Difference between revisions of "2014 AIME I Problems/Problem 7"
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<cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath> | <cmath>\frac{99}{20t}+\frac{t}{20}=\cos\theta</cmath> | ||
if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math> | if we want to maximize <math>\theta</math> we need to minimize <math>\cos\theta</math> | ||
− | , using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t | + | , using AM-GM inequality we get that the minimum value for <math>\cos\theta= 2(\sqrt{\dfrac{99}{20t}\dfrac{t}{20}})=2\sqrt{\dfrac{99}{400}}=\dfrac{\sqrt{99}}{10}</math> |
hence using the identity <math>\tan^2\theta=\sec^2\theta-1</math> | hence using the identity <math>\tan^2\theta=\sec^2\theta-1</math> | ||
we get <math>\tan^2\theta=\frac{1}{99}</math>and our answer is <math>1 + 99 = \boxed{100}</math>. | we get <math>\tan^2\theta=\frac{1}{99}</math>and our answer is <math>1 + 99 = \boxed{100}</math>. |
Revision as of 21:02, 26 March 2014
Problem 7
Let and
be complex numbers such that
and
. Let
. The maximum possible value of
can be written as
, where
and
are relatively prime positive integers. Find
. (Note that
, for
, denotes the measure of the angle that the ray from
to
makes with the positive real axis in the complex plane.
Solution
Let and
. Then,
.
Multiplying both the numerator and denominator of this fraction by gives us:
.
We know that is equal to the imaginary part of the above expression divided by the real part. Let
. Then, we have that:
We need to find a maximum of this expression, so we take the derivative:
Thus, we see that the maximum occurs when . Therefore,
, and
. Thus, the maximum value of
is
, or
, and our answer is
.
Solution 2 (No calculus)
with out the loss of generality one can let lie on the positive x axis and since
is a measure of the angle if
then
and we can see that the question is equivelent to having a triangle
with sides
and
and trying to maximize the angle
using the law of cosines we get:
rearranging:
solving for
we get:
if we want to maximize
we need to minimize
, using AM-GM inequality we get that the minimum value for
hence using the identity
we get
and our answer is
.
See also
2014 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 6 |
Followed by Problem 8 | |
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.