Difference between revisions of "1992 AIME Problems/Problem 13"
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== 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>. |
| + | |||
| + | 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. 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 \ge 0}</math> then plugging in <math>a+n</math> for <math>x</math> gives us <math>{(a+n)^2 \ge 0}</math>. So we can keep increasing the left hand side of our earlier equation until <math>{(a+n)^2 = 0}</math>. We can factor <math>-a^2 -\frac{3200}{9}a +1600 = b^2</math> into <math>-(a +\frac{1600}{9})^2 +1600+(\frac{3200}{9})^2 = b^2</math>. We find <math>b</math>, and plug into <math>9\cdot\frac{1}{2} \cdot b</math>. Thus, the area is <math>9\cdot\frac{1}{2} \cdot \frac{40*41}{9} = 820</math>. | ||
== See also == | == See also == | ||
Revision as of 15:20, 12 March 2007
Problem
Triangle 
has 
and 
. What's the largest area that this triangle can have?
Solution
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 for . This follows directly from the trivial inequality, because if 
then plugging in 
for 
gives us 
. So we can keep increasing the left hand side of our earlier equation until 
. We can factor into 
. We find , and plug into 
. Thus, 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 | ||
