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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| + | ===Solution 3=== | ||
| + | Let <math>A, B</math> be the endpoints of the side with length <math>9</math>. Let <math>\Gamma</math> be the Apollonian Circle of <math>AB</math> with ratio <math>40:41</math>; let this intersect <math>AB</math> at <math>P</math> and <math>Q</math>, where <math>P</math> is inside <math>AB</math> and <math>Q</math> is outside. Then because <math>(A, B; P, Q)</math> describes a harmonic set, <math>AP/AQ=BP/BQ\implies \dfrac{\frac{41}{9}}{BQ+9}=\dfrac{\frac{40}{9}}{BQ}\implies BQ=360</math>. Finally, this means that the radius of <math>\Gamma</math> is <math>\dfrac{360+\frac{40}{9}}{2}=180+\dfrac{20}{9}</math>. | ||
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| + | Since the area is maximized when the altitude to <math>AB</math> is maximized, clearly we want the last vertex to be the highest point of <math>\Gamma</math>, which just makes the altitude have length <math>180+\dfrac{20}{9}</math>. Thus, the area of the triangle is <math>\dfrac{9\cdot \left(180+\frac{20}{9}\right)}{2}=\boxed{820}</math> | ||
== See also == | == See also == | ||
Revision as of 23:52, 25 November 2015
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. 
