Difference between revisions of "1992 AIME Problems/Problem 13"
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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>). | 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=-(a +\frac{1600}{9})^2 +1600+(\frac{1600}{9})^2</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+(\frac{1600}{9})^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}</math>. | <math>\Rightarrow b\le\sqrt{1600+(\frac{1600}{9})^2}=40\sqrt{1+\frac{1600}{81}}=\frac{40}{9}\sqrt{1681}=\frac{40\cdot 41}{9}</math>. | ||
Revision as of 16:44, 8 October 2016
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 
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. 
