Difference between revisions of "1992 AIME Problems/Problem 13"
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== Solution == | == Solution == | ||
+ | ===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>. | 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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Then the area is <math>9\cdot\frac{1}{2} \cdot \frac{40\cdot 41}{9} = \boxed{820}</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>. | ||
== See also == | == See also == |
Revision as of 14:12, 25 March 2009
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 .
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 |