1992 AIME Problems/Problem 13
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, and can be equal with some value of ). MAA is pretty good at generating smooth combinations, so in this case, the AM-GM works; however, always try to double check in math competitions -- the writer of Solution 2 gave us a pretty good example of checking if the AM-GM equality can be obtained. ~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 .
~minor edit by Yiyj1
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 | ||
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