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1992 AIME Problems/Problem 13

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Problem

Triangle has and . What's the largest area that this triangle can have?

Solution

First, consider the triangle in a coordinate system with vertices at , , and .

Applying the distance formula, we see that \frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}. We want to maximize , the height, with being the base. Simplifying gives -a^2 -\frac{3200}{9}a +1600 = b^2. To maximize , we want to maximize . So if we can write: then is the maximum value for . This follows directly from the trivial inequality, because if then plugging in for gives us . So we can keep increasing the left hand side of our earlier equation until . We can factor -a^2 -\frac{3200}{9}a +1600 = b^2 into -(a +\frac{1600}{9})^2 +1600+(\frac{1600}{9})^2 = b^2. We find , and plug into . Thus, the area is 9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820.

See also

1992 AIME (ProblemsResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
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