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

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We need to find a maximum of this expression, so we take the derivative:
 
We need to find a maximum of this expression, so we take the derivative:
  
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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>
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<math>\dfrac{d}{dx} \left (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10} \right) = \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 = \boxed{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>.

Revision as of 17:21, 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, $\dfrac{w - z}{z} = \dfrac{\mathrm{cis}{(\alpha)} - 10\mathrm{cis}{(\beta)}}{10\mathrm{cis}{\beta}}$.

Multiplying both the numerator and denominator of this fraction by $\mathrm{cis}{(-\beta)}$ gives us:

$\dfrac{w - z}{z} = \dfrac{1}{10}\mathrm{cis}{(\alpha - \beta)} - 1 = \dfrac{1}{10}\mathrm{cos}{(\alpha - \beta)} + \dfrac{1}{10}i\mathrm{sin}{(\alpha - \beta)} - 1$.

We know that $\mathrm{tan}{\theta}$ is equal to the imaginary part of the above expression divided by the real part. Let $x = \alpha - \beta$. Then, we have that:

$\mathrm{tan}{\theta} = \dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10}.$

We need to find a maximum of this expression, so we take the derivative:

$\dfrac{d}{dx} \left (\dfrac{\mathrm{sin}{x}}{\mathrm{cos}{x} - 10} \right) = \dfrac{1 - 10\mathrm{cos}{x}}{(\mathrm{cos}{x} - 10)^2}$

Thus, we see that the maximum occurs when $\mathrm{cos}{x} = \dfrac{1}{10}$. Therefore, $\mathrm{sin}{x} = \pm\dfrac{\sqrt{99}}{10}$, and $\mathrm{tan}{\theta} = \pm\dfrac{\sqrt{99}}{99}$. Thus, the maximum value of $\mathrm{tan^2}{\theta}$ is $\dfrac{99}{99^2}$, or $\dfrac{1}{99}$, and our answer is $1 + 99 = \boxed{100}$.

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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