Difference between revisions of "1992 AIME Problems/Problem 13"
(→Comment) |
(→Solution) |
||
Line 19: | Line 19: | ||
===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 { | + | 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 {sqrt {3281}}9</math>. |
====Comment==== | ====Comment==== |
Revision as of 07:15, 11 November 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 --- 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 .
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.