or both of A or B are 60 or 120.
If one of A or B is 120, we have a contradiction, since C must be the largest angle.
Otherwise, if one of A or B is 60, WLOG, assume A = 60, we would have , and thus, cos(3B) and cos(3C) both equal 1, implying , a contradiction to the fact that the sum of the angles of a triangle must be 180 degrees.
CASE 2: If and
Note that , or
Thus , or .
Now we know that , so we can just use the Law of Cosines to get
-Alexlikemath
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 longest side. Final result follows.
-Mathdummy
Solution 4(quick sketch solve)
Let be the unknown side length. By Law of Cosines we have that . We notice that should be negative to optimize so is between and degrees. We also know that the value inside the square root is an integer , so should be an integer. We can then assume that is degrees so . We do this because degrees is a "common" value and it makes the value inside the square root an integer. Plugging this into for we get that it is .
-srisainandan6
Video Solution
https://youtu.be/SLSsGYZ4Ix0?si=_f4Uct20WROmBa8E
~MathProblemSolvingSkills.com
See also
2014 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 11 |
Followed by Problem 13 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
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