Question

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 $\frac{p}{q},$ where $p$ and $q$ are relatively prime positive integers. Find $p + q.$ (Note that $\arg(w),$ for $w \ne 0,$ denotes the measure of the angle that the ray from $0$ to $w$ makes with the positive real axis in the complex plane.)

Answer

Solution:

Since $\frac{w-z}{z} = \frac{w}{z} - 1,$ and $\frac{w}{z}$ can be any complex number of modulus $\frac{1}{10},$ the point $\zeta = \frac{w-z}{z}$ ranges over the circle of radius $\frac{1}{10}$ centered at $-1.$

Because $\tan^2\theta$ is unchanged when $\theta$ shifts by $180^\circ,$ we want the largest angle $\alpha$ that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where $\sin \alpha = \frac{1/10}{1} = \frac{1}{10}.$

Then $\tan^2\theta = \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{1/100}{99/100} = \frac{1}{99},$ so $p + q = 1 + 99 = 100.$

2014 AIME I Problem 7

Solution:

Problem 7 in Other Years