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Difference between revisions of "1992 AIME Problems/Problem 13"
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===Solution 2===
 
===Solution 2===
−
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>.
+
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 {\sqrt {3281}}9</math>.
 +
 
 +
====Comment====
 +
Rigorously, we need to make sure that the equality of the AM-GM inequality is possible to be obtained (in other words, <math>(81^2 - 81x^2)</math> and <math>(81x^2 - 1)</math> can be equal with some value of <math>x</math>). MAA is pretty good at generating smooth combinations, so in this case, the AM-GM works; however, always try to double check in math competitions -- the writer of Solution 2 gave us a pretty good example of checking if the AM-GM equality can be obtained.  ~Will_Dai
  
 
===Solution 3===
 
===Solution 3===
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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
 
We can apply Heron's on this triangle after letting the two sides equal <math>40x</math> and <math>41x</math>. Heron's gives
  
−
<math>\sqrt{(\frac{81x+9}{2})(\frac{81x-9}{2})(\frac{x+9}{2})(\frac{-x+9}{2})}</math>.
+
<math>\sqrt{\left(\frac{81x+9}{2} \right) \left(\frac{81x-9}{2} \right) \left(\frac{x+9}{2} \right) \left(\frac{-x+9}{2} \right)}</math>.
  
 
This can be simplified to
 
This can be simplified to
  
−
<math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>.
+
<math>\frac{9}{4} \cdot \sqrt{(81x^2-1)(81-x^2)}</math>.
  
 
We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.
 
We can optimize the area of the triangle by finding when the derivative of the expression inside the square root equals 0.
Line 42: Line 45:
  
 
<math>\textbf{-RootThreeOverTwo}</math>
 
<math>\textbf{-RootThreeOverTwo}</math>
 +
 +
~minor <math>\LaTeX</math> edit by Yiyj1
 +
 +
===Solution 5 ===
 +
 +
We can start how we did above in solution 4 to get
 +
<math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>.
 +
Then, we can notice the inside is a quadratic in terms of <math>x^2</math>, which is
 +
<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} \cdot \frac{3280}{9} = \boxed{820}</math>
 +
 
== See also ==
 
== See also ==
 
{{AIME box|year=1992|num-b=12|num-a=14}}
 
{{AIME box|year=1992|num-b=12|num-a=14}}

Latest revision as of 22:13, 30 August 2024

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 $81^2 - 81x^2 = 81x^2 - 1\implies x = \frac {\sqrt {3281}}9$.

Comment

Rigorously, we need to make sure that the equality of the AM-GM inequality is possible to be obtained (in other words, $(81^2 - 81x^2)$ and $(81x^2 - 1)$ can be equal with some value of $x$). MAA is pretty good at generating smooth combinations, so in this case, the AM-GM works; however, always try to double check in math competitions -- the writer of Solution 2 gave us a pretty good example of checking if the AM-GM equality can be obtained. ~Will_Dai

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

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

This can be simplified to

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

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 $\boxed{820}$.

$\textbf{-RootThreeOverTwo}$

~minor $\LaTeX$ edit by Yiyj1

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 $\frac{9}{4} \cdot \frac{3280}{9} = \boxed{820}$

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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