Difference between revisions of "2014 AIME I Problems/Problem 7"

(Solution)
(Solution)
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== Solution ==
 
== Solution ==
Let <math>w = \mathrm{cis}{(\alpha)}</math> and <math>z = 10*\mathrm{cis}{(\beta)}</math>.
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Let <math>w = \mathrm{cis}{(\alpha)}</math> and <math>z = 10*\mathrm{cis}{(\beta)}</math>. Then, \dfrac{w - z}{z} = \dfrac{\mathrm{cis}{(\alpha)} - 10*\mathrm{cis}{(\beta)}}{10*\mathrm{cis}{\beta}}.
  
 
== 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 12:05, 15 March 2014

Problem 7

Let $w$ and $z$ be complex numbers such that $|w| = 1$ and $|z| = 10$. Let $\theta = \arg \left(\tfrac{w-z}{z}\right)$. The maximum possible value of $\tan^2 \theta$ can be written as $\tfrac{p}{q}$, where $p$ and $q$ are relatively prime positive integers. Find $p+q$. (Note that $\arg(w)$, for $w \neq 0$, denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.

Solution

Let $w = \mathrm{cis}{(\alpha)}$ and $z = 10*\mathrm{cis}{(\beta)}$. Then, \dfrac{w - z}{z} = \dfrac{\mathrm{cis}{(\alpha)} - 10*\mathrm{cis}{(\beta)}}{10*\mathrm{cis}{\beta}}.

See also

2014 AIME I (ProblemsAnswer KeyResources)
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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