2008 AMC 12A Problems/Problem 24
Contents
[hide]Problem
Triangle has and . Point is the midpoint of . What is the largest possible value of ?
Solution 1
Let . Then , and since and , we have
With calculus, taking the derivative and setting equal to zero will give the maximum value of . Otherwise, we can apply AM-GM:
Thus, the maximum is at .
Solution 2
We notice that is strictly increasing on the interval (if , then it is impossible for ), so we want to maximize .
Consider the circumcircle of and let it meet again at . Any point between and on line is inside this circle, so it follows that . Therefore to maximize , the circumcircle of must be tangent to at . By PoP we find that .
Now our computations are straightforward:
See also
2008 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 23 |
Followed by Problem 25 |
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All AMC 12 Problems and Solutions |
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