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Difference between revisions of "1992 AIME Problems/Problem 13"

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== Solution ==
 
== Solution ==
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First, consider the [[triangle]] in a [[coordinate system]] with [[vertex|vertices]] at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.
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First, consider the [[triangle]] in a [[coordinate system]] with [[vertex|vertices]] at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>. Applying the [[distance formula]], we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</math>.
  
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Applying the [[distance formula]], we see that <math>\frac{ \sqrt{a^2 + b^2} }{ \sqrt{ (a-9)^2 + b^2 } } = \frac{40}{41}</math>.
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We want to maximize <math>b</math>, the height, with <math>9</math> being the base.
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We want to maximize <math>b</math>, the height, with <math>9</math> being the base. Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>. To maximize <math>b</math>, we want to maximize <math>b^2</math>. So if we can write: <math>-(a+n)^2+m=b^2</math> then <math>m</math> is the maximum value for <math>b^2</math>. This follows directly from the [[trivial inequality]], because if <math>{x^2 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>. So we can keep increasing the left hand side of our earlier equation until <math>{(a+n)^2 = 0}</math>. We can factor <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math> into <math>-(a +\frac{1600}{9})^2 +1600+(\frac{1600}{9})^2 = b^2</math>. We find <math>b</math>, and plug into <math>9\cdot\frac{1}{2} \cdot b</math>. Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>.
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 +
Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>.
 +
 
 +
To maximize <math>b</math>, we want to maximize <math>b^2</math>. So if we can write: <math>b^2=-(a+n)^2+m</math>, then <math>m</math> is the maximum value of <math>b^2</math> (this follows directly from the [[trivial inequality]], because if <math>{x^2 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>).
 +
 
 +
<math>b^2=-a^2 -\frac{3200}{9}a +1600=-(a +\frac{1600}{9})^2 +1600+(\frac{1600}{9})^2</math>.
 +
 
 +
<math>\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}</math>.
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Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</math>.
  
 
== See also ==
 
== See also ==

Revision as of 22:01, 15 March 2009

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

See also

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
All AIME Problems and Solutions
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