1992 AIME Problems/Problem 13

Problem

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

Solution

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 $9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}$ .

1992 AIME (Problems • Answer Key • Resources)
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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1992 AIME Problems/Problem 13

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