Difference between revisions of "2014 AIME I Problems/Problem 7"
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Multiplying both the numerator and denominator of this fraction by <math>\mathrm{cis}{(-\beta)}</math> gives us: | Multiplying both the numerator and denominator of this fraction by <math>\mathrm{cis}{(-\beta)}</math> gives us: | ||
− | <math>\dfrac{w - z}{z} = \dfrac{1}{10}\mathrm{cis}{\alpha - \beta} - 1</math>. | + | <math>\dfrac{w - z}{z} = \dfrac{1}{10}\mathrm{cis}{(\alpha - \beta)} - 1</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 12:08, 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:
.
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.