Difference between revisions of "1992 AIME Problems/Problem 13"
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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> | 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> | ||
== 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 00:54, 21 November 2018
Contents
[hide]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 .
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.