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Difference between revisions of "Trivial Inequality"

(→‎Instructive National Math Olympiad Problem)
(→‎USA AIME 1992, Problem 13)
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==USA AIME 1992, Problem 13==
 
==USA AIME 1992, Problem 13==
  
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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?''
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''Triangle <math>ABC</math> has <math>AB</math><math>=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 vertices at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.<br>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 vertices at <math>(0,0)</math>, <math>(9,0)</math>, and <math>(a,b)</math>.<br>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. Simplifying gives <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math>. Recalling that the max/min value of a quadratic is <math>\frac{4ac-b^2}{4a}</math>, we see that <math>\frac{4(-1)(1600) - \left(\frac{3200}{9}\right)^2 }{(-1)(4)} = b^2</math>. Solving for <math>b</math>, <math>b = \frac{40\cdot41}{9}</math> Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>.
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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=b^2</math> then <math>m</math> is the maximum value for <math>b^2</math>.''' This follows directly from the trivial inequality, because if <math>x^2>=0</math>
 +
Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>.
  
 
''Solution credit to: 4everwise''
 
''Solution credit to: 4everwise''
  
 
Note: I am still editing this...
 
Note: I am still editing this...

Revision as of 01:00, 18 June 2006

The Inequality

The trivial inequality states that ${x^2 \ge 0}$ for all x. This is a rather useful inequality for proving that certain quantities are non-negative. The inequality appears to be obvious and unimportant, but it can be a very powerful problem solving technique.

Applications

Maximizing and minimizing quadratic functions

After Completing the square, the trivial inequality can be applied to determine the extrema of a quadratic function.

USA AIME 1992, Problem 13

Triangle $ABC$ has $AB$$=9$ and $BC: AC=40: 41$. What's the largest area that this triangle can have?

Solution: 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: $-(a+n)^2+m=b^2$ then $m$ is the maximum value for $b^2$. This follows directly from the trivial inequality, because if $x^2>=0$ 

Thus, the area is $9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820$.

Solution credit to: 4everwise

Note: I am still editing this...

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