2014 AIME I Problem 7

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

\frac{w-z}{z} = \frac{w}{z} - 1

lies on the circle of radius

\frac{1}{10}

centered at

-1

The argument is extremal when the ray from the origin is tangent to that circle, so the sine of the angle with the real axis is

\frac{1}{10}

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{\frac{1}{10}}{1} = \frac{1}{10}.

Then

\begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{\frac{1}{100}}{\frac{99}{100}} \\ &= \frac{1}{99}, \end{aligned}

so

p + q = 1 + 99 = 100.

Problem 7 in other years