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