Difference between revisions of "2014 AIME I Problems/Problem 7"
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== Problem 7 == | == Problem 7 == | ||
− | Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane | + | Let <math>w</math> and <math>z</math> be complex numbers such that <math>|w| = 1</math> and <math>|z| = 10</math>. Let <math>\theta = \arg \left(\tfrac{w-z}{z}\right) </math>. The maximum possible value of <math>\tan^2 \theta</math> can be written as <math>\tfrac{p}{q}</math>, where <math>p</math> and <math>q</math> are relatively prime positive integers. Find <math>p+q</math>. (Note that <math>\arg(w)</math>, for <math>w \neq 0</math>, denotes the measure of the angle that the ray from <math>0</math> to <math>w</math> makes with the positive real axis in the complex plane) |
== Solution == | == Solution == |
Revision as of 17:17, 3 May 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)
Without 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 |
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