Difference between revisions of "2008 AIME II Problems/Problem 14"
(→Solution 2) |
(→Solution 2) |
||
Line 22: | Line 22: | ||
Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>. | Thus <math>f(\theta) = \frac {a}{b} = \frac {\cos{\theta}}{\sin{\left(\theta + \frac {\pi}{3}\right)}}</math> and we need to maximize this for <math>0 \le \theta \le \frac {\pi}{6}</math>. | ||
− | + | Taking the [[derivative]] shows that <math>-f'(\theta) = \frac {\cos{\frac {\pi}{3}}}{\sin^2{\left(\theta + \frac {\pi}{3}\right)}} \ge 0</math>, so the maximum is at the endpoint <math>\theta = 0</math>. We then get | |
<center><math>\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}</math></center> | <center><math>\rho = \frac {\cos{0}}{\sin{\frac {\pi}{3}}} = \frac {2}{\sqrt {3}}</math></center> | ||
Then, <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3+4=\boxed{007}</math>. | Then, <math>\rho^2 = \frac {4}{3}</math>, and the answer is <math>3+4=\boxed{007}</math>. |
Revision as of 15:36, 5 June 2015
Problem
Let and
be positive real numbers with
. Let
be the maximum possible value of
for which the system of equations
has a solution in
satisfying
and
. Then
can be expressed as a fraction
, where
and
are relatively prime positive integers. Find
.
Solution
Solution 1
Notice that the given equation implies

We have , so
.
Then, notice , so
.
The solution satisfies the equation, so
, and the answer is
.
Solution 2
Consider the points and
. They form an equilateral triangle with the origin. We let the side length be
, so
and
.
Thus and we need to maximize this for
.
Taking the derivative shows that , so the maximum is at the endpoint
. We then get

Then, , and the answer is
.
Solution 3
Consider a cyclic quadrilateral with
, and
. Then
From Ptolemy's Theorem,
, so
Simplifying, we have
.
Note the circumcircle of has radius
, so
and has an arc of
degrees, so
. Let
.
, where both
and
are
since triangle
must be acute. Since
is an increasing function over
,
is also increasing function over
.
maximizes at
maximizes at
. This squared is
, and
.
See also
2008 AIME II (Problems • Answer Key • Resources) | ||
Preceded by Problem 13 |
Followed by Problem 15 | |
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.