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1992 AIME Problems/Problem 13

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Problem

Triangle ABC has AB=9 and BC: AC=40: 41. What's the largest area that this triangle can have?

Solution

Solution 1

First, consider the triangle in a coordinate system with vertices at (0,0), (9,0), and (a,b). 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 b, the height, with 9 being the base.

Simplifying gives -a^2 -\frac{3200}{9}a +1600 = b^2.

To maximize b, we want to maximize b^2. So if we can write: b^2=-(a+n)^2+m, then m is the maximum value of b^2 (this follows directly from the trivial inequality, because if {x^2 \ge 0} then plugging in a+n for x gives us {(a+n)^2 \ge 0}).

b^2=-a^2 -\frac{3200}{9}a +1600=-(a +\frac{1600}{9})^2 +1600+(\frac{1600}{9})^2.

\Rightarrow b\le\sqrt{1600+(\frac{1600}{9})^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}.

Then the area is 9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}.

Solution 2

Let the three sides be 9,40x,41x, so the area is \frac14\sqrt {(81^2 - 81x^2)(81x^2 - 1)} by Heron's formula. By AM-GM, \sqrt {(81^2 - 81x^2)(81x^2 - 1)}\le\frac {81^2 - 1}2, and the maximum possible area is \frac14\cdot\frac {81^2 - 1}2 = \frac18(81 - 1)(81 + 1) = 10\cdot82 = \boxed{820}. This occurs when 81^2 - 81x^2 = 81x^2 - 1\implies x = \frac {4\sqrt {205}}9.

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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