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Difference between revisions of "1992 AIME Problems/Problem 13"
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<math>\textbf{-RootThreeOverTwo}</math>
 
<math>\textbf{-RootThreeOverTwo}</math>
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== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=12|num-a=14}}
 
{{AIME box|year=1992|num-b=12|num-a=14}}

Revision as of 14:36, 2 November 2020

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 ${(a+n)^2 \ge 0}$).

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

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

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 $81^2 - 81x^2 = 81x^2 - 1\implies x = \frac {4\sqrt {205}}9$.

Solution 3

Let $A, B$ be the endpoints of the side with length $9$. Let $\Gamma$ be the Apollonian Circle of $AB$ with ratio $40:41$; let this intersect $AB$ at $P$ and $Q$, where $P$ is inside $AB$ and $Q$ is outside. Then because $(A, B; P, Q)$ describes a harmonic set, $AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360$. Finally, this means that the radius of $\Gamma$ is $\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}$.

Since the area is maximized when the altitude to $AB$ is maximized, clearly we want the last vertex to be the highest point of $\Gamma$, which just makes the altitude have length $180+\dfrac{20}{9}$. Thus, the area of the triangle is $\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}$

Solution 4 (Involves Basic Calculus)

We can apply Heron's on this triangle after letting the two sides equal $40x$ and $41x$. Heron's gives

$\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{-x+9}{2})}$.

This can be simplified to

$\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$.

We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.

We have that $-324x^3+13124x=0$, so $x=\frac{\sqrt{3281}}{9}$.

Plugging this into the expression, we have that the area is $\boxed{820}$.

$\textbf{-RootThreeOverTwo}$

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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