2014 AIME I Problem 7

Attempt Problem 7 of the 2014 AIME I below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2014 AIME I solutions, or check the answer key.

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7.

Let ww and zz be complex numbers such that w=1|w| = 1 and z=10.|z| = 10. Let θ=arg(wzz).\theta = \arg\left(\tfrac{w-z}{z}\right). The maximum possible value of tan2θ\tan^2 \theta can be written as pq,\frac{p}{q}, where pp and qq are relatively prime positive integers. Find p+q.p + q. (Note that arg(w),\arg(w), for w0,w \ne 0, denotes the measure of the angle that the ray from 00 to ww makes with the positive real axis in the complex plane.)

Answer: 100
Concepts:complex numbercircletangent lineoptimization
Difficulty rating: 2560
Solution:

Since wzz=wz1,\frac{w-z}{z} = \frac{w}{z} - 1, and wz\frac{w}{z} can be any complex number of modulus 110,\frac{1}{10}, the point ζ=wzz\zeta = \frac{w-z}{z} ranges over the circle of radius 110\frac{1}{10} centered at 1.-1.

Because tan2θ\tan^2\theta is unchanged when θ\theta shifts by 180,180^\circ, we want the largest angle α\alpha that a ray from the origin to this circle makes with the real axis. The extreme rays are tangent to the circle, where sinα=1/101=110.\sin \alpha = \frac{1/10}{1} = \frac{1}{10}.

Then tan2θ=sin2α1sin2α=1/10099/100=199, \begin{aligned} \tan^2\theta &= \frac{\sin^2\alpha}{1 - \sin^2\alpha} = \frac{1/100}{99/100} \\ &= \frac{1}{99}, \end{aligned} so p+q=1+99=100.p + q = 1 + 99 = 100.

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