Difference between revisions of "1992 AIME Problems/Problem 13"
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===Solution 2=== | ===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>. | 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>. | ||
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| + | ====Comment==== | ||
| + | Rigorously, we need to make sure that the equality of the AM-GM inequality is possible to be obtained (in other words, (81^2 - 81x^2) and (81x^2 - 1) can be equal with some value of x). MAA is pretty good at generating smooth combinations , so in this case, the AM-GM works; but always try to double check in math competitions. ~Will_Dai | ||
===Solution 3=== | ===Solution 3=== | ||
Revision as of 04:17, 12 August 2023
Contents
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 
.
Comment
Rigorously, we need to make sure that the equality of the AM-GM inequality is possible to be obtained (in other words, (81^2 - 81x^2) and (81x^2 - 1) can be equal with some value of x). MAA is pretty good at generating smooth combinations , so in this case, the AM-GM works; but always try to double check in math competitions. ~Will_Dai
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 
.
Solution 5
We can start how we did above in solution 4 to get
.
Then, we can notice the inside is a quadratic in terms of 
, which is
. This is maximized when 
.If we plug it into the equation, we get 
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. 
