Difference between revisions of "1992 AIME Problems/Problem 13"
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<math>\textbf{-RootThreeOverTwo}</math> | <math>\textbf{-RootThreeOverTwo}</math> | ||
+ | |||
+ | ===Solution 5 === | ||
+ | |||
+ | We can start how we did above in solution 4 to get | ||
+ | <math>\frac{9}{4} * \sqrt{(81x^2-1)(81-x^2)}</math>. | ||
+ | Then, we can notice the inside is a quadratic in terms of <math>x^2</math>, which is | ||
+ | <math>-81(x^2)^2+6562x^2-81</math>. This is maximized when <math>x^2 = \frac{3281}{81}</math>.If we plug it into the equation, we get <math>\frac{9}{4} *\frac{9}{4}*\frac{3280}{9} = \boxed{820}</math> | ||
== See also == | == See also == |
Revision as of 15:39, 2 November 2020
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
.
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.