Difference between revisions of "2014 AIME I Problems/Problem 7"
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<math>\dfrac{d}{dx} (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10}) = \dfrac{1 - 10\mathrm{cos}{x}}{(\mathrm{cos}{x} - 10)^2}</math> | <math>\dfrac{d}{dx} (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10}) = \dfrac{1 - 10\mathrm{cos}{x}}{(\mathrm{cos}{x} - 10)^2}</math> | ||
− | Thus, we see that the maximum occurs when <math>\mathrm{cos}{x} = \dfrac{1}{10}</math>. Therefore, <math>\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\mathrm{tan^2}{\theta}</math> is <math>\dfrac{99}{99^2}</math>, or <math>\dfrac{1}{99}</math>, and our answer is <math>1 + 99 = 100</math>. | + | Thus, we see that the maximum occurs when <math>\mathrm{cos}{x} = \dfrac{1}{10}</math>. Therefore, <math>\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}</math>, and <math>\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}</math>. Thus, the maximum value of <math>\mathrm{tan^2}{\theta}</math> is <math>\dfrac{99}{99^2}</math>, or <math>\dfrac{1}{99}</math>, and our answer is <math>1 + 99 = \boxed{100}</math>. |
== See also == | == See also == | ||
{{AIME box|year=2014|n=I|num-b=6|num-a=8}} | {{AIME box|year=2014|n=I|num-b=6|num-a=8}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 12:56, 15 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
.
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.