1992 AIME Problems/Problem 13

Revision as of 00:25, 25 August 2008 by FibonacciFan (talk | contribs) (Solution)

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: $-(a+n)^2+m=b^2$ then $m$ is the maximum value for $b^2$. This follows directly from the trivial inequality, because if ${x^2 \ge 0}$ then plugging in $a+n$ for $x$ gives us ${(a+n)^2 \ge 0}$. So we can keep increasing the left hand side of our earlier equation until ${(a+n)^2 = 0}$. 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 $b$, and plug into $9\cdot\frac{1}{2} \cdot b$. Thus, the area is $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820$.

See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions