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

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== Problem ==
 
== Problem ==
Triangle <math>ABC^{}_{}</math> has <math>AB=9^{}_{}</math> and <math>BC: AC=40: 41^{}_{}</math>. What's the largest area that this triangle can have?
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Triangle <math>ABC</math> has <math>AB=9</math> and <math>BC: AC=40: 41</math>. What's the largest area that this triangle can have?
  
 
== Solution ==
 
== Solution ==
{{solution}}
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===Solution 1===
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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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We want to maximize <math>b</math>, the height, with <math>9</math> being the base.
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Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>.
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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>).
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<math>b^2=-a^2 -\frac{3200}{9}a +1600=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2</math>.
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<math>\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}</math>.
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Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</math>.
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===Solution 2===
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Let the three sides be <math>9,40x,41x</math>, so the area is <math>\frac14\sqrt {(81^2 - 81x^2)(81x^2 - 1)}</math> by Heron's formula. By AM-GM, <math>\sqrt {(81^2 - 81x^2)(81x^2 - 1)}\le\frac {81^2 - 1}2</math>, and the maximum possible area is <math>\frac14\cdot\frac {81^2 - 1}2 = \frac18(81 - 1)(81 + 1) = 10\cdot82 = \boxed{820}</math>. This occurs when <math>81^2 - 81x^2 = 81x^2 - 1\implies x = \frac {4\sqrt {205}}9</math>.
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===Solution 3===
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Let <math>A, B</math> be the endpoints of the side with length <math>9</math>. Let <math>\Gamma</math> be the Apollonian Circle of <math>AB</math> with ratio <math>40:41</math>; let this intersect <math>AB</math> at <math>P</math> and <math>Q</math>, where <math>P</math> is inside <math>AB</math> and <math>Q</math> is outside. Then because <math>(A, B; P, Q)</math> describes a harmonic set, <math>AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360</math>. Finally, this means that the radius of <math>\Gamma</math> is <math>\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}</math>.
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Since the area is maximized when the altitude to <math>AB</math> is maximized, clearly we want the last vertex to be the highest point of <math>\Gamma</math>, which just makes the altitude have length <math>180+\dfrac{20}{9}</math>. Thus, the area of the triangle is <math>\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}</math>
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===Solution 4 (Involves Basic Calculus)===
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We can apply Heron's on this triangle after letting the two sides equal <math>40x</math> and <math>41x</math>. Heron's gives
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<math>\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{-x+9}{2})}</math>.
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This can be simplified to
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<math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>.
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We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.
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We have that <math>-324x^3+13124x=0</math>, so <math>x=\frac{\sqrt{3281}}{9}</math>.
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Plugging this into the expression, we have that the area is <math>\boxed{820}</math>.
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<math>\textbf{-RootThreeOverTwo}</math>
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===Solution 5 ===
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We can start how we did above in solution 4 to get
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<math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>.
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Then, we can notice the inside is a quadratic in terms of <math>x^2</math>, which is
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<math>-81(x^2)^2+6562x^2-81</math>. This is maximized when <math>x^2 = \frac{3281}{81}</math>.If we plug it into the equation, we get <math>\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}</math>
  
 
== See also ==
 
== See also ==
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[[Category:Intermediate Geometry Problems]]
 
[[Category:Intermediate Geometry Problems]]
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{{MAA Notice}}

Revision as of 15:39, 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 $\boxed{820}$.

$\textbf{-RootThreeOverTwo}$

Solution 5

We can start how we did above in solution 4 to get $\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}$. Then, we can notice the inside is a quadratic in terms of $x^2$, which is $-81(x^2)^2+6562x^2-81$. This is maximized when $x^2 = \frac{3281}{81}$.If we plug it into the equation, we get $\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}$

See also

1992 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
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All AIME Problems and Solutions

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