Difference between revisions of "1992 AIME Problems/Problem 13"
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== Problem == | == Problem == | ||
− | Triangle <math>ABC | + | 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 == | ||
− | 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>. | + | ===Solution 1=== |
+ | 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>. | ||
− | + | We want to maximize <math>b</math>, the height, with <math>9</math> being the base. | |
− | 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 | + | |
+ | 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=-\left(a +\frac{1600}{9}\right)^2 +1600+\left(\frac{1600}{9}\right)^2</math>. | ||
+ | |||
+ | <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>. | ||
+ | |||
+ | Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</math>. | ||
+ | |||
+ | ===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>. | ||
+ | |||
+ | ===Solution 3=== | ||
+ | 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>. | ||
+ | |||
+ | 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> | ||
+ | |||
+ | ===Solution 4 (Involves Basic Calculus)=== | ||
+ | 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>. | ||
+ | |||
+ | This can be simplified to | ||
+ | |||
+ | <math>\frac{9}{4} * \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 have that <math>-324x^3+13124x=0</math>, so <math>x=\frac{\sqrt{3281}}{9}</math>. | ||
+ | |||
+ | Plugging this into the expression, we have that the area is <math>\boxed{820}</math>. | ||
+ | |||
+ | <math>\textbf{-RootThreeOverTwo}</math> | ||
+ | |||
+ | ===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} *\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]] | ||
+ | {{MAA Notice}} |
Revision as of 15:39, 2 November 2020
Contents
Problem
Triangle has and . What's the largest area that this triangle can have?
Solution
Solution 1
First, consider the triangle in a coordinate system with vertices at , , and . Applying the distance formula, we see that .
We want to maximize , the height, with being the base.
Simplifying gives .
To maximize , we want to maximize . So if we can write: , then is the maximum value of (this follows directly from the trivial inequality, because if then plugging in for gives us ).
.
.
Then the area is .
Solution 2
Let the three sides be , so the area is by Heron's formula. By AM-GM, , and the maximum possible area is . This occurs when .
Solution 3
Let be the endpoints of the side with length . Let be the Apollonian Circle of with ratio ; let this intersect at and , where is inside and is outside. Then because describes a harmonic set, . Finally, this means that the radius of is .
Since the area is maximized when the altitude to is maximized, clearly we want the last vertex to be the highest point of , which just makes the altitude have length . Thus, the area of the triangle is
Solution 4 (Involves Basic Calculus)
We can apply Heron's on this triangle after letting the two sides equal and . Heron's gives
.
This can be simplified to
.
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 , so .
Plugging this into the expression, we have that the area is .
Solution 5
We can start how we did above in solution 4 to get . Then, we can notice the inside is a quadratic in terms of , which is . This is maximized when .If we plug it into the equation, we get
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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