2014 AIME I Problems/Problem 7

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:

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

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 $1 + 99 = \boxed{100}$ .

2014 AIME I (Problems • Answer Key • Resources)
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2014 AIME I Problems/Problem 7

Problem 7

Solution

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